- #1

RJLiberator

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## Homework Statement

Find the arc length of r(t)= <tsin(t), tcost(t), 3t> from 0 to t to 2pi (inclusive)

## Homework Equations

Integral from 0 to 2pi of the magnitude of r'(t) dt

## The Attempt at a Solution

1. Must find the derivative of the function.

Using the product rule a few times, the derivative becomes

<tcos(t)+sin(t), -tsin(t)+cos(t), 3>

2. Now take the magnitude

Well, this is a bit lengthy, but it simplifies pretty nice with the double angle trig identity. (I hope what I did is correct,

**please check my work in this step**.

(tcos(t)+sin(t))^2 = t^2cos^2t+2tsin(t)*cos(t)+sin^2t

Simplifies to: t^2cos^2t+tsin(2t)+sin^2t

(-tsin(t)+cos(t))^2 = t^2sin^2t+(-2tsin(t)*cos*(t))+cos^2t

simplifies to: t^2sin^2t-tsin(2t)+cos^2t

3^2 = 9

So, when we add everything the double angles cancel out. The cos^2t+sin^2t is equal to 1. We factor a t^2 out of the t^2cos^2t and t^2sin^2t equations and this becomes t^2*1.

We are left with 9+1+t^2

t^2+10, beautiful!

3. Integrate from 0 to 2pi the square root of (t^2+10) dt

Well, here's where I am stuck.

I feel like I've done everything properly up to this point. But this integral cannot be taken easily (at least with calc 3 tools).

**Did I do something wrong? Any hints on the integral?**