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

Question

1075 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-07T08:09:58

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

#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]}