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You CAN Ace Calculus

17calculus > parametrics > calculus

Parametric Equations and Calculus

On this page we cover the most common calculus problems using parametric equations. The two main topics are differentiation and integration.

Search 17Calculus

This table lists the topics on this page with links to jump to the topic you are interested in.

Differentiation

Integration

Slope and Tangent Lines

Area

Surface Area

Higher Order Derivatives

Arc Length

Volume

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

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

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 1

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

solution

Practice 2

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

solution

Practice 3

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

solution

Differentiation - Slope and Tangent Lines

To find the equation of a tangent line to a graph given by a set of parametric equations, we need to be able to find the slope by calculating the derivative \( dy/dx \) using the above theorem.

For horizontal tangent lines, the slope \(dy/dx\) is zero, so we need \( dy/dt = 0 \) and \( dx/dt \neq 0 \). For vertical tangent lines, the slope is undefined, which means that \( dx/dt = 0 \) when \( dy/dt \neq 0 \). In the case where both \( dx/dt = 0\) and \( dy/dt = 0 \) at the same point, we need to handle that case separately, since nothing can be concluded from \( dy/dx = 0/0 \), which is indeterminate.

Once you have found the slope, you can easily find the equation of a tangent line. Go to the Tangent Lines page for more information.

Practice 4

Find the equation of the tangent line to the parametric curve \(x=2t^2+1\), \(y=3t^3+2\) at \(t=1\).

solution

Practice 5

Find the equation of the tangent line to the parametric curve \(x=3(t-\sin(t))\), \(y=3(1-\cos(t))\) at \(t=\pi/2\).

solution

Practice 6

Find the equation of the tangent line to the parametric curve \(x=t\cos(t)\), \(y=t\sin(t)\) at \(t=\pi\).

solution

Practice 7

Find the equation of the tangent line to the parametric curve \(x=3t^2-t\), \(y=\sqrt{t}\) at \(t=4\).

solution

Differentiation - 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 \( \dot{x} = dx/dt \), 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

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

Before we go on, let's work some practice problems with what we have learned so far.

Basic Problems

Practice 8

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

solution

Intermediate Problems

Practice 9

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

solution

Practice 10

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

solution

Practice 11

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

answer

solution

Practice 12

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.

answer

solution

Integration - Area

The area under a smooth curve defined parametrically as \( x = X(t) \) and \( y = Y(t) \) from \( t=t_0\) to \( t=t_1\) can be calculated using the integral

\(\displaystyle{s = \int_{t_0}^{t_1}{Y(t)X'(t) ~ dt}}\)     where \( X'(t) = dX/dt \).

You may also find this written in a shorthand form as

\(\displaystyle{s = \int_{t_0}^{t_1}{y~dx}}\)

In this notation, \( dx = (dx/dt) dt \).

Practice 13

Find the area under the curve \(x=1+e^t\), \(y=t-t^2\) and above the x-axis.

solution

Practice 14

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

answer

solution

Practice 15

Calculate the area under one arc of the parametric curve \(x=r\theta-d\sin\theta\), \(y=r-d\cos\theta\).

solution

Integration - Arc Length

When a smooth curve is defined parametrically as \( x=X(t) \) and \( y = Y(t) \), the arc length between the points \( t=t_0 \) and \( t = t_1 \) can be calculated using the integral

\(\displaystyle{ s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}}\)
Notice we use a small s here to represent the arc length. This is the standard symbol you will see in many textbooks. We reserve a capital S to represent surface area.

Here is a quick video clip going over this equation in more detail.

PatrickJMT - Arc Length

Basic Problems

Practice 16

Compute the arc length of the parametric curve \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

answer

solution

Practice 17

Calculate the arc length of the parametric curve \(x(t)=t\sin(t)\) \(y(t)=t\cos(t)\), \(0\leq t\leq1\).

answer

solution

Practice 18

Calculate the arc length of the parametric curve \(x=1+3t^2\), \(y=4+2t^3\), \(0\leq t\leq1\).

answer

solution

Practice 19

Calculate the arc length of the parametric curve \(x=\abs{6-t}\), \(y=t\), \(0 \leq t \leq 3\).

answer

solution

Intermediate Problems

Practice 20

Find the arc length of the parametric curve \(x=2t\), \(y=(2/3)t^{3/2}\) for \(5\leq t\leq12\).

answer

solution

Practice 21

Calculate the arc length of the parametric curve \(x=t^3\), \(y=t^2\) from \((0,0)\) to \((8,4)\).

answer

solution

Practice 22

Calculate the arc length of the curve \(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\).

answer

solution

Practice 23

Calculate the arc length of the parametric curve \(x=e^t+e^{-t}\), \(y=5-2t\), \(0\leq t\leq3\).

answer

solution

Practice 24

Calculate the arc length of the parametric curve \(x=e^t\cos(t)\), \(y=e^t\sin(t)\), \(0\leq t\leq\pi\).

answer

solution

Integration - Surface Area

To calculate the surface area defined by revolving a parametric curve defined as \( x=X(t)\) and \(y=Y(t)\) from \(t=t_0\) to \(t=t_1\) about the x-axis, we use this integral.

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{ Y(t) \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~ dt}}\)

If you compare this integral to the equation for arc length (in the prevous section) you will see the common factor \( \sqrt{[X'(t)]^2 + [Y'(t)]^2} \)

Some textbooks write the surface integral differently, taking this into account. You may see it written as

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{Y(t) ~ ds}}\)
where \( ds = \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~dt \)

When rotating about the y-axis, the integral we use is

\(\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{X(t) ~ ds}}\)

Notice we use a capital S to represent surface area. This is the standard symbol you will see in many textbooks. We reserve a lowercase s to represent arc length.

Practice 25

Find the surface area of revolution of the parametric curve \(x=3t^2\), \(y=2t^3\), \(0\leq t\leq5\) rotated about the y-axis.

answer

solution

Practice 26

Find the surface area of revolution of the parametric curve \(x=1-t\), \(y=2\sqrt{t}\), \(1 \leq t \leq 4\) rotated about the x-axis.

answer

solution

Integration - Volume

If we have a parametric curve defined as \( x = X(t) \) and \( y = Y(t) \), we can determine the volume of the solid object defined by revolving this curve about an axis. We will limit our curve from \( t=t_0 \) to \( t=t_1 \).

Revolution About the x-axis

   

Revolution About the y-axis

When revolved about the x-axis, the integral is
\(\displaystyle{V_x = \pi \int_{t_0}^{t_1}{y^2 [dx/dt] dt}}\)
This is sometimes written as \(\displaystyle{V_x = \pi \int_{t_0}^{t_1}{y^2 ~ dx}}\) where \( dx = [dx/dt] dt \)

Similar to the x-axis integral, we have
\(\displaystyle{V_y = \pi \int_{t_0}^{t_1}{x^2 [dy/dt] dt} = \pi \int_{t_0}^{t_1}{x^2 ~ dy}}\) where \( dy = [dy/dt] dt \)



Practice 27

Calculate the volume of revolution when the curve \(x=t^3\), \(y=2t^2+1\), \(-1\leq t \leq 1\) is rotated about the x-axis.

answer

solution

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