Answer
(a) The surface area decreases at a rate of $1~cm^2/min$
$\frac{dA}{dt} = -1~cm^2/min$
(b) The unknown is $\frac{dD}{dt}$ where $D$ is the diameter.
(c) We can see a sketch below.
(d) $A = \pi~D^2$
(e) The diameter decreases at a rate of $~~0.016~cm/min$
Work Step by Step
(a) The surface area decreases at a rate of $1~cm^2/min$
$\frac{dA}{dt} = -1~cm^2/min$
(b) The unknown is $\frac{dD}{dt}$ where $D$ is the diameter.
(c) We can see a sketch below.
(d) $r = \frac{D}{2}$
$A = 4\pi~r^2 = 4\pi~(\frac{D}{2})^2 = \pi~D^2$
(e) $A =\pi~D^2$
$\frac{dA}{dt} =2\pi~D~\frac{dD}{dt}$
$\frac{dD}{dt} = \frac{1}{2\pi~D}~\cdot \frac{dA}{dt}$
$\frac{dD}{dt} = \frac{1}{(2\pi)(10~cm)}~\cdot (-1~cm^2/min)$
$\frac{dD}{dt} = -0.016~cm/min$
The diameter decreases at a rate of $~~0.016~cm/min$
