Difference between revisions of "The Calculus of Parametric Equations"

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<strong>Derivatives of Parametric Equations</strong>
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==Derivatives of Parametric Systems==
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Just as we are able to differentiate functions of <math>x</math> , we are able to differentiate <math>x</math> and <math>y</math> , which are functions of <math>t</math> . Consider:
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{{math|size=2em|<math>\begin{align}x&=\sin(t)\\y&=t\end{align}</math>}}
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We would find the derivative of <math>x</math> with respect to <math>t</math> , and the derivative of <math>y</math> with respect to <math>t</math> :
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{{math|size=2em|<math>\begin{align}x'&=\cos(t)\\y'&=1\end{align}</math>}}
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In general, we say that if
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{{math|size=2em|<math>\begin{align}x&=x(t)\\y&=y(t)\end{align}</math>}}
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then:
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{{math|size=2em|<math>\begin{align}x'&=x'(t)\\y'&=y'(t)\end{align}</math>}}
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 +
It's that simple.
 +
 
 +
This process works for any amount of variables.
 +
 
 +
===Slope of Parametric Equations===
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In the above process, <math>x'</math> has told us only the rate at which <math>x</math> is changing, not the rate for <math>y</math> , and vice versa. Neither is the slope.
 +
 
 +
In order to find the slope, we need something of the form <math>\frac{dy}{dx}</math> .
 +
 
 +
We can discover a way to do this by simple algebraic manipulation:
 +
 
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<math>\frac{y'}{x'}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}</math>
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 +
So, for the example in section 1, the slope at any time <math>t</math> :
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<math>\frac{1}{\cos(t)}=\sec(t)</math>
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In order to find a vertical tangent line, set the horizontal change, or <math>x'</math> , equal to 0 and solve.
 +
 
 +
In order to find a horizontal tangent line, set the vertical change, or <math>y'</math> , equal to 0 and solve.
 +
 
 +
If there is a time when both <math>x',y'</math> are 0, that point is called a singular point.
 +
 
 +
===Concavity of Parametric Equations===
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Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.
 +
 
 +
When you have take the derivative of <math>\frac{dy}{dx}</math> in terms of <math>t</math> , you are left with <math>\frac{\frac{d^2y}{dx}}{dt}</math> :
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<math>\frac{d}{dt}\left[\frac{dy}{dx}\right]=\frac{\frac{d^2y}{dx}}{dt}</math> .
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By multiplying this expression by <math>\frac{dt}{dx}</math> , we are able to solve for the second derivative of the parametric equation:
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 +
<math>\frac{\frac{d^2y}{dx}}{dt}\times\frac{dt}{dx}=\frac{d^2y}{dx^2}</math> .
 +
 
 +
Thus, the concavity of a parametric equation can be described as:
 +
 
 +
<math>\frac{d}{dt}\left[\frac{dy}{dx}\right]\times\frac{dt}{dx}</math>
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 +
So for the example in sections 1 and 2, the concavity at any time <math>t</math> :
 +
 
 +
<math>\frac{d}{dt}[\csc(t)]\times\cos(t)=-\csc^2(t)\times\cos(t)</math>
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 +
==Parametric Integration==
 +
Because most parametric equations are given in explicit form, they can be integrated like many other equations. Integration has a variety of applications with respect to parametric equations, especially in kinematics and vector calculus.
 +
 
 +
:{{math|<math>\begin{align}x&=\int x'(t)\mathrm{d}t\\y&=\int y'(t)\mathrm{d}t\end{align}</math>}}
 +
 
 +
So, taking a simple example, with respect to t:
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:{{math|<math>y=\int\cos(t)\mathrm{d}t=\sin(t) + C</math>}}
 +
 
 +
===Arc length===
 +
Consider a function defined by,
 +
:<math>x = f(t)</math>
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:<math>y = g(t)</math>
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 +
Say that <math>f</math> is increasing on some interval, <math>[\alpha, \beta]</math>. Recall, as we have derived in a previous chapter, that the length of the arc created by a function over an interval, <math>[\alpha, \beta]</math>, is given by,
 +
 
 +
:<math>L = \int_\alpha^\beta \sqrt{1 + (f'(x))^2} \mathrm{d}x</math>
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 +
It may assist your understanding, here, to write the above using Leibniz's notation,
 +
 
 +
:<math>L = \int_\alpha^\beta \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x</math>
 +
 
 +
Using the chain rule,
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:<math>\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x}</math>
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We may then rewrite <math>\mathrm{d}x</math>,
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:<math>\frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t</math>
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Hence, <math>L</math> becomes,
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:<math>L = \int_\alpha^\beta \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x}\right)^2} \frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t</math>
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 +
Extracting a factor of <math>\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2</math>,
 +
 
