If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 11 cm.

Answer 1

volume=4pI*r^3 /3
dV/dt=4Pi*r^2 dr/dt
(note, diameter=2r) so ratediamter=2 radius rate.
dV/dt= 3PI(diamter/2)^2 * dDiamter/dt * 1/4

if I did the algebra in my head right, then
rate diatmeter=16 dV/dt * 1/(3PI*diameter^2)

Answer 2

R = Radius

D = Diameter

The surface area of a sphere:

S = 4 π R²

S = 4 π ∙ ( R / 2 )² = 4 π ∙ D² / 4

S = π ∙ D²

dS / dD = 2 π ∙ D

You know:

dS / dt = - 5 cm² / min

d = 11 cm

The derivative by the chain rule:

dS / dD = dS / dt ∙ dt / dD

2 π ∙ D = ( - 5 ) ∙ dt / dD

2 π ∙ 11 = ( - 5 ) ∙ dt / dD

22 π = ( - 5 ) ∙ dt / dD

22 π / ( - 5 ) = dt / dD

- 22 π / 5 = dt / dD

Take reciprocal value:

- 5 / ( 22 π ) = dD / dt

dD / dt = - 5 / ( 22 π ) cm² / min

Answer 3

My typo:

S = 4 π R²

S = 4 π ∙ ( D / 2 )² = 4 π ∙ D² / 4

S = π ∙ D²

dS / dD = 2 π ∙ D