Suppose that the radius r of a right circular cone of fixed height h = 30 cm is increasing at a rate 10 cm/s. How fast is the volume increasing when r = 15? Give your answer correct to two decimal places.

from the information ...

V =30πr^2
dV/dt = 60πr dr/dt
= 60π(15)(10)
= 9000π cm^3/sec
= .... (you do the button-pushing for the required accuracy)

If cone not cylinder

then
V = (1/3) h pi r^2
V = 10 pi r^2
The rest is the same

sorry, I should really read more carefully,

To find how fast the volume of the cone is increasing when the radius is 15 cm, we need to use the chain rule from calculus. Let's start by deriving the formula for the volume of a cone.

The volume of a cone can be calculated using the formula:
V = (1/3)πr²h

Where:
V is the volume of the cone
r is the radius of the cone
h is the height of the cone

Now, let's differentiate the volume equation with respect to time t (since we are looking for the rate of change of volume over time):

dV/dt = (dV/dr)(dr/dt) + (dV/dh)(dh/dt)

Since h is constant at all times (fixed height), the second term (dV/dh)(dh/dt) becomes zero. Therefore, the equation simplifies to:

dV/dt = (dV/dr)(dr/dt)

Let's find each component of this equation separately.

1. Find dV/dr:
To find dV/dr, we need to differentiate V = (1/3)πr²h with respect to r.
dV/dr = (2/3)πrh

2. Find dr/dt:
Given in the problem, the radius r is increasing at a rate of 10 cm/s. Therefore, dr/dt = 10 cm/s.

Now, substitute the known values into the equation:

dV/dt = (2/3)πrh * (dr/dt)

When r = 15 cm, h = 30 cm, dr/dt = 10 cm/s, substitute these values into the equation:

dV/dt = (2/3)π(15)(30) * 10

Calculating this will give us the rate at which the volume is increasing.