The function V=4pir^2 describes the volume of a right circular cylinder of heigh 4ft and radius r feet. Find the (instantaneous) rate of change of the volume with respect to the radius when r=6. Leave the answer in terms of pi.
dV/dr = 8pir*dr/dr
which at r = 6 is 8pi(6)
=48pi
actually the answer is 64pi ft^3/ft.
can you figure out what you did wrong? thanks.
Nope, for the data you gave me the answer is 48pi
this is a very straighforward question,
to have an answer of 64 pi, perhaps you were to find the rate when r = 8.
Check your typing
sorry. yes, r should be equal to 8. Thanks.
To find the instantaneous rate of change of the volume with respect to the radius at a specific point, we need to take the derivative of the volume function with respect to the radius, and then evaluate it at the given point. In this case, the volume function is given as V = 4πr^2, where r is the radius.
To find the derivative of V with respect to r, we use the power rule, which states that if a function is of the form f(x) = x^n, then its derivative is f'(x) = nx^(n-1).
In this case, n = 2 because V = 4πr^2, so we can find the derivative of V with respect to r as follows:
dV/dr = d(4πr^2)/dr
= 8πr
Now, we can evaluate the derivative at the given point r = 6:
dV/dr at r=6 = 8π(6)
= 48π
Therefore, the instantaneous rate of change of the volume with respect to the radius when r = 6 is 48π.