Difference between revisions of "The Calculus of Parametric Equations"

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[https://www.youtube.com/watch?v=qyycNbNJAqg The Derivatvie of Parametric Equations] by James Sousa
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==Derivatives of Parametric Systems==
 +
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:
  
[https://www.youtube.com/watch?v=tnOR99L00Q0 Derivatives of Parametric Equations, Example 1] by James Sousa
+
{{math|size=2em|<math>\begin{align}x&=\sin(t)\\y&=t\end{align}</math>}}
  
[https://www.youtube.com/watch?v=u4xvVwV1ENU Derivatives of Parametric Equations, Example 2] by James Sousa
+
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> :
  
[https://www.youtube.com/watch?v=a0qBBkuroqk Equation of a Tangent Line to a Parametric Curve, Example 1] by James Sousa
+
{{math|size=2em|<math>\begin{align}x'&=\cos(t)\\y'&=1\end{align}</math>}}
  
[https://www.youtube.com/watch?v=l8HTzxaQaOk Equation of a Tangent Line to a Parametric Curve, Example 2] by James Sousa
+
In general, we say that if
  
[https://www.youtube.com/watch?v=0GyRRonZu5E Equation of a Tangent Line to a Parametric Curve, Example 3] by James Sousa
+
{{math|size=2em|<math>\begin{align}x&=x(t)\\y&=y(t)\end{align}</math>}}
  
[https://www.youtube.com/watch?v=gU2LXxKAT3s Determine the Points Where the Tangent Lines are Horizontal or Vertical] by James Sousa
+
then:
  
[https://www.youtube.com/watch?v=MKWsl2W08HE The Second Derivative of Parametric Equations, Part 1] by James Sousa
+
{{math|size=2em|<math>\begin{align}x'&=x'(t)\\y'&=y'(t)\end{align}</math>}}
  
[https://www.youtube.com/watch?v=BIsnZfuGXXk The Second Derivative of Parametric Equations, Part 2] by James Sousa
+
It's that simple.
  
[https://www.youtube.com/watch?v=0D31I0jQMLw The First and Second Derivative of Parametric Equations, Example 1] by James Sousa
+
This process works for any amount of variables.
  
[https://www.youtube.com/watch?v=jUzdYHmh7e0 The First and Second Derivative of Parametric Equations, Example 2] by James Sousa
+
===Slope of Parametric Equations===
 +
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.
  
[https://www.youtube.com/watch?v=nEWt3-i5E8s Area Under Parametric Curves] by James Sousa
+
In order to find the slope, we need something of the form <math>\frac{dy}{dx}</math> .
  
[https://www.youtube.com/watch?v=mF-N3eUpsEY Arc Length in Parametric Form] by James Sousa
+
We can discover a way to do this by simple algebraic manipulation:
  
[https://www.youtube.com/watch?v=s-ma8BER3EM Find the Arc Length of a Parametric Curve, Example 1] by James Sousa
+
<math>\frac{y'}{x'}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}</math>
  
[https://www.youtube.com/watch?v=WM89GkZq-Dw Find the Arc Length of a Parametric Curve, Example 2] by James Sousa
+
So, for the example in section 1, the slope at any time <math>t</math> :
  
[https://www.youtube.com/watch?v=fqA4PrCD5Xk Find the Arc Length of a Loop of a Parametric Curve] by James Sousa
+
<math>\frac{1}{\cos(t)}=\sec(t)</math>
  
[https://www.youtube.com/watch?v=HYQ-KfgdNOE Surface Area of Revolution in Parametric Form] by James Sousa
+
In order to find a vertical tangent line, set the horizontal change, or <math>x'</math> , equal to 0 and solve.
  
[https://www.youtube.com/watch?v=-HQ2zxGX29A Surface Area of Revolution in Parametric Form, Example 1] by James Sousa
+
In order to find a horizontal tangent line, set the vertical change, or <math>y'</math> , equal to 0 and solve.
  
[https://www.youtube.com/watch?v=utu37DbgHvs Surface Area of Revolution in Parametric Form, Example 2] by James Sousa
+
If there is a time when both <math>x',y'</math> are 0, that point is called a singular point.
  
[https://www.youtube.com/watch?v=k5QnaGVk1JI Derivatives of Parametric Equations] by Patrick JMT
+
===Concavity of Parametric Equations===
 +
Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.
  
[https://www.youtube.com/watch?v=RUXgwKoIds4 Derivatives of Parametric Equations, Example 1] by Patrick JMT
+
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> :
  
[https://www.youtube.com/watch?v=azMnzLcdPPo Derivatives of Parametric Equations, Example 2] by Patrick JMT
+
<math>\frac{d}{dt}\left[\frac{dy}{dx}\right]=\frac{\frac{d^2y}{dx}}{dt}</math> .
  
