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17Calculus - Derivatives of Parametric Equations

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Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

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.

How to Ace Calculus: The Streetwise Guide

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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Practice Instructions

Unless otherwise instructed, determine the requested derivative of the given parametric curves. Give your answers in exact, simplified form.

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