# arc length parametric

• October 10th 2008, 08:16 PM
andrewsx
arc length parametric
Find arc length
x=t^2, y=4t^3-1 -1<t<1

dx/dt=2t dy/dt= 12t^2

integral from -1 to 1 sq root (4t^2 +144t^4)dt

then, integral from -1 to 1 2t*sq root (1+36t^2)dt

Please help me find the answer. I know i have to use trig substitution does u=6tantheta?
• October 10th 2008, 09:54 PM
mr fantastic
Quote:

Originally Posted by andrewsx
Find arc length
x=t^2, y=4t^3-1 -1<t<1

dx/dt=2t dy/dt= 12t^2

integral from -1 to 1 sq root (4t^2 +144t^4)dt

then, integral from -1 to 1 2t*sq root (1+36t^2)dt

Please help me find the answer. I know i have to use trig substitution does u=6tantheta?

Make the substitution $u = 1 + 36t^2 \Rightarrow dt = \frac{du}{72t}$.