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)
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