Difference between revisions of "Curves in Space and Vector-Valued Functions"

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== Introduction ==
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If we have a function <math>\vec{f}:\R\to\R^n</math> , we say that <math>\vec{f}</math>'s image (the set <math>\Big\{\vec{f}(t)\Big|t\in\R\Big\}</math> - or some subset of <math>\R</math>) is a ''curve'' in <math>\R^n</math> and <math>\vec{f}</math> is its parametrization.
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Parameterizations are not necessarily unique - for example, <math>\vec{f}(t)=(\cos(t),\sin(t))</math> such that <math>t\in[0,2\pi)</math> is one parametrization of the unit circle, and <math>\vec{g}(t)=(\cos(at),\sin(at))</math> such that <math>t\in\left[0,\frac{2\pi}{a}\right)</math> is a whole family of parameterizations of that circle.
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===Collision and intersection points===
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Say we have two different curves. It may be important to consider
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*points the two curves share - where they ''intersect''
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*intersections which occur for the same value of <math>t</math> - where they ''collide''.
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==== Intersection points ====
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Firstly, we have two parameterizations <math>\vec{f}(t)</math> and <math>\vec{g}(t)</math> , and we want to find out when they intersect, this means that we want to know when the function values of each parametrization are the same. This means that we need to solve
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:<math>\vec{f}(t)=\vec{g}(s)</math>
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because we're seeking the function values independent of the times they intersect.
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For example, if we have <math>\vec{f}(t)=(t,3t)</math> and <math>\vec{g}(t)=(t,t^2)</math> , and we want to find intersection points:
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:<math>\vec{f}(t)=\vec{g}(s)</math>
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:<math>(t,3t)=(s,s^2)</math>
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:<math>t=s\ ,\ 3t=s^2</math>
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with solutions <math>(t,s)=(0,0)\ ,\ (3,9)</math>
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So, the two curves intersect at the points <math>(0,0)\ ,\ (3,9)</math> .
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==== Collision points ====
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However, if we want to know when the points "collide", with <math>\vec{f}(t)</math> and <math>\vec{g}(t)</math> , we need to know when both the function values ''and'' the times are the same, so we need to solve instead
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:<math>\vec{f}(t)=\vec{g}(t)</math>
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For example, using the same functions as before, <math>\vec{f}(t)=(t, 3t)</math> and <math>\vec{g}(t)=(t,t^2)</math> , and we want to find collision points:
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:<math>\vec{f}(t)=\vec{g}(t)</math>
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:<math>(t,3t)=(t,t^2)</math>
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:<math>t=t\ ,\ 3t=t^2</math>
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which gives solutions <math>t=0,3</math>
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So the collision points are <math>(0,0)\ ,\ (3,9)</math> .
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We may want to do this to actually model physical problems, such as in ballistics.
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=== Intersection vector functions ===
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We can also use vector functions to represent the curve of intersection of two surfaces. For example, we want to know the curve of intersection of the cylinder <math>x^2+y^2=1</math> and the plane <math>y+z=2</math>.<blockquote>Vector functions rely on parameterizations, so we can rewrite the equation of the cylinder into: <math>\cos^2 t+\sin^2 t=1</math>, where <math>x=\cos t,y=\sin t</math>.
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From the equation of the plane, we know that <math>z=2-y=2-\sin t</math>. Thus the corresponding vector equation is:
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<math>\mathbf{r}(t)=\langle \cos t,\sin t,2-\sin t\rangle</math></blockquote>
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=== Limits and Continuity ===
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The limit of a vector function <math>\mathbf{r}</math> is defined by taking the limits of its component functions.
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{{Calculus/Def|title=The limit of a vector function|text=There is a vector function <math>\mathbf{r}(t)=\langle f(t),g(t),h(t) \rangle</math>. If <math>\lim_{t \rightarrow a}f(t),\lim_{t \rightarrow a}g(t),\lim_{t \rightarrow a}h(t)</math> exist, then
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:<math>\lim_{t \rightarrow a}\mathbf{r}(t)=\langle \lim_{t \rightarrow a}f(t),\lim_{t \rightarrow a}g(t),\lim_{t \rightarrow a}h(t) \rangle</math>}}
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And the requirement for continuity is also simple:<blockquote>A vector function <math>\mathbf{r}</math> is continuous at <math>a</math> if <math>\lim_{t\rightarrow a}\mathbf{r}(t)=\mathbf{r}(a)</math>.</blockquote>
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== Derivatives and Integrals ==
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=== Differentiation ===
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Recall that the first derivative of a scalar function <math>f(x)</math> is defined as:<blockquote><math>f'(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}</math></blockquote>The first derivative of a vector function <math>\mathbf{r}(t)</math> is defined in much the same way:<blockquote><math>\mathbf{r}'(t)=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}</math></blockquote>We can use this definition to prove that the derivative of a vector function can be presented as the derivative of its component functions.<blockquote><math>\begin{align}
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\mathbf{r}'(t) & = \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}[\mathbf{r}(t+\Delta t)-\mathbf{r}(t)] \quad \text{definition}\\
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& = \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}[\langle f(t+\Delta t),g(t+\Delta t),h(t+\Delta t)\rangle-\langle f(t),g(t),h(t) \rangle] \quad \text{component form}\\
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& = \lim_{\Delta t\rightarrow 0}\langle \frac{f(t+\Delta t)-f(t)}{\Delta t},\frac{g(t+\Delta t)-g(t)}{\Delta t},\frac{h(t+\Delta t)-h(t)}{\Delta t}\rangle \quad\text{vector addition and multiplication}\\
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& = \langle \lim_{\Delta t\rightarrow 0}\frac{f(t+\Delta t)-f(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\frac{g(t+\Delta t)-g(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\frac{h(t+\Delta t)-h(t)}{\Delta t}\rangle\quad\text{definition}\\
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& = \langle f'(t),g'(t),h'(t)\rangle \quad\text{definition}\\
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\end{align}
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</math></blockquote>Thus, using the same method, we can derive the second derivative and so on.
