# Rate of Change of Area of a Circle

• Oct 14th 2009, 11:37 AM
Velvet Love
Rate of Change of Area of a Circle
For time t between 2 and 3 hours the radius circle C is 5t-t^2-6 feet. At time t=2.3 hours the area of the circle is (to within .001) changing at a rate of ?

How do I do this? The answers I get are always wrong.
• Oct 14th 2009, 11:43 AM
e^(i*pi)
Quote:

Originally Posted by Velvet Love
For time t between 2 and 3 hours the radius circle C is 5t-t^2-6 feet. At time t=2.3 hours the area of the circle is (to within .001) changing at a rate of ?

How do I do this? The answers I get are always wrong.

$\displaystyle f(t) = A = \pi r^2 = \pi (5t-t^2-6)^2$

Use the chain rule to differentiate:

$\displaystyle f'(t) = 2\pi \,(5t-t^2-6) \times (5-2t)$

$\displaystyle f'(2.3) = 2\pi \,(10.6-2.3^2-6)(5-4.6) = -1.734$

(note a negative differential is fine- it means the area is getting smaller)
• Oct 14th 2009, 11:46 AM
Arturo_026
.
• Oct 14th 2009, 11:59 AM
Velvet Love
It says the answer is wrong for some reason.
I hate this online homework :/
• Oct 14th 2009, 12:02 PM
Arturo_026
I think you are lookin for dA
so; dA = A' (dt)
• Oct 14th 2009, 12:04 PM
Velvet Love
Sorry, but what does that mean? XD. My teacher doesn't teach us terms or notation for some reason
• Oct 14th 2009, 12:08 PM
Arturo_026
you know that dA/dt is the rate of change with respect to time.

so aproximately, since dA/dt= A' then dA = A' * dt you can use that to aproximate.

so you use the f'(x) the gentleman got and plug 2 for t
then the dt will be .3 since that is the change in time
• Oct 14th 2009, 12:11 PM
Velvet Love
When I plug 2 in for t, I get 0 for the answer.
• Oct 14th 2009, 12:14 PM
Arturo_026