Question

How do you find the arc length of `x=2/3(y-1)^(3/2)` between `1<=y<=4`?

Answer 1

It can be found by `L=int_0^4sqrt{1+(frac{dx}{dy})^2}dy`.

Let us evaluate the above definite integral.

By differentiating with respect to y,
`frac{dx}{dy}=(y-1)^{1/2}`

So, the integrand can be simplified as
`sqrt{1+(frac{dx}{dy})^2}=sqrt{1+[(y-1)^{1/2}]^2}=sqrt{y}=y^{1/2}`

Finally, we have
`L=\int_0^4y^{1/2}dy=[frac{2}{3}y^{3/2}]_0^4=frac{2}{3}(4)^{3/2}-2/3(0)^{3/2}=16/3`

Hence, the arc length is `16/3`.

I hope that this helps.