Calculus help please. For time t between 2 and 3 hours the radius of circle C is5*⁢t-t^2-6 feet......?

a.) At time t=2.8 hours the area of the circle is (to within .001) changing at a rate of?





b.) At time t=2.8 hours the area of the circle is?

Calculus help please. For time t between 2 and 3 hours the radius of circle C is5*⁢t-t^2-6 feet......?
Well, the first thing you want to do, to find the rate of change of the area, is find a function of t which gives the area. If the radius is r(t)= -t^2 + 5t - 6, and the area for a given radius is A(r)= pi * r^2, then the area as a function of time is





A(t)= pi * (-t^2 + 5t - 6)^2





Multiplying out that square gets a little bit messy, but it's not _too_ wild. You end up with (I think)





A(t)= pi(t^4 - 10t^3 + 37t^2 - 60t +36)





At this point solving part B is easier than part A of your question. You just plug in 2.8 to the function here, and get





A(2.8)= (16pi) / 625 (I think.) This is about .080.





Then the rate of change. For this, you need to find the derivative of A(t).





A'(t)= pi(4t^3 - 30t^2 + 74t - 60)





A(2.8)= (-24pi) / (125) square feet per hour. This is about -.603.





I think. I mean, I'm pretty sure.