[https://www.youtube.com/watch?v=7ANhq51wjWM The Second Derivative of Parametric Curves] by Patrick JMT
+
By multiplying this expression by <math>\frac{dt}{dx}</math> , we are able to solve for the second derivative of the parametric equation:
  
[https://www.youtube.com/watch?v=GDLZYp2U9g8 Area Under a Parametric Curve] by Patrick JMT
+
<math>\frac{\frac{d^2y}{dx}}{dt}\times\frac{dt}{dx}=\frac{d^2y}{dx^2}</math> .
  
[https://www.youtube.com/watch?v=5fP443JvTUg Arc Length of a Parametric Curve] by Patrick JMT
+
Thus, the concavity of a parametric equation can be described as:
  
[https://www.youtube.com/watch?v=kf2dZWqLnqE Derivatives of Parametric Equations] by The Organic Chemistry Tutor
+
<math>\frac{d}{dt}\left[\frac{dy}{dx}\right]\times\frac{dt}{dx}</math>
  
[https://www.youtube.com/watch?v=vcZPjuG8GrM Tangent Lines of Parametric Curves] by The Organic Chemistry Tutor
+
So for the example in sections 1 and 2, the concavity at any time <math>t</math> :
  
[https://www.youtube.com/watch?v=jD2RIjbv1Us Horizontal and Vertical Tangent Lines of Parametric Curves] by The Organic Chemistry Tutor
+
<math>\frac{d}{dt}[\csc(t)]\times\cos(t)=-\csc^2(t)\times\cos(t)</math>
  
[https://www.youtube.com/watch?v=93IdxRdd5eg The Second Derivative of Parametric Equations] by The Organic Chemistry Tutor
+
==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.
  
[https://www.youtube.com/watch?v=riEx7TcLfzk Area Under Parametric Curves] by The Organic Chemistry Tutor
+
:{{math|<math>\begin{align}x&=\int x'(t)\mathrm{d}t\\y&=\int y'(t)\mathrm{d}t\end{align}</math>}}
  
[https://www.youtube.com/watch?v=X8N21DrWmjU Arc Length of Parametric Curves] by The Organic Chemistry Tutor
+
So, taking a simple example, with respect to t:
  
[https://www.youtube.com/watch?v=USiKDtILhmc Surface Area of Revolution in Parametric Form] by The Organic Chemistry Tutor
+
:{{math|<math>y=\int\cos(t)\mathrm{d}t=\sin(t) + C</math>}}
  
[https://www.youtube.com/watch?v=g1awTqrKquA The Derivative of a Parametric Curve] by Krista King
+
===Arc length===
 +
Consider a function defined by,
 +
:<math>x = f(t)</math>
 +
:<math>y = g(t)</math>
  
[https://www.youtube.com/watch?v=glt1CtpTOxg Tangent Line to a Parametric Curve] by Krista King
+
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,
  
[https://www.youtube.com/watch?v=wsumI7UeeII The Second Derivative of a Parametric Curve] by Krista King
+
:<math>L = \int_\alpha^\beta \sqrt{1 + (f'(x))^2} \mathrm{d}x</math>
  
[https://www.youtube.com/watch?v=R-SyNQxsWK4 Arc Length of a Parametric Curve] by Krista King
+
It may assist your understanding, here, to write the above using Leibniz's notation,
  
[https://www.youtube.com/watch?v=Kh_17V195ZE] Arc Length of a Parametric Curve by Krista King
+
:<math>L = \int_\alpha^\beta \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x</math>
  
[https://www.youtube.com/watch?v=Rjwi6BnSLnk Surface Area of Revolution in Parametric Form] by Krista King
+
Using the chain rule,
  
[https://www.youtube.com/watch?v=M2blPS7o0SY Surface Area of Revolution of a Parametric Curve Rotated About the y-Axis] by Krista King
+
:<math>\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x}</math>
 +
 
 +
We may then rewrite <math>\mathrm{d}x</math>,
 +
 
 +
:<math>\frac{\mathrm{d}x}{\mathrm{d}t} \mathrm{d}t</math>
 +
 
 +
Hence, <math>L</math> becomes,
 +
 
 +
:<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>
 +
 
 +
Extracting a factor of <math>\left(\frac{\mathrm{d}t}{\mathrm{d}x}\right)^2</math>,
 +
 
 +
:<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>
 +
 
 +
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,
 +
 
 +
:<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>
 +
 
 +
====Example====
 +
Take a circle of radius <math>R</math>, which may be defined with the parametric equations,
 +
 
 +
:<math>x = R\sin\theta</math>
 +
:<math>y = R\cos\theta</math>
 +
 
 +
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>,
 +
 
 +
:<math>x = 0 \implies \theta = \arcsin\left(\frac{0}{R}\right) = 0</math>
 +
:<math>x = R \implies \theta = \arcsin\left(\frac{R}{R}\right) = \arcsin(1) = \frac{\pi}{2}</math>
 +
 
 +
Computing the derivatives of both equations,
 +
 
 +
:<math>\frac{\mathrm{d}x}{\mathrm{d}\theta} = R\cos\theta</math>
 +
:<math>\frac{\mathrm{d}y}{\mathrm{d}\theta} = -R\sin\theta</math>
 +
 
