Answer :
Answer:
The rate of change is [tex]108\pi[/tex] ft^(2)/min
Step-by-step explanation:
The area of a circle is given by the following equation:
[tex]A(t) = \pi r^{2}[/tex]
To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So
[tex]\frac{dA(t)}{dt} = 2\pi r \frac{dr}{dt}[/tex]
We have that:
[tex]\frac{dr}{dt} = 6, r = 9[/tex].
We want to find [tex]\frac{dA}{dt}[/tex]
So
[tex]\frac{dA(t)}{dt} = 2\pi*9*6[/tex]
[tex]\frac{dA}{dt} = 108\pi[/tex]
Since the area is in square feet, the rate of change is in ft^(2)/min.
So the rate of change is [tex]108\pi[/tex] ft^(2)/min
The rate of change of the area at the moment in time is 108π square feet per min.
What is the rate of change?
The net external force acting on a body is zero, so the rate of momentum change is also zero, meaning there is no momentum change.
Given
The radius is changing at the rate of dr/dt = 6 ft/min.
Let A(t) be the area of a circle with radius r(t), at time t in min.
[tex]\rm Area \ of \ circle=\pi r^2[/tex]
Differentiate the function on both sides to find the rate of change with respect to r.
[tex]\rm A(t)=\pi r^2\\\\\dfrac{dA(t)}{dt}=2\pi r \dfrac{dr}{dt}[/tex]
Substitute all the values in the equation
[tex]\rm \dfrac{dA(t)}{dt}=2\pi r \dfrac{dr}{dt}\\\\ \dfrac{dA(t)}{dt}=2\pi \times 9 \times 6\\\\ \dfrac{dA(t)}{dt}=108\pi[/tex]
Hence, the rate of change of the area at the moment in time is 108π square feet per min.
To know more about the rate of change click the link given below.
https://brainly.com/question/15492078