Question

The radius of a right circular cone is increasing at a rate of 5 inches per second and its height is decreasing at a rate of 2 inches per second. At what rate is the volume of the cone changing when the radius is 20 inches and the height is 40 inches?

Answer 1

I'm about the same level as you so I'm not too sure.

r=length of radius in inches
h=height of cone in inches
t=seconds
V=volume

r=5t
h=-2t
V=(1/3)pi(r^2)h

V=(1/3)pi((5t)^2)(-2t)
V=(1/3)pi(25t^2)(-2t)
V=(1/3)pi(-50t^3)
V=(-50pi/3)(t^3)
dV/dt=-50pi(t^2)

dV/dt=(dV/dh)(dh/dt)

Answer 2

the above post was wrong. r=5t+an arbitrary constant. Let me try again. I'll make the constants like this:

r=5t+20
h=-2t+40

Answer 3

\[V=\frac{1}{3}\pi r^2 h\]Take the Total Derivative of the expression above.\[\text{Dt}[V]=\frac{1}{3} \pi r^2 \text{Dt}[h]+\frac{2}{3} h \pi r \text{Dt}[r] \]Replace the variables as follows:\[\{\text{Dt}[r]\to 5, \text{ Dt}[h]\to -2,r\to 20,h\to 40\} \]\[\text{Dt}[V]=2400 \pi \text{ cubic} \text{ inches}/\sec\]

Answer 4

Yes! I got the same thing as robtobey except I got there in a more convoluted manner. Good to know I can do derivatives.