Answer: A. [tex]A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}[/tex]
B. A'(5) = 1.76 cm/s
Step-by-step explanation: Rate of change measures the slope of a curve at a certain instant, therefore, rate is the derivative.
A. Area of a circle is given by
[tex]A=\pi.r^{2}[/tex]
So to find the rate of the area:
[tex]\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}[/tex]
[tex]\frac{dA}{dr} =2.\pi.r[/tex]
Using [tex]r(t)=3-\frac{363}{(t+11)^{2}}[/tex]
[tex]\frac{dr}{dt}=\frac{726}{(t+11)^{3}}[/tex]
Then
[tex]\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}][/tex]
[tex]\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}[/tex]
Multipying and simplifying:
[tex]\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}[/tex]
The rate at which the area is increasing is given by expression [tex]A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}[/tex].
B. At t = 5, rate is:
[tex]A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}[/tex]
[tex]A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}[/tex]
[tex]A'(5)=\frac{2408693760\pi}{4294967296}[/tex]
[tex]A'(5)=1.76268[/tex]
At 5 seconds, the area is expanded at a rate of 1.76 cm/s.
