The area A equals = πr^2 of a circular oil spill changes with the radius. At what rate does the area change with respect to the radius when r equals= 3ft?
This is 9pie right? not 6pie?
yes. correct.
3 times 3 is the radius squared.. Then multiply 9 and pi
To find the rate at which the area changes with respect to the radius, we need to differentiate the area formula with respect to the radius.
The area of a circle is given by the formula A = πr^2, where A is the area and r is the radius.
To find the derivative of A with respect to r, we can use the power rule of differentiation.
Differentiating both sides of the equation A = πr^2 with respect to r, we get:
dA/dr = d(πr^2)/dr.
Using the power rule, which states that for any constant n, the derivative of x^n with respect to x is n*x^(n-1), we can differentiate πr^2 with respect to r:
dA/dr = 2πr^(2-1).
Simplifying this expression, we have:
dA/dr = 2πr.
Now, we need to evaluate this expression when r = 3ft.
Substituting r = 3ft into the expression for dA/dr, we get:
dA/dr = 2π(3) = 6π.
Therefore, when the radius r is equal to 3ft, the rate at which the area changes with respect to the radius is 6π square feet per foot.