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?
the distance traveled does not equal 0.
look at the graph of the particle's path in the attachment.
using symmetry ...
$\displaystyle 2\int_0^1 \sqrt{36t^2 + 9t^4} \, dt$
$\displaystyle 6\int_0^1 t\sqrt{4 + t^2} \, dt$
$\displaystyle u = 4 + t^2$
$\displaystyle du = 2t \, dt$
$\displaystyle 3\int_0^1 2t\sqrt{4 + t^2} \, dt$
$\displaystyle 3\int_4^5 \sqrt{u} \, du$
$\displaystyle 2\left[u^{\frac{3}{2}}\right]_4^5
$
$\displaystyle 2[5\sqrt{5} - 8]$
I used the FTC.
$\displaystylehow did you put it in root form
5^{\frac{3}{2}} = \sqrt{5}^3 = 5\sqrt{5}$
doubled from the start ... did you notice the constant multiple "2" in front of the beginning integral?and wouldnt the distance be twice of that cause there are two graphs???