Part A
Given that the puddle is circular in shape and that the radius of the puddle, in centimeters, at time t, in seconds, is given by the equation [tex]r(t)=\sqrt{t}[/tex].
Then the area of the puddle is given by the area of a circle = [tex]Area=\pi r^2[/tex]
But, given that [tex]r(t)=\sqrt{t}[/tex], then [tex]A(t)=\pi (r(t))^2=\pi(\sqrt{t})^2=\pi t[/tex]
Therefore, the equation for the area of the puddle as a function of t is given by [tex]A(t)=\pi t[/tex]
Part B
The average rate of change of a function f(x) between x = a and x = b is given by [tex] \frac{f(b)-f(a)}{b-a} [/tex].
Thus, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is given by [tex] \frac{A(16)-A(0)}{16-0} = \frac{16\pi-0}{16} = \frac{16\pi}{16} =\pi[/tex]
Therefore, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is π.
Part C
The area of the puddle with respect to the radius is given by [tex]A(r)=\pi r^2[/tex]
Given that [tex]r(t)=\sqrt{t}[/tex], thus when t = 0, [tex]r(0)=\sqrt{0}=0[/tex] and when t = 16, [tex]r(16)=\sqrt{16}=4[/tex]
Thus, the average rate of change of the area of the puddle with respect to the radius between r = 0 and r = 4 is given by
[tex] \frac{A(4)-A(0)}{4-0} = \frac{\pi(4)^2-\pi(0)^2}{4} = \frac{16\pi}{4} =4\pi[/tex]
Therefore, the average rate of change of the area of the puddle with respect to the radius between t = 0 and t = 16 is 4π.
Part D
The circumference of a circle is given by [tex]C=2\pi r[/tex]
Thus, the radius of the puddle in terms of circumference is given by [tex]r= \frac{C}{2\pi} [/tex]
Thus, the area of the puddle with respect to the circumference, C, of the puddle is given by [tex]A(C)=\pi\left( \frac{C}{2\pi} \right)^2= \frac{1}{4\pi} C^2[/tex]
Since, [tex]C=2\pi r[/tex] and [tex]r(t)= \sqrt{t} [/tex], thus when t = 0, r = 0 and C = 0; when t = 16, r = 4 and C = 8π.
Thus the area of the puddle with respect to the circumference, C, of the puddle between C = 0 and C = 8π is given by [tex] \frac{A(8\pi)-A(0)}{8\pi-0} = \frac{ \frac{(8\pi)^2}{4\pi}- \frac{(0)^2}{4\pi} }{8\pi} = \frac{ \frac{64\pi^2}{4\pi} }{8\pi} = \frac{16\pi}{8\pi} =2[/tex]
Therefore, the average rate of change of the area of the puddle with respect to the circumference of the puddle between t = 0 and t = 16 is 2.
