Question

A vertical container with base area of length L and width W is being filled with identical pieces of candy, each with a volume of v and a mass m. Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at a certain rate per unit time dH/dt, at what rate per unit time does the mass of the candies in the container increase?

Answer 1

d volume/d h = base area (draw a picture :)

dv/dt = dv/dh * dh/dt
so
dv/dt = base area * dh/dt

let rho = m/v of candy

d m = dv (rho) so dv = dm/rho

so
(1/rho) dm/dt = base area * dh/dt
or
dm/dt = rho * base area * dh/dt

= (m/v) * L * W * dh/dt

Answer 2

Thanks for the help!
I also suspected this as the answer, but unfortunately it is not correct.

Answer 3

Here's the hint they give:
Write an equation for the total mass in the container in terms of the base area, height x of the candy in the container, mass of each candy, and volume of each candy. Then take a time derivative of each side of the equation.

When they say mass of the container in terms of base area -- do they mean LWH = m/roh??
Also, height of x -- is this the rate dH/dt?

The hint doesn't really help me -- it just confuses me more haha.

Answer 4

Did you ever get the correct answer? Im on this right now...