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

Answer 1

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

Answer 2

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

Answer 3

but now i am doing parametric

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

what confused me is the integral part
i keep getting