## 17Calculus - Derivatives of Parametric Equations

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On this page we get started with applying calculus to parametric equations by covering derivatives of parametric equations and higher order derivatives.
If you want a complete lecture on calculus involving parametric equations, we recommend this video.

### Prof Leonard - Calculus 2 Lecture 10.3: Calculus of Parametric Equations [1hr-34min-18secs]

video by Prof Leonard

Let's start out with a quick video clip giving us an introduction to finding the derivative $$dy/dx$$ if the function $$y=f(x)$$ is given in parametric equations $$x(t)$$ and $$y(t)$$. The theorem is given below.

### PatrickJMT - Parametric Differentiation [1min-7secs]

video by PatrickJMT

Theorem: Parametric Derivative

On a smooth curve given by the equations $$x=X(t)$$ and $$y = Y(t)$$, the slope of the curve at the point $$(x,y)$$ is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0$

### Parametric Derivative Proof

Theorem: Parametric Derivative

On a smooth curve given by the equations $$x=X(t)$$ and $$y = Y(t)$$, the slope of the curve at the point $$(x,y)$$ is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ where } dx/dt \neq 0$

We present two proofs here, a short informal version and a longer, more formal version.

Short Informal Proof
Given $$y(t)$$ and $$x(t)$$, we can write $$y = y(x(t))$$. We want to find $$dy/dx$$.
Using the chain rule on $$y(x(t))$$, we have $$dy/dt = dy/dx \cdot dx/dt$$.
Solving for $$dy/dx$$ we have $$\displaystyle{ dy/dx = \frac{dy/dt}{dx/dt} }$$      [qed]

Longer More Formal Proof
Given the two points $$(x_1,y_1) = (X(t), Y(t))$$ and $$(x_2,y_2) = (X(t+\Delta t), Y(t+\Delta t))$$ and $$\Delta t > 0$$ on a smooth curve, let $$\Delta x = x_2 - x_1 = X(t+\Delta t) - X(t)$$ and $$\Delta y = y_2 - y_1 = Y(t+\Delta t) - Y(t)$$.

As $$\Delta t \to 0$$, we know that $$\Delta x \to 0$$ and we can write

$$\displaystyle{ \frac{dy}{dx} = \lim_{\Delta x \to 0}{\frac{\Delta y}{\Delta x}} = \lim_{\Delta t \to 0}{\frac{Y(t+\Delta t) - Y(t)}{X(t+\Delta t) - X(t)}} }$$

Now we can multiply the numerator and denominator by $$1/\Delta t$$ and use limit laws to give us

$$\begin{array}{rcl} \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\lim_{\Delta t \to 0}{\frac{[Y(t+\Delta t) - Y(t)](1/\Delta t)}{[X(t+\Delta t) - X(t)](1/\Delta t)}} } \\ & = & \displaystyle{ \frac{\lim_{\Delta t \to 0}{[Y(t+\Delta t) - Y(t)]/\Delta t}}{\lim_{\Delta t \to 0}{[X(t+\Delta t) - X(t)]/\Delta t}} } \\ & = & \displaystyle{ \frac{dy/dt}{dx/dt} ~~~~~ \text{ [qed] } } \end{array}$$

Before we go on, let's practice using this theorem.

Practice

Unless otherwise instructed, determine $$dy/dx$$ of the given parametric curves. Give your answers in exact, simplified form.

$$x=t+5\cos(t)$$, $$y=3e^t$$

Problem Statement

Find $$dy/dx$$ of the parametric curve $$x=t+5\cos(t)$$, $$y=3e^t$$.

Solution

### 1373 video

video by PatrickJMT

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$$x=t\sin(t)$$, $$y=t^2+t$$

Problem Statement

Find the derivative of the parametric curve $$x=t\sin(t)$$, $$y=t^2+t$$.

Solution

### 468 video

video by Krista King Math

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$$x=4t+1$$, $$y=t^2+2t$$

Problem Statement

Find $$dy/dx$$ of the parametric curve $$x=4t+1$$, $$y=t^2+2t$$.

