Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 12

How fast is the diameter decreasing when the diameter is 10cm?
a.) State the quantities given in the problem.
b.) State the unknown.
c.) Illustrate a picture of the situation for any time $t$.
d.) Write an expression that relates the quantities.
e.) Solve the problem

a.) Given: $\displaystyle \frac{dA}{dt} = -1 cm^2 / min$, rate of change of the surface area of snowball.

b.) Unknown: $\displaystyle \frac{d(d)}{dt}$ when $d = 10cm$, the rate at which the diameter changes with respect to time

c.)







d.) Let the snowball be a perfect sphere, its surface area is

$A = 4 \pi r^2$ where $r$ is the radius of the sphere

but we are given in terms of its diameter so we make radius in terms of diameter


$
\begin{equation}
\begin{aligned}

d =& 2r ; r = \frac{d}{2}
\\
\\
A =& 4 \pi \left( \frac{d}{2} \right)^2 = \frac{4 \pi d^2}{4} = \pi d^2
\\
\\
A =& \pi d^2
\\
\\
\frac{dA}{dt} =& \pi \cdot \frac{d}{d(d)} (d^2) \frac{d(d)}{dt}
\\
\\
\frac{dA}{dt} =& \pi \cdot 2d \left( \frac{d(d)}{dt} \right)
\\
\\
\frac{dA}{dt} =& 2 \pi d \left( \frac{d(d)}{dt } \right)
\\
\\
-1 =& 2 \pi (10) \left( \frac{d(d)}{dt} \right)
\\
\\

& \boxed{\displaystyle \frac{d(d)}{dt} = \frac{-1}{20 \pi} cm/min, \text{ negative values means the diameter is decreasing } }

\end{aligned}
\end{equation}
$