The area A = πr2 of a circular oil spill changes with the radius. At what rate does the area change with respect to the radius when r = 3ft?
dA/dr=2PI r Put in r=3ft, calculate dA/dr
To find the rate at which the area changes with respect to the radius when r = 3 ft, we need to find the derivative of the area function with respect to the radius. In this case, the area function is A = πr^2.
Step 1: Differentiate the area function with respect to the radius.
To find the derivative of A with respect to r, we can use the power rule of differentiation. For a function of the form f(x) = ax^n, where a is a constant, we can differentiate it as f'(x) = anx^(n-1).
In our case, A = πr^2, so differentiating with respect to r, we get dA/dr = d(πr^2)/dr = 2πr.
Step 2: Substitute the given value of r.
Since we want to find the rate of change when r = 3 ft, we substitute r = 3 into the derivative we found in Step 1.
dA/dr = 2πr
dA/dr = 2π(3)
dA/dr = 6π
Therefore, when r = 3 ft, the rate at which the area changes with respect to the radius is 6π square feet per foot.
To find the rate at which the area of the oil spill changes with respect to the radius, we can use the concept of differentiation.
Given that the area (A) of a circular oil spill is given by the formula A = πr^2, we can differentiate both sides of the equation with respect to the radius (r) using the power rule of differentiation.
dA/dr = d/dx (πr^2)
Using the power rule, we differentiate each term:
dA/dr = 2πr
Now we need to evaluate when r = 3ft. Substituting this value into the derived equation, we have:
dA/dr = 2π(3)
Simplifying further:
dA/dr = 6π
Therefore, when the radius is 3ft, the rate at which the area changes with respect to the radius is 6π square feet per foot.