Question

Water drips into a circular puddle such that the radius of the puddle, in centimeters, at time t, in seconds, is given by the equation r(t) = root t. a. Write an equation for the area of the puddle as a function of t. b. What is the AROC of the area of the puddle with respect to time between t = 0 and t = 16? c. What is the AROC of the area of the puddle with respect to the radius between t = 0 and t = 16? d. What is the AROC of the area of the puddle with respect to the circumference of the puddle between t = 0 and t = 16?

Answer 1

a. The equation for the area of the puddle as a function of time is A(t) = πt. b. The AROC of the area of the puddle with respect to time between t = 0 and t = 16 is π. c. The AROC of the area of the puddle with respect to the radius between t = 0 and t = 16 is π√t. d. The AROC of the area of the puddle with respect to the circumference between t = 0 and t = 16 is r(t)/(2π).

Step-by-step explanation:

a. The area of a circular puddle is given by the equation A(t) = π(r(t))^2, where r(t) is the radius of the puddle at time t. Using the given equation for r(t), we can substitute it into the formula for the area to get A(t) = π(root t)^2. Simplifying further, we have A(t) = πt.

b. The average rate of change (AROC) of the area of the puddle with respect to time between t = 0 and t = 16 can be found by calculating A(t) at time t = 16 and subtracting A(t) at time t = 0, and then dividing by the change in time: (A(16) - A(0))/(16 - 0). Using the equation A(t) = πt, we can calculate A(16) = π(16) = 16π and A(0) = π(0) = 0, so the AROC is (16π - 0)/(16 - 0) = π.

c. The AROC of the area of the puddle with respect to the radius can be found by calculating A(r) at radius r = √t and subtracting A(r) at radius r = 0, and then dividing by the change in radius: (A(√t) - A(0))/(√t - 0). Using the equation A(r) = πr^2, we can calculate A(√t) = π(√t)^2 = πt and A(0) = π(0)^2 = 0, so the AROC is (πt - 0)/(√t - 0) = π√t.

d. The AROC of the area of the puddle with respect to the circumference can be found by calculating A(c) at circumference c = 2πr(t) and subtracting A(c) at circumference c = 0, and then dividing by the change in circumference: (A(2πr(t)) - A(0))/(2πr(t) - 0). Using the equation A(c) = π(c/(2π))^2 = (c/(2π))^2, we can calculate A(2πr(t)) = (2πr(t)/(2π))^2 = r(t)^2 and A(0) = (0/(2π))^2 = 0, so the AROC is (r(t)^2 - 0)/(2πr(t) - 0) = r(t)/(2π).

Answer 2

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
`r(t)=√(t)`.

Then the area of the puddle is given by the area of a circle =
`Area=\pi r^2`
But, given that
`r(t)=√(t)`, then
`A(t)=\pi (r(t))^2=\pi(√(t))^2=\pi t`

Therefore, the equation for the area of the puddle as a function of t is given by
`A(t)=\pi t`



Part B

The average rate of change of a function f(x) between x = a and x = b is given by
`(f(b)-f(a))/(b-a)`.

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
`(A(16)-A(0))/(16-0) = (16\pi-0)/(16) = (16\pi)/(16) =\pi`

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
`A(r)=\pi r^2`

Given that
`r(t)=√(t)`, thus when t = 0,
`r(0)=√(0)=0` and when t = 16,
`r(16)=√(16)=4`

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


`(A(4)-A(0))/(4-0) = (\pi(4)^2-\pi(0)^2)/(4) = (16\pi)/(4) =4\pi`

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
`C=2\pi r`

Thus, the radius of the puddle in terms of circumference is given by
`r= (C)/(2\pi)`

Thus, the area of the puddle with respect to the circumference, C, of the puddle is given by
`A(C)=\pi\left( (C)/(2\pi) \right)^2= (1)/(4\pi) C^2`

Since,
`C=2\pi r` and
`r(t)= √(t)`, 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
`(A(8\pi)-A(0))/(8\pi-0) = ( ((8\pi)^2)/(4\pi)- ((0)^2)/(4\pi) )/(8\pi) = ( (64\pi^2)/(4\pi) )/(8\pi) = (16\pi)/(8\pi) =2`

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.