# Frustrum rate of change

• Mar 5th 2007, 12:50 AM
Shaun Gill
Frustrum rate of change
A shock wave is advancing in the shape of a frustrum of a right circular cone. The upper radius, x of the frustrum is increasing at the rate of 2m per sec, the lower radius, y is increasing at the rate of 3m per sec, and the height, z is increasing at the rate of 1.5m per sec. Using the chain rule find the rate of increase of the volume, V of the frustrum at the moment when x = 10m, y = 16m and z = 8m.
• Mar 5th 2007, 04:17 AM
Soroban
Hello, Shaun!

Quote:

A shock wave is advancing in the shape of a frustrum of a right circular cone.
The upper radius, x of the frustrum is increasing at the rate of 2m/sec,
the lower radius, y is increasing at the rate of 3m/sec,
and the height, z is increasing at the rate of 1.5m/sec.

Using the chain rule find the rate of increase of the volume, V of the frustrum
at the moment when x = 10m, y = 16m and z = 8m.

The volume of a frustum of a cone (or pyramid) is given by:
. . . . . . . . . . . . . . . _____
. . V .= .(h/3)(B1 + √B1·B2 + B2)

where h is the height and B1, B2 are the areas of the two bases.

Our problem has: .B1 = πx², B2 = πy², h = z.
. . . . . . . . . . . . . . . . . . .______
Then: .V .= .(z/3)(πx² + √πx²·πy² + πy²)

. .or: . V .= .(π/3)z(x² + xy + y²)

Can you finish it now?
Differentiate with respect to time (product rule and chain rule)
. . then plug in the given values.

• Mar 5th 2007, 04:29 AM
ticbol
Quote:

Originally Posted by Shaun Gill
A shock wave is advancing in the shape of a frustrum of a right circular cone. The upper radius, x of the frustrum is increasing at the rate of 2m per sec, the lower radius, y is increasing at the rate of 3m per sec, and the height, z is increasing at the rate of 1.5m per sec. Using the chain rule find the rate of increase of the volume, V of the frustrum at the moment when x = 10m, y = 16m and z = 8m.

The volume V of the frustrum is
V = (pi/3)z[x^2 +y^2 +xy]
Differentiate both sides with respect to time t,
dV/dt = (pi/3){(dz/dt)[x^2 +y^2 +xy] +z[2x(dx/dt) +2y(dy/dt) +x(dy/dt) +y(dx/dt)}
Substituting all the givens,
dV/dt = (pi/3){(1.5)[10^2 +16^2 +10*16] +(8)[2(10)(2) +2(16)(3) +10(3) +16(2)]}
dV/dt = (pi/3){774 +1584}
dV/dt = 786pi

Therefore, at that instant, the volume is increasing at the rate of 786pi cubic meters per second. -------------answer.