For time t between 2 and 3 hours the radius of circle C is 5*⁢t-t^2-6 feet....?

Question

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

Answer 1

A = pi * r^2
A = pi * (5t - t^2 -6)^2
differentiate both side wrt t
dA/dt = 2 * pi * (5t - t^2 -6)(5 -2t)
plugin in the value of t
dA/dt = 2*3.14*(5*2.8 - 2.8^2 -6)(5 - 2*2.8) = -0.60288
where DA/dt is rate at which the area changes.

Answer 2

Sorry but I don't know what ⁢ means.