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{{Calculus/Def|title=Derivatives of a vector function|text=There is a vector function <math>\mathbf{r}(t)=\langle f(t),g(t),h(t) \rangle</math>. The first derivative of this vector function is:
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:<math>\mathbf{r}'(t)=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}= \langle f'(t),g'(t),h'(t)\rangle</math>}}So the <math>n</math>th-order derivative should look like this:<blockquote><math>\mathbf{r}^{(n)}(t)=\langle f^{(n)}(t),g^{(n)}(t),h^{(n)}(t)\rangle</math></blockquote>
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=== Differentiation rules ===
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Just like real-valued functions, there are some differentiation rules in the world of vector functions. The factor that makes vector differentiation rules slight more complicated is the product rule because there are two kinds of multiplication in vectors: dot product and cross product.
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{{Calculus/Def|title=Differentiation rules for vector functions|text=Suppose <math>\mathbf{u},\mathbf{v}</math> are differentiable vector functions and <math>f</math> is a real-valued function. Then
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# <math>\frac{d}{dt}[\mathbf{u}(t)+\mathbf{v}(t)]=\mathbf{u}'(t)+\mathbf{v}'(t)\quad</math> (addition)
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# <math>\frac{d}{dt}[f(t)\mathbf{u}(t)]=f'(t)\mathbf{u}(t)+f(t)\mathbf{u}'(t)\quad</math> (scalar multiplication)
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# <math>\frac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]=\mathbf{u}'(t)\cdot\mathbf{v}(t)+\mathbf{u}(t)\cdot\mathbf{v}'(t)\quad</math> (dot product)
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# <math>\frac{d}{dt}[\mathbf{u}(t)\times\mathbf{v}(t)]=\mathbf{u}'(t)\times\mathbf{v}(t)+\mathbf{u}(t)\times\mathbf{v}'(t)\quad</math> (cross product)
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# <math>\frac{d}{dt}[\mathbf{u}(f(t))]=f'(t)\mathbf{u}'(f(t))\quad</math> (chain rule)}}
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Naturally, we will prove that those rules are correct. Let us assume that <math>\mathbf{u}(t)=\langle f_1(t),f_2(t),f_3(t)\rangle</math> and <math>\mathbf{v}(t)=\langle g_1(t),g_2(t),g_3(t)\rangle</math>.
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 +
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Rule 1: the addition rule<blockquote><math>\begin{align}
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\frac{d}{dt}[\mathbf{u}(t)+\mathbf{v}(t)] & = \frac{d}{dt}[\langle f_1(t),f_2(t),f_3(t)\rangle+\langle g_1(t),g_2(t),g_3(t)\rangle]\quad \text{component form} \\
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& = \frac{d}{dt}\langle f_1(t)+g_1(t),f_2(t)+g_2(t),f_3(t)+g_3(t)\rangle \quad \text{vector addition}\\
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& = \langle \frac{d}{dt}(f_1(t)+g_1(t)),\frac{d}{dt}(f_2(t)+g_2(t)),\frac{d}{dt}(f_3(t)+g_3(t))\rangle \quad\text{distribution}\\
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& = \langle f_1'(t)+g_1'(t),f_2'(t)+g_2'(t),f_3'(t)+g_3'(t)\rangle\quad\text{addition rule for real-valued functions}\\
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& = \langle f_1'(t),f_2'(t),f_3'(t)\rangle+\langle g_1'(t),g_2'(t),g_3'(t)\rangle\quad\text{vector addition}\\
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& = \mathbf{u}'(t)+\mathbf{v}'(t)\quad\text{vector form}\\
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\end{align}</math></blockquote>Rule 2: the scalar multiplication rule<blockquote><math>\begin{align}
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\frac{d}{dt}[f(t)\mathbf{u}(t)] & = \frac{d}{dt}[f(t)\begin{pmatrix} f_1(t)\\f_2(t)\\f_3(t)\end{pmatrix}]\quad\text{component form}\\
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& = \frac{d}{dt}\begin{pmatrix} f(t)f_1(t)\\f(t)f_2(t)\\f(t)f_3(t)\end{pmatrix}\quad\text{scalar multiplication}\\
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& = \begin{pmatrix}
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\frac{d}{dt}(f(t)f_1(t))\\
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\frac{d}{dt}(f(t)f_2(t))\\
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\frac{d}{dt}(f(t)f_3(t))\\
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\end{pmatrix}\quad\text{distribution}\\
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& = \begin{pmatrix}
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f'(t)f_1(t)+f(t)f_1'(t)\\
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f'(t)f_2(t)+f(t)f_2'(t)\\
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f'(t)f_3(t)+f(t)f_3'(t)\\
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\end{pmatrix}\quad\text{multiplication rule for real-valued functions}\\
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& = \begin{pmatrix} f'(t)f_1(t)\\f'(t)f_2(t)\\f'(t)f_3(t)\end{pmatrix}+
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\begin{pmatrix} f(t)f_1'(t)\\f(t)f_2'(t)\\f(t)f_3'(t)\end{pmatrix}\quad\text{vector addition}\\
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& = f'(t)\begin{pmatrix} f_1(t)\\f_2(t)\\f_3(t)\end{pmatrix}+f(t)\begin{pmatrix} f_1'(t)\\f_2'(t)\\f_3'(t)\end{pmatrix}\quad\text{scalar multiplication}\\
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& = f'(t)\mathbf{u}(t)+f(t)\mathbf{u}'(t)\quad\text{vector form}\\
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\end{align}</math></blockquote>Rule 3: the dot product rule<blockquote><math>\begin{align}
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\frac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)] & = \frac{d}{dt}[\langle f_1(t),f_2(t),f_3(t)\rangle \cdot \langle g_1(t),g_2(t),g_3(t)\rangle]\quad\text{component form}\\
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& = \frac{d}{dt}(f_1(t)g_1(t)+f_2(t)g_2(t)+f_3(t)g_3(t))\quad\text{dot product}\\
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& = \frac{d}{dt}\sum_{i=1}^3f_i(t)g_i(t)\quad\text{simplification}\\
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& = \sum_{i=1}^3\frac{d}{dt}f_i(t)g_i(t)\quad\text{distribution}\\
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& = \sum_{i=1}^3[f_i'(t)g_i(t)+f_i(t)g_i'(t)]\quad\text{multiplication rule for real-valued functions}\\
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& = \sum_{i=1}^3f_i(t)g_i(t)+\sum_{i=1}^3f_i(t)g_i'(t)\quad\text{vector addition}\\
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& = \mathbf{u}'(t)\cdot\mathbf{v}(t)+\mathbf{u}(t)\cdot\mathbf{v}'(t)\quad\text{vector form}\\
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\end{align}</math></blockquote>Rule 4: the cross product rule<blockquote><math>\begin{align}
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\frac{d}{dt}[\mathbf{u}(t)\times\mathbf{v}(t)] & = \frac{d}{dt}[\langle f_1(t),f_2(t),f_3(t)\rangle\times\langle g_1(t),g_2(t),g_3(t)\rangle]\quad\text{component form}\\
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& = \frac{d}{dt}\begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\
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f_1(t) & f_2(t) & f_3(t) \\
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g_1(t) & g_2(t) & g_3(t) \\
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\end{vmatrix}\quad\text{cross product}\\
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& = \frac{d}{dt}\begin{pmatrix} f_2(t)g_3(t)-f_3(t)g_2(t)\\
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f_3(t)g_1(t)-f_1(t)g_3(t)\\
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f_1(t)g_2(t)-f_2(t)g_1(t)\\
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\end{pmatrix}\quad\\
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& = \begin{pmatrix} \frac{d}{dt}(f_2(t)g_3(t)-f_3(t)g_2(t))\\
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\frac{d}{dt}(f_3(t)g_1(t)-f_1(t)g_3(t))\\
