Question

How do you find an equation for the function `f'(x)=-2xsqrt(8-x^2)` whose graph passes through the point (2,7)?

Answer 1

`f(x) = 1/3(2(8-x^2)^(3/2)+5)`. The cycloid-like graph for f is inserted.

Explanation:

`f(x) = -2 int x sqrt((2sqrt2)^2-x^2) dx`,

and substituting

`x = 2sqrt2 sin theta`, so that

`dx = 2sqrt2 cos theta`,

`f(sin theta ) = -32sqrt2 int cos^2theta sin theta d theta`

`=32sqrt 2 int cos^2theta d (cos theta)`

`=32sqrt2[1/3 cos^3theta] + C`

`=32/3sqrt2(1-sin^2theta)^(3/2)+C`

Reverting to x,

`f(x)=32sqrt2/3(1-x^2/8)^(3/2)+C`

As f = 7, at x = 2,

`f(2)=7=32/3sqrt2+C=16/3+C`, giving

`C=5/3`. So,

`f(x) = 1/3(2(8-x^2)^(3/2)+5)`

graph{y-5/3-2/3(8-x^2)^(3/2)=0 [-4, 4, -25, 25]}