17Calculus Parametric Equations - Arc Length

17Calculus

On this page we discuss how to calculate the arc length of curves described by parametric equations.

Equations

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 $s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}$

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 [47secs]

video by PatrickJMT

See the resources section for links to pages where these equations are derived.

Practice

Unless otherwise instructed, calculate the arc length of these parametric curves on the given intervals. Give your answers in exact form.

$$x = 2+6t^2$$, $$y = 5+4t^3$$; $$0 \leq t \leq \sqrt{8}$$

Problem Statement

Calculate the arc length of the parametric curve $$x = 2+6t^2$$, $$y = 5+4t^3$$; $$0 \leq t \leq \sqrt{8}$$

Solution

The Organic Chemistry Tutor - 3505 video solution

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$$x = 9t^2$$, $$y = 9t-3t^3$$; $$0 \leq t \leq 2$$

Problem Statement

Calculate the arc length of the parametric curve $$x = 9t^2$$, $$y = 9t-3t^3$$; $$0 \leq t \leq 2$$

Solution

The Organic Chemistry Tutor - 3506 video solution

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$$x(t)=3t^2-9$$, $$y(t)=t^3-3t$$; $$0\leq t\leq 3$$

Problem Statement

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

$$s = 36$$

Problem Statement

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

Solution

Dr Chris Tisdell - 2001 video solution

video by Dr Chris Tisdell

$$s = 36$$

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$$x = 3t + 1$$, $$y = 4 - t^2$$; $$-2 \leq t \leq 0$$

Problem Statement

Find the arc length of the parametric curve $$x = 3t + 1$$, $$y = 4 - t^2$$ for $$-2 \leq t \leq 0$$.

Solution

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$$x(t) = t \sin(t)$$, $$y(t) = t \cos(t)$$; $$0 \leq t \leq1$$

Problem Statement

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

Solution

Krista King Math - 483 video solution

video by Krista King Math

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$$x = 1+3t^2$$, $$y = 4+2t^3$$; $$0 \leq t \leq 1$$

Problem Statement

Calculate the arc length of the parametric curve $$x = 1+3t^2$$, $$y = 4+2t^3$$; $$0 \leq t \leq 1$$

Solution

PatrickJMT - 1377 video solution

video by PatrickJMT

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$$x = |6-t|$$, $$y = t$$; $$0 \leq t \leq 3$$

Problem Statement

Calculate the arc length of the parametric curve $$x = |6-t|$$, $$y = t$$; $$0 \leq t \leq 3$$

Solution

PatrickJMT - 1379 video solution

video by PatrickJMT

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$$x = 2t$$, $$y = (2/3)t^{3/2}$$; $$5 \leq t \leq 12$$

Problem Statement

Find the arc length of the parametric curve $$x = 2t$$, $$y = (2/3)t^{3/2}$$; $$5 \leq t \leq 12$$

Solution

Krista King Math - 58 video solution

video by Krista King Math

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$$x = 8t^{3/2}$$, $$y = 3 + (8-t)^{3/2}$$; $$0 \leq t \leq 4$$

Problem Statement

Find the arc length of the parametric curve $$x = 8t^{3/2}$$, $$y = 3 + (8-t)^{3/2}$$ for $$0 \leq t \leq 4$$.

Solution

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$$x = t^3$$, $$y = t^2$$ from $$(0,0)$$ to $$(8,4)$$

Problem Statement

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

Solution

Dr Chris Tisdell - 469 video solution

video by Dr Chris Tisdell

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$$x(\theta)=a\cos^3(\theta)$$, $$y(\theta)=a\sin^3(\theta)$$; $$0\leq\theta\leq2\pi$$

Problem Statement

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

Solution

Upon initial inspection, this video does not seem to go with the problem statement. However, if you eliminate the parameter $$\theta$$ from the equation, you will get the equation that he starts with in the video.

Dr Chris Tisdell - 1380 video solution

video by Dr Chris Tisdell

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$$x = e^t+e^{-t}$$, $$y = 5-2t$$; $$0 \leq t \leq 3$$

Problem Statement

Calculate the arc length of the parametric curve $$x = e^t+e^{-t}$$, $$y = 5-2t$$; $$0 \leq t \leq 3$$

Solution

PatrickJMT - 1378 video solution

video by PatrickJMT

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$$x = e^t \cos(t)$$, $$y = e^t \sin(t)$$; $$0 \leq t \leq \pi$$

Problem Statement

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

Solution

PatrickJMT - 1381 video solution

video by PatrickJMT

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A particle travels along a path defined by the parametric equations $$x = 4\sin(t/4)$$, $$y = 1 - 2\cos^2(t/4)$$; $$-52\pi \leq t \leq 34\pi$$. Determine the total distance the particle travels and compare this to the length of the parametric curve itself.

Problem Statement

A particle travels along a path defined by the parametric equations $$x = 4\sin(t/4)$$, $$y = 1 - 2\cos^2(t/4)$$; $$-52\pi \leq t \leq 34\pi$$. Determine the total distance the particle travels and compare this to the length of the parametric curve itself.

Solution

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Set up, but do not evaluate, an integral that gives the length of the parametric curve given by the parametric equations $$x = 2 + t^2$$, $$y = e^t \sin(2t)$$; $$0 \leq t \leq 3$$ .

Problem Statement

Set up, but do not evaluate, an integral that gives the length of the parametric curve given by the parametric equations $$x = 2 + t^2$$, $$y = e^t \sin(2t)$$; $$0 \leq t \leq 3$$ .

Solution

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Set up, but do not evaluate, an integral that gives the length of the parametric curve given by the parametric equations $$x = \cos^3(2t)$$, $$y = \sin(1-t^2)$$; $$-3/2 \leq t \leq 0$$ .

Problem Statement

Set up, but do not evaluate, an integral that gives the length of the parametric curve given by the parametric equations $$x = \cos^3(2t)$$, $$y = \sin(1-t^2)$$; $$-3/2 \leq t \leq 0$$ .

Solution