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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.

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

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

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 derivatives of parametric equations.

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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 \(dy/dx\) 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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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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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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Practice Instructions

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

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