Hello, Shaun!

The volume of a frustum of a cone (or pyramid) is given by:A shock wave is advancing in the shape of a frustrum of a right circular cone.

The upper radius,xof the frustrum is increasing at the rate of 2m/sec,

the lower radius,yis increasing at the rate of 3m/sec,

and the height,zis increasing at the rate of 1.5m/sec.

Using the chain rule find the rate of increase of the volume,Vof the frustrum

at the moment when x = 10m, y = 16m and z = 8m.

. . . . . . . . . . . . . . . _____

. . 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.