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\frac{d}{dt}(f_1(t)g_2(t)-f_2(t)g_1(t))\\
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\end{pmatrix}\quad\text{distribution}\\
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& = \begin{pmatrix} f_2'(t)g_3(t)+f_2(t)g_3'(t)-f_3'(t)g_2(t)-f_3(t)g_2'(t)\\
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f_3'(t)g_1(t)+f_3(t)g_1'(t)-f_1'(t)g_3(t)-f_1(t)g_3'(t)\\
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f_2'(t)g_3(t)+f_2(t)g_3'(t)-f_3'(t)g_2(t)-f_3(t)g_2'(t)\\
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\end{pmatrix}\quad\text{multiplication rule for real-valued functions}\\
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& = \begin{pmatrix} f_2'(t)g_3(t)-f_3'(t)g_2(t)+f_2(t)g_3'(t)-f_3(t)g_2'(t)\\
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f_3'(t)g_1(t)-f_1'(t)g_3(t)+f_3(t)g_1'(t)-f_1(t)g_3'(t)\\
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f_2'(t)g_3(t)-f_3'(t)g_2(t)+f_2(t)g_3'(t)-f_3(t)g_2'(t)\\
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\end{pmatrix}\quad\text{rearrangement}\\
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& = \begin{pmatrix} f_2'(t)g_3(t)-f_3'(t)g_2(t)\\f_3'(t)g_1(t)-f_1'(t)g_3(t)\\f_2'(t)g_3(t)-f_3'(t)g_2(t)\\\end{pmatrix}+
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\begin{pmatrix} f_2(t)g_3'(t)-f_3(t)g_2'(t)\\f_3(t)g_1'(t)-f_1(t)g_3'(t)\\f_2(t)g_3'(t)-f_3(t)g_2'(t)\\\end{pmatrix}\quad\text{vector addtion}\\
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& = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\
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f_1'(t) & f_2'(t) & f_3'(t) \\
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g_1(t) & g_2(t) & g_3(t) \\\end{vmatrix}+\begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\
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f_1(t) & f_2(t) & f_3(t) \\
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g_1'(t) & g_2'(t) & g_3'(t) \\
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\end{vmatrix}\quad\text{cross product}\\
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& = \begin{pmatrix} f_1'(t) \\ f_2'(t) \\ f_3'(t)\\\end{pmatrix}\times\begin{pmatrix} g_1(t) \\ g_2(t) \\ g_3(t) \\\end{pmatrix}+
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\begin{pmatrix} f_1(t) \\ f_2(t) \\ f_3(t)\\\end{pmatrix}\times\begin{pmatrix} g_1'(t) \\ g_2'(t) \\ g_3'(t) \\\end{pmatrix}\quad\\
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& = \mathbf{u}'(t)\times\mathbf{v}(t)+\mathbf{u}(t)\times\mathbf{v}'(t)\quad\text{vector form}\\
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\end{align}</math></blockquote>Rule 5: the chain rule<blockquote><math>\begin{align}
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\frac{d}{dt}[\mathbf{u}(f(t))] & = \frac{d}{dt}\begin{pmatrix} f_1(f(t)) \\ f_2(f(t)) \\ f_3(f(t)) \\\end{pmatrix}\quad\text{component form}\\
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& = \begin{pmatrix} \frac{d}{dt}f_1(f(t)) \\ \frac{d}{dt}f_2(f(t)) \\ \frac{d}{dt}f_3(f(t)) \\\end{pmatrix}\quad\text{distribution}\\
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& = \begin{pmatrix} f'(t)f_1'(f(t)) \\ f'(t)f_2'(f(t)) \\ f'(t)f_3'(f(t)) \\\end{pmatrix}\quad \text{chain rule for real-valued functions}\\
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& = f'(t)\begin{pmatrix} f_1'(f(t)) \\ f_2'(f(t)) \\ f_3'(f(t)) \\\end{pmatrix}\quad\text{scalar multiplication}\\
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& = f'(t)\mathbf{u}'(f(t))\quad\text{vector form}\\
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\end{align}</math></blockquote><math>\blacksquare</math>
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==== Angle between curves ====
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[[Image:Inner-product-angle.png|right|thumb]]
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We can then formulate the concept of the ''angle'' between two curves by considering the angle between the two tangent vectors. If two curves, parametrized by <math>\vec{f_1}</math> and <math>\vec{f_2}</math> intersect at some point, which means that
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:<math>\vec{f_1}(s)=\vec{f_2}(t)=c</math>
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the angle between these two curves at <math>c</math> is the angle between the tangent vectors <math>\vec{f_1'}(s)</math> and <math>\vec{f_2'}(t)</math>, using the dot product, is given by
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:<math>\arccos\left(\frac{\vec{f_1'}(s)\cdot\vec{f_2'}(t)}{\Big\|\vec{f_1'}(s)\Big\|\Big\|\vec{f_2'}(t)\Big\|}\right)</math>
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=== Integration ===
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Similar to real-valued functions, the definite integral of a vector function <math>\mathbf{r}(t)</math> is defined as:<blockquote><math>\begin{align}
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\int_a^b\mathbf{r}(t)dt & = \lim_{n\rightarrow\infty}\sum_{i=1}^n\mathbf{r}(t_i^*)\Delta t \\
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& = \lim_{n\rightarrow\infty}\biggl[\biggl(\sum_{i=1}^nf(t_i^*)\Delta t\biggr)\mathbf{i}+\biggl(\sum_{i=1}^ng(t_i^*)\Delta t\biggr)\mathbf{j}+\biggl(\sum_{i=1}^nh(t_i^*)\Delta t\biggr)\mathbf{k}\biggr]\\
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& = \biggl(\int_a^bf(t)dt\biggr)\mathbf{i}+\biggl(\int_a^bg(t)dt\biggr)\mathbf{j}+\biggl(\int_a^bh(t)dt\biggr)\mathbf{k}
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\end{align} </math></blockquote>We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:
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{{Calculus/Def|title=The Fundamental Theorem of Calculus (including vector functions)|text=:<math>\int_a^b\mathbf{r}(t)dt=\mathbf{R}(t)\biggr]_a^b=\mathbf{R}(b)-\mathbf{R}(a)</math>}}
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For indefinite integrals, the definition is:<blockquote><math>\int\mathbf{r}(t)dt=\biggl(\int f(t)dt\biggr)\mathbf{i}+\biggl(\int g(t)dt\biggr)\mathbf{j}+\biggl(\int h(t)dt\biggr)\mathbf{k}+\mathbf{C} </math></blockquote>
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=== Arc length ===
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Recall that in Chapter {{Calculus/map page|Parametric Integration}}, we deduced the length of a curve with parametric equations <math>\begin{cases} x=f(t) \\ y=g(t) \end{cases} </math>, <math>a\le t\le b </math> should be:<blockquote><math>L=\int_a^b\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2}dt </math></blockquote>Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.
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{{Calculus/Def|title=Arc length of a space curve|text=If the curve has the vector equation <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a\le t\le b</math>, or, equivalently, the parametric equations <math>x=f(t),y=g(t),z=h(t)</math>, where <math>f',g',h'</math> are continuous, then the length of the curve from <math>t=a</math> to <math>t=b</math> is:
 +
:<math>L=\int_a^b\sqrt{[f'(t)]^2+[g'(t)]^2+[h'(t)]^2}dt=\int_a^b\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2+\biggl(\frac{dx}{dz}\biggr)^2}dt</math>}}
 +
For those who prefer simplicity, the formula can be rewritten into:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math> or <math>\quad\frac{dL}{dt}=|\mathbf{r}'(t)|</math></blockquote>
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=== Reparametrization ===