Solution

### 1372 video

video by PatrickJMT

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Higher Order Derivatives

In order to determine concavity of a graph and other information, you will need higher order derivatives. We list two of them below from which you can extract a pattern.

Second Derivative

$$\displaystyle{ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \frac{dy}{dx} \right] = \frac{d\left[ dy/dx \right]/dt}{dx/dt} }$$

You may also find the notation $$\dot{x} = dx/dt$$ and $$\dot{y} = dy/dt$$, in which case the above derivative can be written $$\displaystyle{ \frac{d^2y}{dx^2} = \frac{\dot{x} \ddot{y} - \dot{y} \ddot{x}}{\dot{x}^3} }$$

Third Derivative

$$\displaystyle{ \frac{d^3y}{dx^3} = \frac{d}{dx}\left[ \frac{d^2y}{dx^2} \right] = \frac{d\left[ d^2y/dx^2 \right] /dt}{dx/dt} }$$

Practice

Basic

Find the second derivative of the parametric equations $$x=t^2+t$$, $$y=2t-1$$.

Problem Statement

Find the second derivative of the parametric equations $$x=t^2+t$$, $$y=2t-1$$.

Solution

### 56 video

video by Krista King Math

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Intermediate

Find the second derivative of the parametric curve $$x=t^3+t$$, $$y=t^5+1$$.

Problem Statement

Find the second derivative of the parametric curve $$x=t^3+t$$, $$y=t^5+1$$.

Solution

### 1374 video

video by PatrickJMT

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Find the second derivative of the parametric curve $$x = t - t^3$$, $$y = 2t + 5$$.

Problem Statement

Find the second derivative of the parametric curve $$x = t - t^3$$, $$y = 2t + 5$$.

Solution

### 1375 video

video by PatrickJMT

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Find the horizontal tangent points to the curve $$x=1-t, y=t^2$$ and determine the concavity at those points.

Problem Statement

Find the horizontal tangent points to the curve $$x=1-t, y=t^2$$ and determine the concavity at those points.

Final Answer

$$(1,0)$$, concave up

Problem Statement

Find the horizontal tangent points to the curve $$x=1-t, y=t^2$$ and determine the concavity at those points.

Solution

How do you know when you have a horizontal tangent? When the slope is zero.
$$dx/dt=-1, dy/dt=2t$$
$$\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{-1} }$$
$$dy/dx = 0$$ when $$-2t=0 \to t=0$$
When $$t=0$$, $$(x,y) = (1,0)$$.

To determine concavity at that point, we need to calculate the second derivative.
$$\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{-2}{-1} = 2 }$$
Since the second derivative is positive everywhere, the curve is concave up.

Final Answer

$$(1,0)$$, concave up

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For the parametric curve $$x=\sqrt{t}, y=3t-1$$, find the derivative and the slope at the point $$t=1$$. Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Problem Statement

For the parametric curve $$x=\sqrt{t}, y=3t-1$$, find the derivative and the slope at the point $$t=1$$. Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Final Answer

$$dy/dx=6\sqrt{t}, m=6, y=6x-4$$, concave up

Problem Statement

For the parametric curve $$x=\sqrt{t}, y=3t-1$$, find the derivative and the slope at the point $$t=1$$. Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

Solution

$$dx/dt=(1/2)t^{-1/2}, dy/dt = 3$$
$$\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{(1/2)t^{-1/2}} = 6\sqrt{t} }$$
slope at $$t=1, m=6\sqrt{1} = 6$$
for the equation of tangent line, we need the $$(x,y)$$ point at $$t=1$$, so $$(\sqrt{1},3-1)=(1,2)$$
$$y-y_1=m(x-x_1) \to y-2=6(x-1) \to y=6x-4$$
$$\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{6(1/2)t^{-1/2}}{(1/2)t^{-1/2}} = 6 }$$
Since the second derivative is positive at $$t=1$$ (and everywhere else for that matter), the curve is concave up.

Final Answer

$$dy/dx=6\sqrt{t}, m=6, y=6x-4$$, concave up

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### Related Topics and Links

 Wikipedia - Parametric Derivative Wikipedia - Arc Length Pauls Online Notes - Arc Length with Parametric Equations Wikipedia - Solid of Revolution

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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