the volume V of a right circular cone of height h=8 feet and radius r feet is V=V(r)=2pier^2. If r is changing, find the instantaneous rate of change of the volume V with respect to the radius r at r=4.
PI, not pie!
It's a Greek letter, not a dessert.
iv v = 2πr^2,
dv/dr = 4πr
now just plug in r=4
However, for a cone of height 8,
v = 8/3 πr^2
If that's what you want, make the necessary change to the above logic.
To find the instantaneous rate of change of the volume V with respect to the radius r, we can take the derivative of the volume function V(r) with respect to r and evaluate it at a specific value of r.
Given that the volume function V(r) is defined as V(r) = 2πer^2, where h = 8 feet is the height and r is the radius of the cone, we can differentiate V(r) with respect to r.
First, let's find the derivative of V(r) using the power rule of differentiation:
dV/dr = d/dx (2πer^2)
= 2πe * d/dx (r^2)
= 2πe * 2r
Simplifying the derivative, we get:
dV/dr = 4πer
Now that we have the derivative, we can find the instantaneous rate of change of V with respect to r at r = 4 by evaluating dV/dr at r = 4:
dV/dr (at r = 4) = 4πe * 4
= 16πe
Therefore, the instantaneous rate of change of V with respect to the radius r at r = 4 is 16πe.