## 17Calculus Parametric Equations - Arc Length

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On this page we discuss integration and arc length of parametric 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

$$\displaystyle{ s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}}$$
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 [47secs]

video by PatrickJMT

Practice

Unless otherwise instructed, calculate the arc length of these parametric curves on the given intervals.

Basic

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

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

### 3505 video

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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

### 3506 video

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

$$36$$

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

### 2001 video

video by Dr Chris Tisdell

$$36$$

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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 $$x(t) = t \sin(t)$$ $$y(t) = t \cos(t)$$ on the interval $$0 \leq t \leq1$$.

arc length = $$\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }$$

Problem Statement

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

Solution

### 483 video

video by Krista King Math

arc length = $$\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }$$

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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

### 1377 video

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

### 1379 video

video by PatrickJMT

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Intermediate

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

Problem Statement

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

Solution

### 58 video

video by Krista King Math

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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

### 469 video

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 θ from the equation, you will get the equation that he starts with in the video.

### 1380 video

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

### 1378 video

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

### 1381 video

video by PatrickJMT

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### parametric calculus 17calculus youtube playlist

You CAN Ace Calculus

 Wikipedia - Parametric Derivative Wikipedia - Arc Length Pauls Online Notes - Arc Length with Parametric Equations Wikipedia - Solid of Revolution

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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