# Arc length of vector-valued function; am I starting right?

1. Homework Statement
Given: R(t)= <(1/2)t^2, (4/3)t^(3/2), 2*sqrt(3)t>
Find: Arc length function s(t) where t_0 =0

2. Homework Equations
Is this the correct formula?
∫[0,t] sqrt( derivative^2 + derivative^2 +derivative^2) dt
∫[0,t] sqrt(t^2 + 4t + 12) dt

3. The Attempt at a Solution

∫[0,t] sqrt(t^2 + 4t + 12) dt

I am not looking for anyone to do my homework, but before I start a messy integral I just want to know if:
A) I am starting correctly, and
B) if their is an easier way to do this (or strategy)

Thanks.

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$$s=\int_a^b|r'(t)|dt$$

Your work is correct so far! Keep going ... what's your next step?

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I apreciate your encouraging tone rocophysics; I was about to quit for the night.

∫[0,t] sqrt(t^2 + 4t + 12) dt
∫[0,t] sqrt((t+2)^2 + 8) dt
u=t+2 du=1
∫[0,t] sqrt((u)^2 + 8) du

then maybe trigonometric substitution.

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$$\int_0^t\sqrt{t^2+4t+12}dt$$

$$\int_0^t\sqrt{(t+2)^2+8}dt$$

Go ahead and do trig sub after this step ...

$$t+2=\sqrt 8\tan\theta$$
$$dt=\sqrt 9\sec^{2}\theta d\theta$$

But don't forget to change your limits ...

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I tried but I know there is at least one mistake somewhere; here's my attempt (each line is numbered 1-14):
http://www.sudokupuzzles.net/IMG_0030.jpg [Broken]

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You forgot about your constant of 8 (line 10), and it should be to both your Integrals, not just one.

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