Review question

# If we increase $r$ by a small amount, what happens to $A$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5329

## Solution

A closed circular cylinder has height $16$ in. and radius $r$ in. The total surface area is $A$ sq. in.

Prove that $\begin{equation*} \frac{dA}{dr} = 4\pi(r+8). \end{equation*}$

The surface of a closed circular cylinder can be split into three parts, the two discs at the end, and the rectangular central section.

The two discs each have area $\pi r^2$.

The central section is a curled up rectangle whose length is the height of the cylinder ($16$), and whose width is the circumference of the circle ($2\pi r$).

Hence the area of the rectangle is $32 \pi r$, so that $\begin{equation*} A = 2 \pi r^2 + 32 \pi r. \end{equation*}$ As a result, $\begin{equation*} \frac{dA}{dr} = 4\pi r + 32\pi = 4\pi(r+8). \end{equation*}$

Use this result to calculate an approximation for the increase in area when the radius increases from $4$ to $4.02$ in., the height remaining constant. You may leave the answer in terms of $\pi$.

If we increase $r$ by a small amount $\delta r$, we increase the surface area $A$ by a small amount $\delta A$.

Moreover, we can say $\dfrac{\delta A}{\delta r} \approx \dfrac{dA}{dr}$ evaluated at that point, and when $r = 4, \dfrac{dA}{dr} = 48\pi$.

So $\dfrac{\delta A}{\delta r} =\dfrac{\delta A}{0.02} \approx 48 \pi$, which tells us that $\delta A \approx 0.96\pi$.