Question

How do you find the length of the curve for `y=2x^(3/2)` for (0, 4)?

Answer 1

`S=1/6((37)^(3/2)-1)`

Explanation:

Given equation is `f(x)=2x^(3/2)`.
We are given the task to find the length of the curve of the given equation in the interval `(0,4)`.

The equation to find length of the curve is `S=\int_0^4\sqrt{1+f'(x)^2}dx`
So, the derivative of the given equation will be
`f'(x)=2d/dx(x^(3/2))=2*3/2*x^(3/2-1)=3*x^(1/2)`

So substituting for `f'(x)`, `S=\int_0^4\sqrt{1+9x}dx`
Taking `9x=t\impliesdx=dt/9` and that at `x=0\impliest=0` and `x=4\impliest=36`
So the given integral becomes `S=1/9\int_0^36\sqrt{1+t}dt`
So, `S=cancel{3}^1/2*1/\cancel{9}^3(1+t)^(3/2)|_0^36`
Applying limits and totalling, I get the above answer.