 +
Which means that the arc length is given by,
 +
 
 +
:<math>L = \int_0^{\frac{\pi}{2}} \sqrt{(-R\sin\theta)^2 + R^2\cos^2\theta}\mathrm{d}\theta</math>
 +
 
 +
By the Pythagorean identity,
 +
 
 +
:<math>L = R\int_0^{\frac{\pi}{2}} \mathrm{d}\theta = R\frac{\pi}{2}</math>
 +
 
 +
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>.
 +
 
 +
==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=tnOR99L00Q0 Derivatives of Parametric Equations, Example 1] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=u4xvVwV1ENU Derivatives of Parametric Equations, Example 2] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=a0qBBkuroqk Equation of a Tangent Line to a Parametric Curve, Example 1] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=l8HTzxaQaOk Equation of a Tangent Line to a Parametric Curve, Example 2] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=0GyRRonZu5E Equation of a Tangent Line to a Parametric Curve, Example 3] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=gU2LXxKAT3s Determine the Points Where the Tangent Lines are Horizontal or Vertical] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=MKWsl2W08HE The Second Derivative of Parametric Equations, Part 1] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=BIsnZfuGXXk The Second Derivative of Parametric Equations, Part 2] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=0D31I0jQMLw The First and Second Derivative of Parametric Equations, Example 1] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=jUzdYHmh7e0 The First and Second Derivative of Parametric Equations, Example 2] by James Sousa, Math is Power 4U
 +
 
 +
* [https://www.youtube.com/watch?v=k5QnaGVk1JI Derivatives of Parametric Equations] by Patrick JMT
 +
* [https://www.youtube.com/watch?v=RUXgwKoIds4 Derivatives of Parametric Equations, Example 1] by Patrick JMT
 +
* [https://www.youtube.com/watch?v=azMnzLcdPPo Derivatives of Parametric Equations, Example 2] by Patrick JMT
 +
* [https://www.youtube.com/watch?v=7ANhq51wjWM The Second Derivative of Parametric Curves] by Patrick JMT
 +
 
 +
* [https://www.youtube.com/watch?v=g1awTqrKquA The Derivative of a Parametric Curve] by Krista King
 +
* [https://www.youtube.com/watch?v=glt1CtpTOxg Tangent Line to a Parametric Curve] by Krista King
 +
* [https://www.youtube.com/watch?v=wsumI7UeeII The Second Derivative of a Parametric Curve] by Krista King
 +
 
 +
* [https://www.youtube.com/watch?v=kf2dZWqLnqE Derivatives of Parametric Equations] by The Organic Chemistry Tutor
 +
* [https://www.youtube.com/watch?v=vcZPjuG8GrM Tangent Lines of Parametric Curves] by The Organic Chemistry Tutor
 +
* [https://www.youtube.com/watch?v=jD2RIjbv1Us Horizontal and Vertical Tangent Lines of Parametric Curves] by The Organic Chemistry Tutor
 +
* [https://www.youtube.com/watch?v=93IdxRdd5eg The Second Derivative of Parametric Equations] by The Organic Chemistry Tutor
 +
 
 +
 
 +
===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=GDLZYp2U9g8 Area Under a Parametric Curve] by Patrick JMT
 +
 
 +
* [https://www.youtube.com/watch?v=riEx7TcLfzk Area Under Parametric Curves] by The Organic Chemistry Tutor
 +
 
 +
 
 +
===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=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=WM89GkZq-Dw Find the Arc Length of a Parametric Curve, Example 2] by James Sousa, Math is Power 4U
 +
* [https://www.youtube.com/watch?v=fqA4PrCD5Xk Find the Arc Length of a Loop of a Parametric Curve] by James Sousa, Math is Power 4U
 +
 
 +
* [https://www.youtube.com/watch?v=5fP443JvTUg Arc Length of a Parametric Curve] by Patrick JMT
 +
 
 +
* [https://www.youtube.com/watch?v=R-SyNQxsWK4 Arc Length of a Parametric Curve] by Krista King
 +
* [https://www.youtube.com/watch?v=Kh_17V195ZE Arc Length of a Parametric Curve] by Krista King
 +
 
 +
* [https://www.youtube.com/watch?v=X8N21DrWmjU Arc Length of Parametric Curves] by The Organic Chemistry Tutor
 +
 
 +
 
 +
===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=-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=utu37DbgHvs Surface Area of Revolution in Parametric Form, Example 2 (x-axis)] by James Sousa, Math is Power 4U
 +
 
 +
* [https://www.youtube.com/watch?v=Rjwi6BnSLnk Surface Area of Revolution in Parametric Form] by Krista King
 +
* [https://www.youtube.com/watch?v=M2blPS7o0SY Surface Area of Revolution of a Parametric Curve Rotated About the y-Axis] by Krista King
 +
 
 +
* [https://www.youtube.com/watch?v=USiKDtILhmc Surface Area of Revolution in Parametric Form] by The Organic Chemistry Tutor

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