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.

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- Oct 14th 2009, 12:37 PMVelvet LoveRate 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, 12:43 PMe^(i*pi)
- Oct 14th 2009, 12:46 PMArturo_026
.

- Oct 14th 2009, 12:59 PMVelvet Love
It says the answer is wrong for some reason.

I hate this online homework :/ - Oct 14th 2009, 01:02 PMArturo_026
I think you are lookin for dA

so; dA = A' (dt) - Oct 14th 2009, 01:04 PMVelvet Love
Sorry, but what does that mean? XD. My teacher doesn't teach us terms or notation for some reason

- Oct 14th 2009, 01:08 PMArturo_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, 01:11 PMVelvet Love
When I plug 2 in for t, I get 0 for the answer.

- Oct 14th 2009, 01:14 PMArturo_026
do you have the answer?

- Oct 14th 2009, 01:15 PMVelvet Love
With the info you gave me, I get the answer to be 0, because when i plug in 2 for t, 5t - t^2 -6 = 0.