## 17Calculus Vector Functions - Arc Length, Arc Length Function and The Arc Length Parameter

##### 17Calculus

For the following discussion, we will consider a parameterized curve defined by the vector function $$\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }$$ which is traversed once on the continuous interval $$a \leq t \leq b$$.

Arc Length

The equation for finding the arc length of a curve is $$\displaystyle{ L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} ~dt } }$$. In shorthand notation, you will often see $$\displaystyle{ L = \int_{a}^{b}{ \| \vec{r}'(t) \| ~dt } }$$ where $$\displaystyle{ \| \vec{r}'(t) \| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} }$$. Of course, we are assuming this is a smooth, continuous curve.

Okay, time for some practice problems.

Practice

Unless otherwise instructed, find the arc length on the given interval. Give your answer in exact form.

$$\vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle$$, $$[-10,10]$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle$$ on the interval $$[-10,10]$$.

$$20\sqrt{29}$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle$$ on the interval $$[-10,10]$$.

Solution

### PatrickJMT - 699 video solution

video by PatrickJMT

$$20\sqrt{29}$$

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$$\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k}$$, $$0\leq t \leq 1$$

Problem Statement

Find the arc length of $$\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k}$$ for $$0\leq t \leq 1$$.

$$(e^2-1)/e$$

Problem Statement

Find the arc length of $$\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k}$$ for $$0\leq t \leq 1$$.

Solution

### Krista King Math - 2042 video solution

video by Krista King Math

$$(e^2-1)/e$$

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$$\vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle$$, $$0 \leq t \leq \pi$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle$$ for $$0 \leq t \leq \pi$$

$$2\pi\sqrt{2}$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle$$ for $$0 \leq t \leq \pi$$

Solution

### MIP4U - 2043 video solution

video by MIP4U

$$2\pi\sqrt{2}$$

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$$\vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle$$, $$[0, \pi]$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle$$ on the interval $$[0, \pi]$$.

$$3$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle$$ on the interval $$[0, \pi]$$.

Solution

### MIP4U - 2044 video solution

video by MIP4U

$$3$$

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$$\vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle$$, $$-5\leq t \leq 8$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle$$ for $$-5\leq t \leq 8$$.

$$65\sqrt{17}$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle$$ for $$-5\leq t \leq 8$$.

Solution

### MIP4U - 2045 video solution

video by MIP4U

$$65\sqrt{17}$$

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$$\vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle$$, $$t \in [1,2]$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle$$, $$t \in [1,2]$$.

$$3/2+2\ln 2$$

Problem Statement

Find the arc length of $$\vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle$$, $$t \in [1,2]$$.

Solution

### Dr Chris Tisdell - 2046 video solution

video by Dr Chris Tisdell

$$3/2+2\ln 2$$

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