Given:
Rate of increase in ripple radius, [tex]\frac{dr}{dt}[/tex] = 2 ft/sec
Solution:
Area of circle = [tex]\pi r^{2}[/tex]
Differentiating w.r.t 'r', we get:
[tex]\frac{dA}{dr}[/tex] = 2[tex]\pi r[/tex]
Rate of change in area is given by:
[tex]\frac{dA}{dt}[/tex] = [tex]\frac{dA}{dr}[/tex]. [tex]\frac{dr}{dt}[/tex]
(by chain rule)
[tex]\frac{dA}{dt}[/tex] = 2[tex]\pi r[/tex] . [tex]\frac{dr}{dt}[/tex]
when radius, r = 5ft
[tex]\frac{dA}{dt}[/tex] = 2[tex]\pi 5[/tex] . 2 = 20[tex]\pi[/tex]
Now,
when r = 10 ft
[tex]\frac{dA}{dt}[/tex] = 2[tex]\pi r[/tex] . [tex]\frac{dr}{dt}[/tex]
[tex]\frac{dA}{dt}[/tex] = 2[tex]\pi 10[/tex] . 2 = 40[tex]\pi[/tex]
