Question

If a snowball melts so that its surface area decreases at a rate of 0.2cm^2/min, find the rate at which the diameter decreases when the diameter is 9cm.

Answer 1

Surface area of a sphere is 4pi*r^2 i believe

Answer 2

Okay so: S = 4*pi*(d/2)^2 = pi*d
dS/dt = pi*dd/dt

Answer 3

wait typo

Answer 4

S = pi*d^2
dS/dt = 2pi*d*dd/dt
-0.2 = 2pi*9*dd/dt

Answer 5

-0.2 = 18pi*(dd/dt)
dd/dt = -0.2/(18pi)

Answer 6

\[V=\frac{4}{3}\pi r^3 \implies S.A.=\frac{dV}{dr}=4\pi r^2\]
The radius is half the diameter:
\[\S.A.=\pi D^2 \implies \frac{d (S.A)}{dt}=2 \pi D \frac{d D}{dt} \implies \frac{d D}{dt}=\frac{1}{2\pi D}\frac{d (S.A)}{dt}\]
Just plug in.