Difference between revisions of "Motion in Space"
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| + | == 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> | ||
| + | |||
| + | |||
| + | ==Resources== | ||
* [https://openstax.org/books/calculus-volume-3/pages/3-4-motion-in-space Motion in Space], OpenStax | * [https://openstax.org/books/calculus-volume-3/pages/3-4-motion-in-space Motion in Space], OpenStax | ||
Revision as of 09:23, 6 October 2021
Contents
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}} |
Resources
- Motion in Space, OpenStax