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:<math>L = \int_\alpha^\beta \sqrt{\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t</math>
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As <math>f</math> is increasing on <math>[\alpha,\beta]</math>, <math>\sqrt{\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2} = \frac{\mathrm{d}t}{\mathrm{d}x}</math>, and hence we may write our final expression for <math>L</math> as,
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:<math>\int_\alpha^\beta \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \mathrm{d}t</math>
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====Example====
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Take a circle of radius <math>R</math>, which may be defined with the parametric equations,
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:<math>x = R\sin\theta</math>
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:<math>y = R\cos\theta</math>
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 +
As an example, we can take the length of the arc created by the curve over the interval <math>[0,R]</math>. Writing in terms of <math>\theta</math>,
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:<math>x = 0 \implies \theta = \arcsin\left(\frac{0}{R}\right) = 0</math>
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:<math>x = R \implies \theta = \arcsin\left(\frac{R}{R}\right) = \arcsin(1) = \frac{\pi}{2}</math>
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Computing the derivatives of both equations,
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:<math>\frac{\mathrm{d}x}{\mathrm{d}\theta} = R\cos\theta</math>
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:<math>\frac{\mathrm{d}y}{\mathrm{d}\theta} = -R\sin\theta</math>
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 +
Which means that the arc length is given by,
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:<math>L = \int_0^{\frac{\pi}{2}} \sqrt{(-R\sin\theta)^2 + R^2\cos^2\theta}\mathrm{d}\theta</math>
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 +
By the Pythagorean identity,
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:<math>L = R\int_0^{\frac{\pi}{2}} \mathrm{d}\theta = R\frac{\pi}{2}</math>
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One can use this result to determine the perimeter of a circle of a given radius. As this is the arc length over one "quadrant", one may multiply <math>L</math> by 4 to deduce the perimeter of a circle of radius <math>R</math> to be <math>2\pi R</math>.
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==Resources==
 +
===Derivatives of Parametric Equations===
  
 
* [https://www.youtube.com/watch?v=qyycNbNJAqg The Derivatvie of Parametric Equations] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=qyycNbNJAqg The Derivatvie of Parametric Equations] by James Sousa, Math is Power 4U
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<strong>Area Under Parametric Curves</strong>
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===Area Under Parametric Curves===
 
* [https://www.youtube.com/watch?v=nEWt3-i5E8s Area Under Parametric Curves] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=nEWt3-i5E8s Area Under Parametric Curves] by James Sousa, Math is Power 4U
  
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<strong>Arc Length of Parametric Curves</strong>
+
===Arc Length of Parametric Curves===
 
* [https://www.youtube.com/watch?v=mF-N3eUpsEY Arc Length in Parametric Form] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=mF-N3eUpsEY Arc Length in Parametric Form] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=s-ma8BER3EM Find the Arc Length of a Parametric Curve, Example 1] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=s-ma8BER3EM Find the Arc Length of a Parametric Curve, Example 1] by James Sousa, Math is Power 4U
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<strong>Surface Area of Revolution in Parametric Form</strong>
+
===Surface Area of Revolution in Parametric Form===
 
* [https://www.youtube.com/watch?v=HYQ-KfgdNOE Surface Area of Revolution in Parametric Form] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=HYQ-KfgdNOE Surface Area of Revolution in Parametric Form] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=-HQ2zxGX29A Surface Area of Revolution in Parametric Form, Example 1 (y-axis)] by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=-HQ2zxGX29A Surface Area of Revolution in Parametric Form, Example 1 (y-axis)] by James Sousa, Math is Power 4U

Latest revision as of 16:46, 10 October 2021

Derivatives of Parametric Systems

Just as we are able to differentiate functions of , we are able to differentiate and , which are functions of . Consider:

We would find the derivative of with respect to , and the derivative of with respect to  :

In general, we say that if

then:

It's that simple.

This process works for any amount of variables.

Slope of Parametric Equations

In the above process, has told us only the rate at which is changing, not the rate for , and vice versa. Neither is the slope.

In order to find the slope, we need something of the form .

We can discover a way to do this by simple algebraic manipulation:

So, for the example in section 1, the slope at any time  :

In order to find a vertical tangent line, set the horizontal change, or , equal to 0 and solve.

In order to find a horizontal tangent line, set the vertical change, or , equal to 0 and solve.

If there is a time when both are 0, that point is called a singular point.

Concavity of Parametric Equations

Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.

When you have take the derivative of in terms of , you are left with  :

.

By multiplying this expression by , we are able to solve for the second derivative of the parametric equation:

.

Thus, the concavity of a parametric equation can be described as:

So for the example in sections 1 and 2, the concavity at any time  :

Parametric Integration

Because most parametric equations are given in explicit form, they can be integrated like many other equations. Integration has a variety of applications with respect to parametric equations, especially in kinematics and vector calculus.

So, taking a simple example, with respect to t:

Arc length

Consider a function defined by,

Say that is increasing on some interval, . Recall, as we have derived in a previous chapter, that the length of the arc created by a function over an interval, , is given by,

It may assist your understanding, here, to write the above using Leibniz's notation,

Using the chain rule,

We may then rewrite ,

Hence, becomes,

Extracting a factor of ,

As is increasing on , , and hence we may write our final expression for as,

Example

Take a circle of radius , which may be defined with the parametric equations,

As an example, we can take the length of the arc created by the curve over the interval . Writing in terms of ,

Computing the derivatives of both equations,

Which means that the arc length is given by,

By the Pythagorean identity,

One can use this result to determine the perimeter of a circle of a given radius. As this is the arc length over one "quadrant", one may multiply by 4 to deduce the perimeter of a circle of radius to be .

Resources

Derivatives of Parametric Equations


Area Under Parametric Curves


Arc Length of Parametric Curves


Surface Area of Revolution in Parametric Form