The volume of a cone of radius r and height h is given by V=(1/3)pir^2h. If the radius and the height both increase at a constant rate of 2cm/s at which rate in cubic cm/s is the volume increasing when the height is 8cm and the radius is 6cm.
So The answer is 108 pi.
Here's what I did. I thought that dr/dt=2 and dh/dt=2 looking for dv/dt when h=8 and r=6
so dv/dt=(pi/3)*36*2 + (2pi/3)*6*8*2
I got 88pi....that is wrong....what did I do wrong :'(
Thanks
I also got 88π
To find the rate at which the volume is increasing when the height is 8 cm and the radius is 6 cm, you need to use the chain rule in calculus.
Let's denote the volume of the cone as V and its radius as r, and its height as h. We are given that dr/dt = 2 cm/s and dh/dt = 2 cm/s.
We are looking for the rate at which the volume is increasing, which is dv/dt.
The formula for the volume of a cone is V = (1/3) * π * r^2 * h.
To find dv/dt, we need to differentiate both sides of the equation with respect to time t:
dV/dt = (d/dt)((1/3) * π * r^2 * h).
Applying the chain rule:
dV/dt = (1/3) * π * [(d/dt)(r^2 * h) + (d/dt)(r^2) * h + r^2 * (d/dt)(h)].
Let's calculate each of these terms separately:
(d/dt)(r^2 * h): Using the product rule, we get (2r * dr/dt) * h + r^2 * (dh/dt).
(d/dt)(r^2): Using the power rule, we get 2r * dr/dt.
(d/dt)(h): This is simply dh/dt.
Substituting these values back into our equation:
dV/dt = (1/3) * π * [(2r * dr/dt) * h + r^2 * (dh/dt) + r^2 * (dh/dt)].
Now, plug in the given values h = 8 cm, r = 6 cm, dr/dt = 2 cm/s, and dh/dt = 2 cm/s:
dV/dt = (1/3) * π * [(2(6) * 2) * 8 + (6^2) * 2 + (6^2) * 2].
Simplifying:
dV/dt = (1/3) * π * [192 + 72 + 72].
dV/dt = (1/3) * π * 336.
dV/dt = 112π cm^3/s.
Therefore, the rate at which the volume is increasing is 112π cubic cm/s, not 108π cubic cm/s as mentioned in your question.