# Rate of Change in surface area

Question: $A$ sunflower grows so that its diameter increases at a constant rate of $0.1$ mm per hour until it reaches $200$ mm. Find the rate of increase in the surface area of the flower when its diameter is $50$ mm.

How do I get the answer of $7.85 {(\mathrm{mm})^2}$ per hour?

Start with the formula relating its surface area (probably assumed to be the area of a circle) to its diameter. Take the derivative with respect to time, and don't forget the chain rule! Plug in the values you know, and solve for the one you don't.

Let $\frac{\mathrm{d}A}{\mathrm{d}t}=$ Rate of change of surface area.

Let $\color{red}{\frac{\mathrm{d}r}{\mathrm{d}t}}=$ Rate of change of radius$=0.05$mm per hour.

We require $\frac{\mathrm{d}A}{\mathrm{d}t}$ and by the chain rule $\frac{\mathrm{d}A}{\mathrm{d}t}=\color{blue}{\frac{\mathrm{d}A}{\mathrm{d}r}}\times \color{red}{\frac{\mathrm{d}r}{\mathrm{d}t}}$

We already know the value of the part marked red, so we need to find a value for the part marked blue. To do this, we need a connection between the surface area $A$ and the radius $r$.

Can you see what to do next?

Assuming the sunflower shape is a perfect circle we have $A=\pi r^2\implies \frac{\mathrm{d}A}{\mathrm{d}r}=2 \pi r$. Insertion of these values into $\frac{\mathrm{d}A}{\mathrm{d}t} \implies \frac{\mathrm{d}A}{\mathrm{d}t}=2\pi r \times 0.05=0.1\pi r$. Now since we want the rate of increase when $r=25$mm: $\frac{\mathrm{d}A}{\mathrm{d}t}=0.1\pi \times 25=2.5\pi\approx \color{green}{7.85{(\mathrm{mm})}^2}$ $\color{green}{\mathrm{per}}$ $\color{green}{\mathrm{hour}}$.