A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what r

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 11 cm. (The answer is a positive number).

Diameter is decreasing at rate of 0.4 cm/min
So radius is decreasing at rate of 0.2 cm/min ------> dr/dt = -0.2

When diameter is 11 cm -----> r = 5.5

V = 4/3 π r³

Differentiate both sides with respect to t:
dV/dt = 4π r² dr/dt
dV/dt = 4π (5.5)² (-0.2)
dV/dt = -76.0265

Note that negative dV/dt indicates that volume is increasing at a negative rate, or decreasing at a positive rate.

Volume is decreasing at a rate of 76.0265 cm³/min

--------------------

Note that jdash01's solution above uses diameter instead of radius.
He should get same result, except he makes small error in formula of volume:
V = 4/3 π r³ = 4/3 π (D/2)³ = 4/3 π D³/8 = 1/6 π D³ (not 1/12 π D³)
So using this method, with 1/6 instead of 1/12, he would twice 38.01 = 76.02

Ματπmφm

This is a sphere so with a volume V =(4/3) pi*r^3 or V = (1/12) pi*D^3
Given:
dD/dt = - 0.4 cm/min
D = 11 cm
dV/dt=?

V = (1/12) pi*D^3
dV/dt = (1/4) pi*D^2 dD

substitute:::

dV/dt = (1/4) pi*(11^2) * (- 0.4)
= -38.01 cc/min

so volume is decreasing at a rate of 38.01 cubic cm per minute.

=))

Do the question, if you get stuck, come back, and I will help you. Otherwise you just sound retarded by asking me this very elementary calculus question that YOU NEED to do in order to successfully graduate from college of any type.

===4.4 min: .4cm/minx11min=4.4 cm