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- Multivariable Calculus
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Get Started Now- Intro Lesson: a6:16
- Intro Lesson: b13:10
- Lesson: 17:42
- Lesson: 23:57
- Lesson: 37:09
- Lesson: 49:07
- Lesson: 59:35

Given a vector function $r(t)= <f(t),g(t),h(t)>$, we can find the arc length of it on the interval $a \leq t \leq b$ by calculating:

$L = \int^b_a ||r'(t)||dt$

$s(t) = \int^t_0 ||r'(u)||du$

Where $s$ is the length or distance travelled on the curve in terms of $t$. We usually want to find this if we are looking for $r(t(s))$, which tells us where a point is located on the curve.- Introduction
**Arc Length with Vector Functions Overview:**a)__Arc Length__- Length of a vector function
- Example of finding the length

b)__Arc Length Function/Why is it Useful?__- $s(t)\to$ The distance travelled on the curve from 0 to $t$
- Example of calculating $s(t)$ and $r(t(s))$

- 1.
**Finding the Arc Length**

Determine the length of the vector function on the given interval:$r(t) = \lt 2 + 3t, t^2, \frac{4\sqrt{3}}{3} t^{\frac{3}{2}} \gt \;\; 0 \leq t \leq 1$

- 2.Determine the length of the vector function on the given interval:
$r(t) = (3+4t)i + (2t-3)j+(5-t)k \;\; 2 \leq t \leq 3$

- 3.
**Finding the Arc Length Function**

Determine the arc length function for the given vector function$r(t) = \lt 2t, \frac{1}{3} t^3 , t^2 \gt$

- 4.Determine the arc length function for the given vector function
$r(t) = \lt t^2, 2t^2, \frac{1}{3} t^3 \gt$

- 5.
**Finding a Specific Point on a Curve**

After traveling a distance of $\sqrt{2} \pi$, determine where we are on the vector function $r(t)= \lt \cos t, \sin t, t\gt$.