Related Rates Problem: The radius of a spherical watermelon is growing at a constant rate of 2 centimeters per week. The thickness of the rind is always one tenth of the radius. The volume of the rind is growing at the rate ? cubic centimeters per week at the end of the fifth week. Assume that the radius is initially zero. ...?

The volume of a sphere is V = 4/3 x pi x r^3

The rind volume is the difference of two spheres. The radius of the inner sphere volume is, r - .1 r or .9 r. So the volume of the rind is

V = 4/3 x pi x r^3 - 4/3 x pi (.9 r )^3

V = 4/3 x pi x r^3 - 4/3 x pi x (.9^3) x r^3

Implicitly differentiating both sides with respect to t yields:

V' = ( 4 x pi x r^2 x r' ) - ( 4 x pi x (.9^3) x r^2 x r' )

This equation define the rate of change of the rind volume with respect to time.

As defined in the problem, r' = 2

As well, after 5 weeks the radius will be r = 2 * 5 = 10

V' = (4 x pi x 10^2 x 2) - (4 x pi x (.9^3) x 10 x 2)

V' = (800 x pi) - (800 x pi x (.729))

V' = (800 x pi) (1 - .729)

V' = (800 x pi)(.271)

V' = 681.09 cubic centimeters

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