hi all, its not the calculus im stuck on but the function $\displaystyle A$ in the following question

A circular pipe has outer diameter $\displaystyle \text{4 cm}$ and thickness $\displaystyle \text{t cm}$.

a) Show that the area of cross-section, $\displaystyle \text{A }cm^2$, is given by $\displaystyle A = \pi(4t-t^2)$

b) Find the rate of increase of $\displaystyle A$ with respect to $\displaystyle t$ when $\displaystyle t=\frac{1}{4}$ and when $\displaystyle t=\frac{1}{2}$, leaving $\displaystyle \pi$ in the answer.

As far as i can see

$\displaystyle r=\frac{4-t}{2}$

and

$\displaystyle A=\pi(4-2t+\frac{1}{4}t^2)$