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Suppose that there is a vector function which describes the displacement of a particle with respect to time and has the equation:<blockquote><math>\mathbf{r}(t)=\cos t\ \mathbf{i}+\sin t\ \mathbf{j}+t\ \mathbf{k}  </math></blockquote>However, for some reason, we do not want to know the displacement of this particle with respect to time. Instead, we want to know its displacement with respect to its traveled distance (<math>s </math>) from <math>(1,0,0) </math> in the direction of increasing <math>t </math>. In order to do so, we need to find a way to describe time as a function of distance. In other words, we need to find <math>t=t(s) </math>. We can use the formula for the arc length to establish the relationship between time and distance because arc length in this case describes the distance traveled by the particle.
 +
 +
 +
Before we start calculating, we need to introduce the arc length function.
 +
{{Calculus/Def|title=Arc length function|text=Suppose that a piecewise-smooth curve with a vector function <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a \le t \le b</math>, and the curve is traversed exactly once as <math>t</math> increases from <math>a</math> to <math>b</math>. The arc length function <math>s(t)</math> is:
 +
:<math>s(t)=\int_a^t\sqrt{\biggl(\frac{dx}{du}\biggr)^2+\biggl(\frac{dy}{du}\biggr)^2+\biggl(\frac{dz}{du}\biggr)^2}du</math>}}
 +
According to the definition, the arc length function for our curve <math>\mathbf{r}(t) </math> should be<blockquote><math>s(t)=\int_0^t|\mathbf{r}'(u)|du=\int_0^t\sqrt{(-\sin u)^2+(\cos u)^2+1^2}du=\sqrt{2}u\bigg]_0^t=\sqrt{2}t </math>
 +
 +
Note that <math>a=0 </math> because the initial point <math>(1,0,0) </math> corresponds to the parameter value <math>t=0 </math>. Since it is in the direction of increasing <math>t </math>, the integration direction should be from <math>0 </math> to <math>t </math>.
 +
 +
<math>t(s)=\frac{s}{\sqrt{2}} </math></blockquote>Then we substitute the original function and get the answer:<blockquote><math>\mathbf{r}(s)=\cos \frac{s}{\sqrt{2}}\ \mathbf{i}+\sin \frac{s}{\sqrt{2}}\ \mathbf{j}+\frac{s}{\sqrt{2}}\ \mathbf{k}  </math></blockquote>Reparametrization has important applications in real life because sometimes we want to know a value with respect to different variables. In this case, instead of describing the path of a particle with respect to time, we described its path with respect to its distance, which will be very useful in certain situations.
 +
 +
== Curvature ==
 +
 +
=== Terminology ===
 +
Before we start discussing curvature, there are some important vectors and concepts we need to be at least aware of.
 +
 +
==== The unit tangent vector ====
 +
In the differentiation section of this chapter, we discussed the derivatives of a vector function. We know that <math>\mathbf{v}(t)=\mathbf{r}'(t)</math> at <math>t=t_0</math> is tangent to the curve <math>\mathbf{r}(t)</math> at <math>t=t_0</math>. <math>\mathbf{r}'(t)</math> is called the tangent vector. The unit tangent vector, however, eliminates the aspect of magnitude because it is defined as:<blockquote><math>\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}</math></blockquote>As we can see, the magnitude of the unit tangent vector is always <math>1</math>. We can imagine that <math>\mathbf{r}(t)</math> as the displacement of a particle with respect to time. So, the unit tangent vector can be perceived as the direction of the velocity of the particle with respect to time. It can also be perceived as the direction of the tangential acceleration of the particle with respect to time. We will discuss motion in space in the next section, but this is a useful method to intuitively understand some vectors.
 +
 +
==== The unit normal vector ====
 +
The unit normal vector is defined as<blockquote><math>\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}</math></blockquote>The unit normal is orthogonal to the unit tangent because since <math>|\mathbf{T}(t)|=1</math>, we can get that:<blockquote><math>\frac{d}{dt}|\mathbf{T}(t)|^2=0=\frac{d}{dt}[\mathbf{T}(t)\cdot\mathbf{T}(t)]=2\mathbf{T}'(t)\cdot\mathbf{T}(t)</math>
 +
 +
 +
<math>\Leftrightarrow \mathbf{T}'(t)\cdot\mathbf{T}(t)=0</math></blockquote>This means that <math>\mathbf{T}'(t)</math> is orthogonal to <math>\mathbf{T}(t)</math>. Therefore, <math>\mathbf{N}(t)</math> is orthogonal to <math>\mathbf{T}(t)</math>. We can imagine that the unit normal vector is the direction of the normal acceleration of the particle with respect to time.
 +
 +
==== The binormal vector ====
 +
The binormal vector is defined as<blockquote><math>\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)</math></blockquote>The binormal vector is perpendicular to both the unit tangent and the unit normal because of the properties of the cross product. The magnitude of the binormal is always 1 because<blockquote><math>|\mathbf{B}(t)|=|\mathbf{T}(t)||\mathbf{N}(t)|\sin\theta=1</math></blockquote>
 +
 +
==== The normal plane, osculating plane, and the osculating circle ====
 +
 +
* The normal plane is the plane determined by the normal and binormal vectors <math>\mathbf{N}\text{ and }\mathbf{B}</math>. The normal plane consists of all lines that are orthogonal to the tangent vector <math>\mathbf{T}</math>.
 +
* The osculating plane is the plane determined by the unit tangent and unit normal <math>\mathbf{T}\text{ and }\mathbf{N}</math>. It is the plane that comes closest to containing the part of the curve near a point where <math>t=t_0</math>.
 +
* The osculating circle is the circle that lies in the osculating plane towards the direction of <math>\mathbf{N}</math> with a radius <math>r=\frac{1}{\kappa}</math> (the inverse of the curvature, which we will immediately discuss after this). It best describes how the curve behaves near the point where <math>t=t_0</math> because it shares the same tangent, normal, and curvature at that point.
 +
These concepts are very important in the branch of differential geometry and in its applications to the motion of spacecraft.
 +
 +
=== Curvature ===
 +
The curvature of a curve at a given point is a measure of how quickly the curve changes direction at that point. We define it to be the magnitude of the rate of change of the unit tangent with respect to arc length. We use arc length so that the curvature will be independent of parametrization.<blockquote>Suppose that a space curve has vector function <math>\mathbf{r}</math>, unit tangent vector <math>\mathbf{T}</math>, and arc length <math>s</math>. The curvature of this curve is: <math>\kappa=\biggl|\frac{d\mathbf{T}(s)}{ds}\biggr|</math>.</blockquote>There are two other ways to express the curvature. We can express curvature in terms of <math>t</math> instead of <math>s</math> by utilizing the chain rule (recall that <math>\frac{dL}{dt}=|\mathbf{r}'(t)|</math>):<blockquote><math>\kappa=\biggl|\frac{d\mathbf{T}}{ds}\biggr|=\Biggl|\frac{\frac{d\mathbf{T}}{ds}\frac{ds}{dt}}{\frac{ds}{dt}}\Biggr|=\Biggl|\frac{\frac{d\mathbf{T}(t)}{dt}}{\frac{ds}{dt}}\Biggr|=\frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}</math></blockquote>The third way is more complicated to deduce, but it is often more convenient to apply because it only requires <math>\mathbf{r}(t)</math> and its derivatives.<blockquote><math>\kappa(t)=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}</math></blockquote>And now for the proof for this theorem:<blockquote>According to the definition of the unit tangent vector, we know that <math>\mathbf{r}'=|\mathbf{r}'|\mathbf{T}</math>. So the second derivative of <math>\mathbf{r}</math> should be:<blockquote><math>\begin{align}
 +
\mathbf{r}'' & = (|\mathbf{r}'|\ \mathbf{T})' \\
 +
& = |\mathbf{r}''|\ \mathbf{T}+|\mathbf{r}'|\ \mathbf{T}' \quad\text{the product rule}\\
 +
\end{align}</math></blockquote>Now we calculate <math>\mathbf{r}'\times\mathbf{r}''</math>.<blockquote><math>\begin{align}
 +
\mathbf{r}'\times\mathbf{r}'' & = \bigl(|\mathbf{r}'|\mathbf{T}\bigr)\times\bigl(|\mathbf{r}''|\ \mathbf{T}+|\mathbf{r}'|\ \mathbf{T}' \bigr)\quad\text{substitution} \\
 +
& = \bigl(|\mathbf{r}'|\mathbf{T}\times|\mathbf{r}''|\mathbf{T}\bigr)+\bigl(|\mathbf{r}'|\mathbf{T}\times|\mathbf{r}'|\mathbf{T}'\bigr)\quad\text{distribution} \\
 +
