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

1 Answer
Feb 7, 2016

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.