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
To find the rate at which the volume is increasing, you need to use the chain rule of differentiation.
First, let's find the expressions for dr/dt and dh/dt when h = 8cm and r = 6cm.
Given that both the radius and height increase at a constant rate of 2 cm/s, we have:
dr/dt = 2 cm/s
dh/dt = 2 cm/s
Now, let's differentiate the volume function with respect to time (t):
V = (1/3) * pi * r^2 * h
To apply the chain rule, we need to differentiate each variable separately, and mulitply by their respective rates of change:
dV/dt = (1/3) * pi * (2r(dr/dt)h + r^2(dh/dt))
Substituting the given values:
dV/dt = (1/3) * pi * (2(6)(2)8 + 6^2(2))
Simplifying further:
dV/dt = (1/3) * pi * (192 + 72)
dV/dt = (1/3) * pi * 264
dV/dt = 88 * pi
So, the rate at which the volume is increasing when the height is 8cm and the radius is 6cm is 88pi cubic cm/s.
It seems you made an error in multiplying the terms while simplifying the expression. The correct value should be 88pi, not 108pi.