## 17Calculus - Derivatives of Parametric Equations

##### 17Calculus

On this page we get started with applying calculus to parametric equations by covering derivatives of parametric equations and higher order derivatives.

### Resources

Topics You Need To Understand For This Page

basics of parametric equations

Related Topics and Links

Wikipedia - Parametric Derivative

Wikipedia - Arc Length

Pauls Online Notes - Arc Length with Parametric Equations

Wikipedia - Solid of Revolution

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 going on, let's work some practice problems.

Practice - First Derivative

Unless otherwise instructed, calculate the first derivative of these parametric curves.

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

Problem Statement

Find $$dy/dx$$ where $$y=f(x)$$ is given by the parametric equations $$x=t+5\cos(t)$$, $$y=3e^t$$ .

Solution

### PatrickJMT - 1373 video solution

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

### Krista King Math - 468 video solution

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

### PatrickJMT - 1372 video solution

video by PatrickJMT

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Higher Order Derivatives

In order to determine concavity of a graph and other information, we 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

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

Let's calculate some higher order derivatives of parametric equations.

Practice - Higher Order Derivatives

Unless otherwise instructed, calculate the first and second derivatives of these parametric equations.

Find $$\dfrac{d^2y}{dx^2}$$ of $$y=f(x)$$ where $$x=t^2+t$$, $$y=2t-1$$

Problem Statement

Find the second derivative $$\dfrac{d^2y}{dx^2}$$ where $$y=f(x)$$ is given by the parametric equations $$x=t^2+t$$, $$y=2t-1$$ .

Solution

### Krista King Math - 56 video solution

video by Krista King Math

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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

### PatrickJMT - 1374 video solution

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

### PatrickJMT - 1375 video solution

video by PatrickJMT

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$$x = 4t^3 - t^2 + 7t$$, $$y = t^4 - 6$$

Problem Statement

Find $$dy/dx$$ and $$d^2y/dx^2$$ where $$y=f(x)$$ is given by the parametric equations $$x = 4t^3 - t^2 + 7t$$, $$y = t^4 - 6$$ .

Solution

The solution can be found on this page.

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$$x = e^{-7t} + 2$$, $$y = 6e^{2t} + e^{-3t} - 4t$$

Problem Statement

Find $$dy/dx$$ and $$d^2y/dx^2$$ where $$y=f(x)$$ is given by the parametric equations $$x = e^{-7t} + 2$$, $$y = 6e^{2t} + e^{-3t} - 4t$$ .

Solution

The solution can be found on this page.

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