# Thread: simple geometry applied to calculus

1. ## simple geometry applied to calculus

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

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

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

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

As far as i can see

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

and

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

2. sorry i realise

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

so

$A=\pi(4-4t+t^2)$

3. correct ...

the inner cross-sectional area of the pipe is $A = \pi(2-t)^2$

4. hi skeeter, thats what i dont understand the question states $A = \pi(4t-t^2)$

5. Originally Posted by sammy28
hi skeeter, thats what i dont understand the question states $A = \pi(4t-t^2)$
I do not agree with that equation if the problem was written as posted.

6. thanks skeeter

the question is in error then.

7. just thought of this ... the cross-section the problem wants is the cross-section of the physical pipe itself (not the inside).

A = outer area of pipe - inner area of pipe

$A = \pi \cdot 2^2 - \pi(2-t)^2$

$A = 4\pi - \pi(4 - 4t + t^2)$

$A = 4\pi - 4\pi + 4\pi t - \pi t^2$

$A = \pi(4t - t^2)$

8. Ah so thats what they are talking about. thanks skeeter.
All makes sense now.