Find the arc length of the curve
x= (1/3)t^3
y= (1/2)t^2
from zero is less and equal to t and t is greater than or equal to 1

Question

Find the arc length of the curve
x= (1/3)t^3
y= (1/2)t^2
from zero is less and equal to t and t is greater than or equal to 1

anonymous · Answer

$$x=\frac{ 1 }{ 3 }t ^{3}$$ $$y=\frac{ 1 }{ 2 }t ^{2}$$

ybarrap · Answer

The Formula for Arclength is:
$$
\large
s=\int _{{a}}^{{b}}{\sqrt {1+[f'(x)]^{2}}}\,dx.
$$
Have you used it?

anonymous · Answer

but now i am doing parametric

ybarrap · Answer

Ok, so you'll need this one. Have you used it?
$$
\large
s=\int _{{a}}^{{b}}{\sqrt  {[X'(t)]^{2}+[Y'(t)]^{2}}}\,dt.
$$

ybarrap · Answer

1st determine
$$
\large {
\cfrac{dx(t)}{dt}	ext{&}
\cfrac{dy(t)}{dt}\}
$$

anonymous · Answer

i did
dx/dt= t^2
dy/dt= t

anonymous · Answer

what confused me is the integral part
i keep getting |dw:1392081373932:dw|