Question #c84c8

1 Answer
Oct 29, 2017

(dr)/(dt) = 1/(5pi) (cm)/s

Explanation:

To find the rate of increase of the radius, we need an expression that we can use which will result in a derivative of radius with respect to time. Since we are first given a rate of volume change of 40 (cm^3)/s, it seems like a natural starting point to use the volume of a sphere formula.

Take derivatives of all terms with respect to time t:

V = 4/3pi*r^3

(dV)/(dt) = 4/3pi*(3r^2)(dr)/(dt)

(dV)/(dt) = 4pir^2(dr)/(dt)

Now, we know from the first sentence than (dV)/(dt) = 40, and by careful observation of the last line above we note that we're looking right at the formula for the surface area of the sphere (4pir^2)! This means the quantity 4pir^2 = 200pi, and we can substitute both of these values in the final line and solve for the rate of change of the radius:

(dV)/(dt) = 4pir^2(dr)/(dt)

40 (cm^3)/s = (200pi cm^2)(dr)/(dt)

(40 cm^3//s)/(200pi cm^2) = (dr)/(dt)

(dr)/(dt) = 1/(5pi) (cm)/s