Question

A spherical snowball is melting in such a way that its diameter is at rate of 0.3cm/min. At what rate is the volume of the snowball decreasing when the diameter is 14 cm?

Rate of change of volume = ___ cm^3/min

Answer 1

Okay, I'll try it.

Answer 2

I tried (pi)(1/6)(0.3)^3, but it didn't work.

Answer 3

It's 2/3, not 4/24

Answer 4

Okay, we get:
D=14
dD/dt=0.3
Thus,
R=7
dR/dt=0.15.

Answer 5

Then, since V=4/3*pi*r^3, we have dV/dt=4/3*pi*3r^2*dr/dt=4/3*pi*3(7)^2*(0.15).

Answer 6

But remember, it has to be...NEGATIVE! I did it! Hooray!

Answer 7

The volume of a sphere is

\(V = \dfrac{4}{3}\pi r^3 \)

Since \(r = \dfrac{D}{2}\), we get


\(V = \dfrac{4}{3}\pi (\dfrac{D}{2})^3\)

\(V = \dfrac{4 \pi}{3} \cdot \dfrac{D^3}{8} \)

\(V = \dfrac{4\pi D^3}{24} \)

\(V = \dfrac{\pi D^3}{6} \)

Now we take the derivative of both sides:

\(\dfrac{V}{dt} = \dfrac{3\pi D^2}{6} \cdot \dfrac{dD}{dt} \)

\(\dfrac{V}{dt} = \dfrac{\pi D^2}{2} \cdot \dfrac{dD}{dt} \)

Now substitute D = 14 cm and dD/dt = -0.3 cm/min.

\(\dfrac{dV}{dt} = \dfrac{\pi (7~cm)^2}{2} \cdot \dfrac{-0.3~cm}{min} \)