The volume V or a cone (V = 1/3 π r² h) is increasing at a rate of 28π cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12π cubic units and the radius is increasing at 0.5 unit per second.
a) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base? (I got 3π)
b) At the instant when the radius of the cone is 3 units, what is the rate of change of its height h?
c)At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height h?
For A I got 3pi
For B I got 8
What do i do for C? and what answer should I get?
To find the instantaneous rate of change of the area of the base with respect to its height (dA/dh), we can use implicit differentiation. Let's use the formula for the volume of a cone to relate the variables:
V = (1/3) * π * r^2 * h
To differentiate both sides of the equation with respect to h, we treat r as a constant because we are only interested in the change of A with respect to h:
dV/dh = (1/3) * π * (2 * r * dr/dh * h + r^2)
Now, let's substitute the given values at the instant when the radius of the cone is 3 units:
dV/dh = (1/3) * π * (2 * 3 * 0.5 * h + 3^2)
Simplifying the equation:
dV/dh = (1/3) * π * (3h + 9)
We're looking for the instantaneous rate of change, so we need to evaluate dV/dh at the given instant when the radius is 3 units. Let's substitute h = 8 into the equation:
dV/dh = (1/3) * π * (3 * 8 + 9)
= (1/3) * π * (24 + 9)
= (1/3) * π * 33
= 11π
Therefore, at the instant when the radius of the cone is 3 units, the instantaneous rate of change of the area of its base with respect to its height is 11π.