& = |\mathbf{r}'||\mathbf{r}''|(\mathbf{T}\times\mathbf{T})+|\mathbf{r}'|^2(\mathbf{T}\times\mathbf{T}')\quad\text{rearrangement} \\
 +
& = |\mathbf{r}'|^2(\mathbf{T}\times\mathbf{T}') \quad\text{realizing that }\mathbf{T}\times\mathbf{T}=0 \\
 +
\end{align}</math></blockquote>And then we calculate <math>|\mathbf{r}'\times\mathbf{r}''|</math>.<blockquote><math>\begin{align}
 +
|\mathbf{r}'\times\mathbf{r}''| & = |\mathbf{r}'|^2|\mathbf{T}\times\mathbf{T}'|\quad\text{substitution} \\
 +
& = |\mathbf{r}'|^2|\mathbf{T}||\mathbf{T}'|\sin\theta \quad\text{the magnitude for the cross product} \\
 +
& = |\mathbf{r}'|^2|\mathbf{T}'| \quad\text{when you realize }|\mathbf{T}|=1\text{ and }\mathbf{T}\perp\mathbf{T}' \\
 +
\end{align}</math></blockquote>We rearrange the equation into:<blockquote><math>|\mathbf{T}'|=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^2}</math></blockquote>Since <math>\kappa=\frac{|\mathbf{T}'|}{|\mathbf{r}'|}</math>, we can substitute <math>|\mathbf{T}'|</math> with <math>\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^2}</math> and get: <math>\kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}</math></blockquote>Here is a little summary on ways to calculate the curvature.<center>
 +
{| class="wikitable"
 +
!Definition
 +
!With parametrization with respect to <math>\mathbf{t}</math> 
 +
!In terms of <math>\mathbf{r}(t)</math> and its derivatives
 +
|-
 +
|align=center|<math>\kappa=\biggl|\frac{d\mathbf{T}}{ds}\biggr|</math>
 +
|align=center|<math>\kappa=\frac{|\mathbf{T}'|}{|\mathbf{r}'|}</math>
 +
|align=center|<math>\kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}</math>
 +
|}</center>
 +
 +
== Motion in space ==
 +
 +
=== Velocity and acceleration ===
 +
Remember in 2-dimensional calculus, we mentioned that a particle with displacement function <math>f(x)</math> has velocity <math>f'(x)</math> and acceleration <math>f''(x)</math>. In vector functions, the definition is basically the same. Suppose a particle moves through space so that its position vector at time <math>t</math> is <math>\mathbf{r}(t)</math>, its velocity function and acceleration function are:<blockquote><math>\mathbf{v}(t)=\lim_{h\rightarrow 0}\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}=\mathbf{r}'(t)\quad</math> and <math>\quad\mathbf{a}(t)=\lim_{h\rightarrow 0}\frac{\mathbf{v}(t+h)-\mathbf{v}(t)}{h}=\mathbf{v}'(t)=\mathbf{r}''(t)</math></blockquote>Simply put: <math>\mathbf{r}''(t)=\mathbf{v}'(t)=\mathbf{a}(t)</math>.
 +
 +
 +
The speed of the particle ignores the direction. It is the magnitude of the velocity vector: <math>|\mathbf{v}(t)|</math>. The distance traveled by the particle from <math>t=a \text{ to } b</math> is <math>\int_a^b|\mathbf{v}(t)|dt</math>, which is also the formula for the arc length.
 +
 +
 +
With the help of the Fundamental Theorem of Calculus, we can deduce the velocity function and position function when we know the particle's acceleration.<blockquote><math>\mathbf{v}(t)=\mathbf{v}(t_0)+\int_{t_0}^t\mathbf{a}(u)du\quad</math> and <math>\quad\mathbf{r}(t)=\mathbf{r}(t_0)+\int_{t_0}^t\mathbf{v}(u)du</math></blockquote>
 +
 +
=== Tangential and normal acceleration ===
 +
We can split the acceleration vector into two components: the tangential acceleration <math>a_T</math> and the normal acceleration <math>a_N</math>. The tangential acceleration faces the same direction as the unit tangent vector (<math>\mathbf{T}</math>), and the normal acceleration faces the same direction as the unit normal vector (<math>\mathbf{N}</math>). Since both <math>\mathbf{T}</math> and <math>\mathbf{N}</math> are unit vectors, the acceleration vector can be written as the sum of two vectors:<blockquote><math>\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}</math></blockquote>Our goal is to figure out how to describe the two components.<blockquote>Recall that <math>\mathbf{T}=\frac{\mathbf{v}}{|\mathbf{v}|}</math>, thus<blockquote><math>\mathbf{v}=|\mathbf{v}|\mathbf{T}</math></blockquote>Now we differentiate both sides of the equation,<blockquote><math>\begin{align}
 +
\mathbf{v}' & = \bigl(|\mathbf{v}|\mathbf{T}\bigr)' \\
 +
& = |\mathbf{v}|'\mathbf{T}+|\mathbf{v}|\mathbf{T}'\\
 +
\end{align}</math></blockquote>So we get <math>\mathbf{a}=|\mathbf{v}|'\mathbf{T}+|\mathbf{v}|\mathbf{T}'</math>. <blockquote>Recall that <math>\kappa=\frac{\big|\mathbf{T}'\big|}{\big|\mathbf{r}'\big|}=\frac{\big|\mathbf{T}'\big|}{\big|\mathbf{v}\big|}</math>, thus <math>\big|\mathbf{T}'\big|=\kappa|\mathbf{v}|</math>.</blockquote><blockquote>Recall that <math>\mathbf{N}=\frac{\mathbf{T}'}{\big|\mathbf{T}'\big|}</math>, thus <math>\mathbf{T}'=\big|\mathbf{T}'\big|\mathbf{N}=\kappa|\mathbf{v}|\mathbf{N}</math>.</blockquote>We substitute <math>\mathbf{T}'</math> for <math>\kappa|\mathbf{v}|\mathbf{N}</math> to yield: <math>\mathbf{a}=|\mathbf{v}|'\mathbf{T}+\kappa|\mathbf{v}|^2\mathbf{N}</math></blockquote>That leaves us with:<blockquote><math>a_T=|\mathbf{v}|'\quad</math> and <math>\quad a_N=\kappa|\mathbf{v}|^2</math></blockquote>
 +
 +
<math>\blacksquare</math>
 +
 +
Of course, it will be more convenient if those components can be written in terms if <math>\mathbf{r}(t)</math> and its derivatives. Suppose that <math>\theta</math> is the angle between <math>\mathbf{T}</math> and <math>\mathbf{a}</math>, then we can write <math>a_T,a_N</math> like this:<blockquote><math>\begin{align}
 +
a_T & = |\mathbf{a}(t)|\cos\theta \\
 +
& = \frac{|\mathbf{v}(t)||\mathbf{a}(t)|\cos\theta}{|\mathbf{v}(t)|} \quad\text{algebraic manipulation}\\
 +
& = \frac{\mathbf{v}(t)\cdot\mathbf{a}(t)}{|\mathbf{v}(t)|}\quad\text{dot product}\\
 +
& = \frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r}'(t)|}\quad\text{in terms of }\mathbf{r}(t)\text{ and its derivatives}\\
 +
\end{align}</math><math>\begin{align}
 +
a_N & = |\mathbf{a}(t)|\sin\theta \\
 +
& = \frac{|\mathbf{v}(t)||\mathbf{a}(t)|\sin\theta}{|\mathbf{v}(t)|} \quad\text{algebraic manipulation}\\
 +
& = \frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{|\mathbf{v}(t)|}\quad\text{cross product}\\
 +
& = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|}\quad\text{in terms of }\mathbf{r}(t)\text{ and its derivatives}\\
 +
\end{align}</math></blockquote>
 +
 +
<math>\blacksquare</math>
 +
 +
To sum it up:<blockquote><math>a_T=|\mathbf{v}|'=\frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r}'(t)|}\quad</math> and <math>\quad a_N=\kappa|\mathbf{v}|^2= \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|}</math></blockquote>
 +
 +
== Expansion ==
 +
We only discussed vector functions with three variables <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle</math>. How about expanding our understanding of vector functions into <math>n</math> variables? Suppose we have a curve with vector function:<blockquote><math>\mathbf{r}(t)=
 +
\begin{pmatrix}
 +
f_1(t) \\ f_2(t) \\ \vdots \\ f_n(t) \\
 +
\end{pmatrix}</math></blockquote>
 +
 +
=== Limits ===
 +
The limit of a vector function is defined as:<blockquote><math>\lim_{t\rightarrow c} \mathbf{r}(t)=
 +
\begin{pmatrix}
 +
\lim_{t\rightarrow c} f_1(t) \\ \lim_{t\rightarrow c} f_2(t) \\ \vdots \\ \lim_{t\rightarrow c} f_n(t) \\
 +
\end{pmatrix}</math></blockquote>
 +
 +
=== Differentiation and integration ===
 +
The derivative of a vector function is defined as:<blockquote><math>\mathbf{r}'(t)=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}</math>, thus <math>\quad\mathbf{r}'(t)=
 +
\begin{pmatrix}
 +
f_1'(t) \\ f_2'(t) \\ \vdots \\ f_n'(t) \\
 +
\end{pmatrix}</math></blockquote>All differentiation rules apply.
 +
 +
 +
We can expand integration into:<blockquote><math>\int\mathbf{r}(t)dt=
 +
\begin{pmatrix}
 +
\int f_1(t)dt \\ \int f_2(t)dt \\ \vdots \\ \int f_n(t)dt \\
 +
\end{pmatrix}
 +
+\mathbf{C} </math></blockquote>Then, the arc length will become:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math>or <math>\quad L=\int_a^b \sqrt{[f_1'(t)]^2+[f_2'(t)]^2+\cdots+[f_n'(t)]^2}dt </math></blockquote>{{BookCat}}
 +
 +
==Resources==
 
* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/13%3A_Vector-Valued_Functions/13.1%3A_Vector-Valued_Functions_and_Space_Curves Vector-Valued Functions and Space Curves], Mathematics LibreTexts
 
* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/13%3A_Vector-Valued_Functions/13.1%3A_Vector-Valued_Functions_and_Space_Curves Vector-Valued Functions and Space Curves], Mathematics LibreTexts

Revision as of 13:53, 1 October 2021

Introduction

If we have a function , we say that 's image (the set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big\{\vec{f}(t)\Big|t\in\R\Big\}} - or some subset of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \R} ) is a curve in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \R^n} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}} is its parametrization.

Parameterizations are not necessarily unique - for example, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=(\cos(t),\sin(t))} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\in[0,2\pi)} is one parametrization of the unit circle, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{g}(t)=(\cos(at),\sin(at))} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\in\left[0,\frac{2\pi}{a}\right)} is a whole family of parameterizations of that circle.

Collision and intersection points

Say we have two different curves. It may be important to consider

  • points the two curves share - where they intersect
  • intersections which occur for the same value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} - where they collide.

Intersection points

Firstly, we have two parameterizations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{g}(t)} , and we want to find out when they intersect, this means that we want to know when the function values of each parametrization are the same. This means that we need to solve

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=\vec{g}(s)}

because we're seeking the function values independent of the times they intersect.

For example, if we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=(t,3t)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{g}(t)=(t,t^2)} , and we want to find intersection points:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=\vec{g}(s)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (t,3t)=(s,s^2)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=s\ ,\ 3t=s^2}

with solutions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (t,s)=(0,0)\ ,\ (3,9)}

So, the two curves intersect at the points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,0)\ ,\ (3,9)} .

Collision points

However, if we want to know when the points "collide", with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{g}(t)} , we need to know when both the function values and the times are the same, so we need to solve instead

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=\vec{g}(t)}

For example, using the same functions as before, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=(t, 3t)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{g}(t)=(t,t^2)} , and we want to find collision points:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f}(t)=\vec{g}(t)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (t,3t)=(t,t^2)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t\ ,\ 3t=t^2}

which gives solutions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0,3} So the collision points are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,0)\ ,\ (3,9)} .

We may want to do this to actually model physical problems, such as in ballistics.

Intersection vector functions

We can also use vector functions to represent the curve of intersection of two surfaces. For example, we want to know the curve of intersection of the cylinder Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+y^2=1} and the plane Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y+z=2} .

Vector functions rely on parameterizations, so we can rewrite the equation of the cylinder into: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos^2 t+\sin^2 t=1} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\cos t,y=\sin t} .


From the equation of the plane, we know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z=2-y=2-\sin t} . Thus the corresponding vector equation is:


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)=\langle \cos t,\sin t,2-\sin t\rangle}

Limits and Continuity

The limit of a vector function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}} is defined by taking the limits of its component functions. Template:Calculus/Def

And the requirement for continuity is also simple:

A vector function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}} is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{t\rightarrow a}\mathbf{r}(t)=\mathbf{r}(a)} .

Derivatives and Integrals

Differentiation

Recall that the first derivative of a scalar function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}}

The first derivative of a vector function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} is defined in much the same way:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}'(t)=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}}

We can use this definition to prove that the derivative of a vector function can be presented as the derivative of its component functions.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{r}'(t) & = \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}[\mathbf{r}(t+\Delta t)-\mathbf{r}(t)] \quad \text{definition}\\ & = \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}[\langle f(t+\Delta t),g(t+\Delta t),h(t+\Delta t)\rangle-\langle f(t),g(t),h(t) \rangle] \quad \text{component form}\\ & = \lim_{\Delta t\rightarrow 0}\langle \frac{f(t+\Delta t)-f(t)}{\Delta t},\frac{g(t+\Delta t)-g(t)}{\Delta t},\frac{h(t+\Delta t)-h(t)}{\Delta t}\rangle \quad\text{vector addition and multiplication}\\ & = \langle \lim_{\Delta t\rightarrow 0}\frac{f(t+\Delta t)-f(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\frac{g(t+\Delta t)-g(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\frac{h(t+\Delta t)-h(t)}{\Delta t}\rangle\quad\text{definition}\\ & = \langle f'(t),g'(t),h'(t)\rangle \quad\text{definition}\\ \end{align} }

Thus, using the same method, we can derive the second derivative and so on. Template:Calculus/DefSo the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th-order derivative should look like this:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}^{(n)}(t)=\langle f^{(n)}(t),g^{(n)}(t),h^{(n)}(t)\rangle}

Differentiation rules

Just like real-valued functions, there are some differentiation rules in the world of vector functions. The factor that makes vector differentiation rules slight more complicated is the product rule because there are two kinds of multiplication in vectors: dot product and cross product. Template:Calculus/Def Naturally, we will prove that those rules are correct. Let us assume that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{u}(t)=\langle f_1(t),f_2(t),f_3(t)\rangle} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}(t)=\langle g_1(t),g_2(t),g_3(t)\rangle} .


Rule 1: the addition rule

Rule 2: the scalar multiplication rule

Rule 3: the dot product rule

Rule 4: the cross product rule

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d}{dt}[\mathbf{u}(t)\times\mathbf{v}(t)] & = \frac{d}{dt}[\langle f_1(t),f_2(t),f_3(t)\rangle\times\langle g_1(t),g_2(t),g_3(t)\rangle]\quad\text{component form}\\ & = \frac{d}{dt}\begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ f_1(t) & f_2(t) & f_3(t) \\ g_1(t) & g_2(t) & g_3(t) \\ \end{vmatrix}\quad\text{cross product}\\ & = \frac{d}{dt}\begin{pmatrix} f_2(t)g_3(t)-f_3(t)g_2(t)\\ f_3(t)g_1(t)-f_1(t)g_3(t)\\ f_1(t)g_2(t)-f_2(t)g_1(t)\\ \end{pmatrix}\quad\\ & = \begin{pmatrix} \frac{d}{dt}(f_2(t)g_3(t)-f_3(t)g_2(t))\\ \frac{d}{dt}(f_3(t)g_1(t)-f_1(t)g_3(t))\\ \frac{d}{dt}(f_1(t)g_2(t)-f_2(t)g_1(t))\\ \end{pmatrix}\quad\text{distribution}\\ & = \begin{pmatrix} f_2'(t)g_3(t)+f_2(t)g_3'(t)-f_3'(t)g_2(t)-f_3(t)g_2'(t)\\ f_3'(t)g_1(t)+f_3(t)g_1'(t)-f_1'(t)g_3(t)-f_1(t)g_3'(t)\\ f_2'(t)g_3(t)+f_2(t)g_3'(t)-f_3'(t)g_2(t)-f_3(t)g_2'(t)\\ \end{pmatrix}\quad\text{multiplication rule for real-valued functions}\\ & = \begin{pmatrix} f_2'(t)g_3(t)-f_3'(t)g_2(t)+f_2(t)g_3'(t)-f_3(t)g_2'(t)\\ f_3'(t)g_1(t)-f_1'(t)g_3(t)+f_3(t)g_1'(t)-f_1(t)g_3'(t)\\ f_2'(t)g_3(t)-f_3'(t)g_2(t)+f_2(t)g_3'(t)-f_3(t)g_2'(t)\\ \end{pmatrix}\quad\text{rearrangement}\\ & = \begin{pmatrix} f_2'(t)g_3(t)-f_3'(t)g_2(t)\\f_3'(t)g_1(t)-f_1'(t)g_3(t)\\f_2'(t)g_3(t)-f_3'(t)g_2(t)\\\end{pmatrix}+ \begin{pmatrix} f_2(t)g_3'(t)-f_3(t)g_2'(t)\\f_3(t)g_1'(t)-f_1(t)g_3'(t)\\f_2(t)g_3'(t)-f_3(t)g_2'(t)\\\end{pmatrix}\quad\text{vector addtion}\\ & = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ f_1'(t) & f_2'(t) & f_3'(t) \\ g_1(t) & g_2(t) & g_3(t) \\\end{vmatrix}+\begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ f_1(t) & f_2(t) & f_3(t) \\ g_1'(t) & g_2'(t) & g_3'(t) \\ \end{vmatrix}\quad\text{cross product}\\ & = \begin{pmatrix} f_1'(t) \\ f_2'(t) \\ f_3'(t)\\\end{pmatrix}\times\begin{pmatrix} g_1(t) \\ g_2(t) \\ g_3(t) \\\end{pmatrix}+ \begin{pmatrix} f_1(t) \\ f_2(t) \\ f_3(t)\\\end{pmatrix}\times\begin{pmatrix} g_1'(t) \\ g_2'(t) \\ g_3'(t) \\\end{pmatrix}\quad\\ & = \mathbf{u}'(t)\times\mathbf{v}(t)+\mathbf{u}(t)\times\mathbf{v}'(t)\quad\text{vector form}\\ \end{align}}

