Question

Sarah is filling ice cream cones with shaving cream to hand out to trick or treaters. The cones have a largest radius of 3 centimeters and a depth of 13 centimeters. She fills them at a rate of 5 cubic centimeters per second. (She works carefully so that shaving cream is compact and even) At what rate is the level of the shaving cream changing when it is still 3 centimeters from the top of the cone?

Answer 1

The trick is that the change in volume is the surface area times the change in height.
dV/dt = pi r^2 dh/dt
here dV/dt is constant = 5 cm^3/s
so all you need is pi r^2 at depth of 11 cm

Answer 2

depth of 10 cm I mean :)

Answer 3

Traditional approach,
at a time of t seconds, let the height of cream be h, and the radius of the cream be r
by similar shapes,
h/r = 13/3
3h = 13r ---> h = 13r/3

V = (1/3)π r^2 h
= (1/3)π r^2 (13r/3)
= (13/9)π r^3
dV/dt = 13/3 π r^2 dr/dt
5 = 13/3 π r^2 dr/dt
15/(13πr^2) = dr/dt

when h = 10, r = 30/13
dr/dt = 15/(13π(900/169)
= 13/(60π)


"level of the shaving cream changing" ?? I will interpret that as how fast is the area of the surface changing.

A = πr^2
dA/dt = 2π r dr/dt
so when r = 30/13

dA/dt = 2π(30/13)(13/(60π))
= 1 cm^2 / s