# Math Help - arclength

1. ## arclength

find the distance travelled by a particle moving on the curve

x(t) = (3t^2, t^3) -1<=t<=1

i got it into the form

L = integral from -1 to 1 t*sqrt(36+9t^2) dt

but how do i integrate this?

2. You don't even need to integrate that. The function which is being integrated over that symmetric interval is odd, whereat the integral is zero.

3. so the distance traveled by the particle is zero then???

and can you show me how o evaluate that integral anyways you can remove the limits and make it an indefinite integral if you want...

4. the distance traveled does not equal 0.
look at the graph of the particle's path in the attachment.

using symmetry ...

$2\int_0^1 \sqrt{36t^2 + 9t^4} \, dt$

$6\int_0^1 t\sqrt{4 + t^2} \, dt$

$u = 4 + t^2$

$du = 2t \, dt$

$3\int_0^1 2t\sqrt{4 + t^2} \, dt$

$3\int_4^5 \sqrt{u} \, du$

$2\left[u^{\frac{3}{2}}\right]_4^5
$

$2[5\sqrt{5} - 8]$

5. how did you do that last step when you changed to 2(5*sqrt(5) - 8)
how did you put it in root form
and wouldnt the distance be twice of that cause there are two graphs???

6. Originally Posted by razorfever
how did you do that last step when you changed to 2(5*sqrt(5) - 8)
I used the FTC.

how did you put it in root form
$

5^{\frac{3}{2}} = \sqrt{5}^3 = 5\sqrt{5}$

and wouldnt the distance be twice of that cause there are two graphs???
doubled from the start ... did you notice the constant multiple "2" in front of the beginning integral?

7. Originally Posted by razorfever

L = integral from -1 to 1 t*sqrt(36+9t^2) dt
Ahhh, here's you mistake: when factored out the $t,$ you didn't consider that it's actually $|t|,$ and then we need to split the original integral into two ones.

8. oh thanks for telling me that ... it was the final piece of the puzzle for me