Rule 5: the chain rule

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d}{dt}[\mathbf{u}(f(t))] & = \frac{d}{dt}\begin{pmatrix} f_1(f(t)) \\ f_2(f(t)) \\ f_3(f(t)) \\\end{pmatrix}\quad\text{component form}\\ & = \begin{pmatrix} \frac{d}{dt}f_1(f(t)) \\ \frac{d}{dt}f_2(f(t)) \\ \frac{d}{dt}f_3(f(t)) \\\end{pmatrix}\quad\text{distribution}\\ & = \begin{pmatrix} f'(t)f_1'(f(t)) \\ f'(t)f_2'(f(t)) \\ f'(t)f_3'(f(t)) \\\end{pmatrix}\quad \text{chain rule for real-valued functions}\\ & = f'(t)\begin{pmatrix} f_1'(f(t)) \\ f_2'(f(t)) \\ f_3'(f(t)) \\\end{pmatrix}\quad\text{scalar multiplication}\\ & = f'(t)\mathbf{u}'(f(t))\quad\text{vector form}\\ \end{align}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Angle between curves

Inner-product-angle.png

We can then formulate the concept of the angle between two curves by considering the angle between the two tangent vectors. If two curves, parametrized by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f_1}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f_2}} intersect at some point, which means that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f_1}(s)=\vec{f_2}(t)=c}

the angle between these two curves at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is the angle between the tangent vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f_1'}(s)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{f_2'}(t)} , using the dot product, is given by

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \arccos\left(\frac{\vec{f_1'}(s)\cdot\vec{f_2'}(t)}{\Big\|\vec{f_1'}(s)\Big\|\Big\|\vec{f_2'}(t)\Big\|}\right)}

Integration

Similar to real-valued functions, the definite integral of a vector function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} is defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_a^b\mathbf{r}(t)dt & = \lim_{n\rightarrow\infty}\sum_{i=1}^n\mathbf{r}(t_i^*)\Delta t \\ & = \lim_{n\rightarrow\infty}\biggl[\biggl(\sum_{i=1}^nf(t_i^*)\Delta t\biggr)\mathbf{i}+\biggl(\sum_{i=1}^ng(t_i^*)\Delta t\biggr)\mathbf{j}+\biggl(\sum_{i=1}^nh(t_i^*)\Delta t\biggr)\mathbf{k}\biggr]\\ & = \biggl(\int_a^bf(t)dt\biggr)\mathbf{i}+\biggl(\int_a^bg(t)dt\biggr)\mathbf{j}+\biggl(\int_a^bh(t)dt\biggr)\mathbf{k} \end{align} }

We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:

Template:Calculus/Def

For indefinite integrals, the definition is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\mathbf{r}(t)dt=\biggl(\int f(t)dt\biggr)\mathbf{i}+\biggl(\int g(t)dt\biggr)\mathbf{j}+\biggl(\int h(t)dt\biggr)\mathbf{k}+\mathbf{C} }

Arc length

Recall that in Chapter Template:Calculus/map page, we deduced the length of a curve with parametric equations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} x=f(t) \\ y=g(t) \end{cases} } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\le t\le b } should be:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_a^b\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2}dt }

Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.

Template:Calculus/Def

For those who prefer simplicity, the formula can be rewritten into:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_a^b|\mathbf{r}'(t)|dt\quad } or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad\frac{dL}{dt}=|\mathbf{r}'(t)|}

Reparametrization

Suppose that there is a vector function which describes the displacement of a particle with respect to time and has the equation:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)=\cos t\ \mathbf{i}+\sin t\ \mathbf{j}+t\ \mathbf{k} }

However, for some reason, we do not want to know the displacement of this particle with respect to time. Instead, we want to know its displacement with respect to its traveled distance (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s } ) from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,0,0) } in the direction of increasing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t } . In order to do so, we need to find a way to describe time as a function of distance. In other words, we need to find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t(s) } . We can use the formula for the arc length to establish the relationship between time and distance because arc length in this case describes the distance traveled by the particle.


Before we start calculating, we need to introduce the arc length function. Template:Calculus/Def

According to the definition, the arc length function for our curve Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t) } should be

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s(t)=\int_0^t|\mathbf{r}'(u)|du=\int_0^t\sqrt{(-\sin u)^2+(\cos u)^2+1^2}du=\sqrt{2}u\bigg]_0^t=\sqrt{2}t }

Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=0 } because the initial point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,0,0) } corresponds to the parameter value Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0 } . Since it is in the direction of increasing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t } , the integration direction should be from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 } to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t } .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t(s)=\frac{s}{\sqrt{2}} }

Then we substitute the original function and get the answer:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(s)=\cos \frac{s}{\sqrt{2}}\ \mathbf{i}+\sin \frac{s}{\sqrt{2}}\ \mathbf{j}+\frac{s}{\sqrt{2}}\ \mathbf{k} }

Reparametrization has important applications in real life because sometimes we want to know a value with respect to different variables. In this case, instead of describing the path of a particle with respect to time, we described its path with respect to its distance, which will be very useful in certain situations.

Curvature

Terminology

Before we start discussing curvature, there are some important vectors and concepts we need to be at least aware of.

The unit tangent vector

In the differentiation section of this chapter, we discussed the derivatives of a vector function. We know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}(t)=\mathbf{r}'(t)} at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t_0} is tangent to the curve Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t_0} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}'(t)} is called the tangent vector. The unit tangent vector, however, eliminates the aspect of magnitude because it is defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}}

As we can see, the magnitude of the unit tangent vector is always Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . We can imagine that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} as the displacement of a particle with respect to time. So, the unit tangent vector can be perceived as the direction of the velocity of the particle with respect to time. It can also be perceived as the direction of the tangential acceleration of the particle with respect to time. We will discuss motion in space in the next section, but this is a useful method to intuitively understand some vectors.

The unit normal vector

The unit normal vector is defined as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}}

The unit normal is orthogonal to the unit tangent because since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{T}(t)|=1} , we can get that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dt}|\mathbf{T}(t)|^2=0=\frac{d}{dt}[\mathbf{T}(t)\cdot\mathbf{T}(t)]=2\mathbf{T}'(t)\cdot\mathbf{T}(t)}


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Leftrightarrow \mathbf{T}'(t)\cdot\mathbf{T}(t)=0}

This means that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}'(t)} is orthogonal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}(t)} . Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}(t)} is orthogonal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}(t)} . We can imagine that the unit normal vector is the direction of the normal acceleration of the particle with respect to time.

The binormal vector

The binormal vector is defined as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)}

The binormal vector is perpendicular to both the unit tangent and the unit normal because of the properties of the cross product. The magnitude of the binormal is always 1 because

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{B}(t)|=|\mathbf{T}(t)||\mathbf{N}(t)|\sin\theta=1}

The normal plane, osculating plane, and the osculating circle

  • The normal plane is the plane determined by the normal and binormal vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}\text{ and }\mathbf{B}} . The normal plane consists of all lines that are orthogonal to the tangent vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}} .
  • The osculating plane is the plane determined by the unit tangent and unit normal Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}\text{ and }\mathbf{N}} . It is the plane that comes closest to containing the part of the curve near a point where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t_0} .
  • The osculating circle is the circle that lies in the osculating plane towards the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}} with a radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\frac{1}{\kappa}} (the inverse of the curvature, which we will immediately discuss after this). It best describes how the curve behaves near the point where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=t_0} because it shares the same tangent, normal, and curvature at that point.

These concepts are very important in the branch of differential geometry and in its applications to the motion of spacecraft.

Curvature

The curvature of a curve at a given point is a measure of how quickly the curve changes direction at that point. We define it to be the magnitude of the rate of change of the unit tangent with respect to arc length. We use arc length so that the curvature will be independent of parametrization.

Suppose that a space curve has vector function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}} , unit tangent vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}} , and arc length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} . The curvature of this curve is: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\biggl|\frac{d\mathbf{T}(s)}{ds}\biggr|} .

There are two other ways to express the curvature. We can express curvature in terms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} instead of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} by utilizing the chain rule (recall that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dL}{dt}=|\mathbf{r}'(t)|} ):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\biggl|\frac{d\mathbf{T}}{ds}\biggr|=\Biggl|\frac{\frac{d\mathbf{T}}{ds}\frac{ds}{dt}}{\frac{ds}{dt}}\Biggr|=\Biggl|\frac{\frac{d\mathbf{T}(t)}{dt}}{\frac{ds}{dt}}\Biggr|=\frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}}

The third way is more complicated to deduce, but it is often more convenient to apply because it only requires Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} and its derivatives.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa(t)=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}}

And now for the proof for this theorem:

According to the definition of the unit tangent vector, we know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}'=|\mathbf{r}'|\mathbf{T}} . So the second derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}} should be:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{r}'' & = (|\mathbf{r}'|\ \mathbf{T})' \\ & = |\mathbf{r}''|\ \mathbf{T}+|\mathbf{r}'|\ \mathbf{T}' \quad\text{the product rule}\\ \end{align}}

Now we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}'\times\mathbf{r}''} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{r}'\times\mathbf{r}'' & = \bigl(|\mathbf{r}'|\mathbf{T}\bigr)\times\bigl(|\mathbf{r}''|\ \mathbf{T}+|\mathbf{r}'|\ \mathbf{T}' \bigr)\quad\text{substitution} \\ & = \bigl(|\mathbf{r}'|\mathbf{T}\times|\mathbf{r}''|\mathbf{T}\bigr)+\bigl(|\mathbf{r}'|\mathbf{T}\times|\mathbf{r}'|\mathbf{T}'\bigr)\quad\text{distribution} \\ & = |\mathbf{r}'||\mathbf{r}''|(\mathbf{T}\times\mathbf{T})+|\mathbf{r}'|^2(\mathbf{T}\times\mathbf{T}')\quad\text{rearrangement} \\ & = |\mathbf{r}'|^2(\mathbf{T}\times\mathbf{T}') \quad\text{realizing that }\mathbf{T}\times\mathbf{T}=0 \\ \end{align}}

And then we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{r}'\times\mathbf{r}''|} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} |\mathbf{r}'\times\mathbf{r}''| & = |\mathbf{r}'|^2|\mathbf{T}\times\mathbf{T}'|\quad\text{substitution} \\ & = |\mathbf{r}'|^2|\mathbf{T}||\mathbf{T}'|\sin\theta \quad\text{the magnitude for the cross product} \\ & = |\mathbf{r}'|^2|\mathbf{T}'| \quad\text{when you realize }|\mathbf{T}|=1\text{ and }\mathbf{T}\perp\mathbf{T}' \\ \end{align}}

We rearrange the equation into:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{T}'|=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^2}}

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{|\mathbf{T}'|}{|\mathbf{r}'|}} , we can substitute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{T}'|} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^2}} and get: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}}

Here is a little summary on ways to calculate the curvature.

Definition With parametrization with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{t}} In terms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} and its derivatives
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\biggl|\frac{d\mathbf{T}}{ds}\biggr|} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{|\mathbf{T}'|}{|\mathbf{r}'|}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}}

Motion in space

Velocity and acceleration

Remember in 2-dimensional calculus, we mentioned that a particle with displacement function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} has velocity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)} and acceleration Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f''(x)} . In vector functions, the definition is basically the same. Suppose a particle moves through space so that its position vector at time Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} , its velocity function and acceleration function are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}(t)=\lim_{h\rightarrow 0}\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}=\mathbf{r}'(t)\quad} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad\mathbf{a}(t)=\lim_{h\rightarrow 0}\frac{\mathbf{v}(t+h)-\mathbf{v}(t)}{h}=\mathbf{v}'(t)=\mathbf{r}''(t)}

Simply put: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}''(t)=\mathbf{v}'(t)=\mathbf{a}(t)} .


The speed of the particle ignores the direction. It is the magnitude of the velocity vector: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{v}(t)|} . The distance traveled by the particle from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=a \text{ to } b} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b|\mathbf{v}(t)|dt} , which is also the formula for the arc length.


With the help of the Fundamental Theorem of Calculus, we can deduce the velocity function and position function when we know the particle's acceleration.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}(t)=\mathbf{v}(t_0)+\int_{t_0}^t\mathbf{a}(u)du\quad} and

Tangential and normal acceleration

We can split the acceleration vector into two components: the tangential acceleration and the normal acceleration Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_N} . The tangential acceleration faces the same direction as the unit tangent vector (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}} ), and the normal acceleration faces the same direction as the unit normal vector (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}} ). Since both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}} are unit vectors, the acceleration vector can be written as the sum of two vectors:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}}

Our goal is to figure out how to describe the two components.

Recall that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}=\frac{\mathbf{v}}{|\mathbf{v}|}} , thus

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}=|\mathbf{v}|\mathbf{T}}

Now we differentiate both sides of the equation,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{v}' & = \bigl(|\mathbf{v}|\mathbf{T}\bigr)' \\ & = |\mathbf{v}|'\mathbf{T}+|\mathbf{v}|\mathbf{T}'\\ \end{align}}

So we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{a}=|\mathbf{v}|'\mathbf{T}+|\mathbf{v}|\mathbf{T}'} .

Recall that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{\big|\mathbf{T}'\big|}{\big|\mathbf{r}'\big|}=\frac{\big|\mathbf{T}'\big|}{\big|\mathbf{v}\big|}} , thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \big|\mathbf{T}'\big|=\kappa|\mathbf{v}|} .

Recall that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{N}=\frac{\mathbf{T}'}{\big|\mathbf{T}'\big|}} , thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}'=\big|\mathbf{T}'\big|\mathbf{N}=\kappa|\mathbf{v}|\mathbf{N}} .

We substitute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}'} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa|\mathbf{v}|\mathbf{N}} to yield: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{a}=|\mathbf{v}|'\mathbf{T}+\kappa|\mathbf{v}|^2\mathbf{N}}

That leaves us with:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_T=|\mathbf{v}|'\quad} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad a_N=\kappa|\mathbf{v}|^2}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Of course, it will be more convenient if those components can be written in terms if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)} and its derivatives. Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} is the angle between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{T}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{a}} , then we can write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_T,a_N} like this:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} a_T & = |\mathbf{a}(t)|\cos\theta \\ & = \frac{|\mathbf{v}(t)||\mathbf{a}(t)|\cos\theta}{|\mathbf{v}(t)|} \quad\text{algebraic manipulation}\\ & = \frac{\mathbf{v}(t)\cdot\mathbf{a}(t)}{|\mathbf{v}(t)|}\quad\text{dot product}\\ & = \frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r}'(t)|}\quad\text{in terms of }\mathbf{r}(t)\text{ and its derivatives}\\ \end{align}} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} a_N & = |\mathbf{a}(t)|\sin\theta \\ & = \frac{|\mathbf{v}(t)||\mathbf{a}(t)|\sin\theta}{|\mathbf{v}(t)|} \quad\text{algebraic manipulation}\\ & = \frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{|\mathbf{v}(t)|}\quad\text{cross product}\\ & = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|}\quad\text{in terms of }\mathbf{r}(t)\text{ and its derivatives}\\ \end{align}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

To sum it up:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_T=|\mathbf{v}|'=\frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r}'(t)|}\quad} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad a_N=\kappa|\mathbf{v}|^2= \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|}}

Expansion

We only discussed vector functions with three variables Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle} . How about expanding our understanding of vector functions into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} variables? Suppose we have a curve with vector function:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}(t)= \begin{pmatrix} f_1(t) \\ f_2(t) \\ \vdots \\ f_n(t) \\ \end{pmatrix}}

Limits

The limit of a vector function is defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{t\rightarrow c} \mathbf{r}(t)= \begin{pmatrix} \lim_{t\rightarrow c} f_1(t) \\ \lim_{t\rightarrow c} f_2(t) \\ \vdots \\ \lim_{t\rightarrow c} f_n(t) \\ \end{pmatrix}}

Differentiation and integration

The derivative of a vector function is defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{r}'(t)=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}} , thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad\mathbf{r}'(t)= \begin{pmatrix} f_1'(t) \\ f_2'(t) \\ \vdots \\ f_n'(t) \\ \end{pmatrix}}

All differentiation rules apply.


We can expand integration into:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\mathbf{r}(t)dt= \begin{pmatrix} \int f_1(t)dt \\ \int f_2(t)dt \\ \vdots \\ \int f_n(t)dt \\ \end{pmatrix} +\mathbf{C} }

Then, the arc length will become:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_a^b|\mathbf{r}'(t)|dt\quad } or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad L=\int_a^b \sqrt{[f_1'(t)]^2+[f_2'(t)]^2+\cdots+[f_n'(t)]^2}dt }

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