Question

CEMENT IS POURED SO THAT IT CONTINUOUSLY FORMS A CONICAL PILE,THE HEIGHT OF W/H IS TWICE THE RADIUS OF THE BASE,IF THE CEMENT IS BEING POURED AT THE RATE OF 12 CUBIC FEET PER SECOND ,HOW FAST IS THE HEIGHT OF THE PILE CHANGING WHEN IT IS 4 FEET HEIGH?

Answer 1

Once you master the math, you can work on the spelling!...

we know h = 2r, so r = h/2

v = 1/3 pi r^2 h = 1/12 pi h^3

dv/dt = 1/4 pi h^2 dh/dt

So, now just plug in the numbers you have for h and dv/dt to find dh/dt.

Answer 2

Or, you can use the product rule:

v = 1/3 pi r^2 h
dv/dt = 1/3 pi (2rh dr/dt + r^2 dh/dt)

Then you have to fill in the other quantities to get dh/dt.

Note that since r = h/2, dr/dt = 1/2 dh/dt

Answer 3

v = (1/3) pi r^2 h

but r = h/2 so dr = (1/2) dh

v = (1/3) pi * 2 r^3

dv/dt = (2) pi r^2 dr/dt

dv/dt = pi r^2 dh/dt

Note: this is just the surface area * dh/dt

Answer 4

Oh, yeah. I like it. I think I pooh-poohed that derivation on another problem, thinking you had short-circuited the product rule, but this time I studied it more closely.

Of <u>course</u> dv is just a thin circle on top of the cone (pi r^2) times the thickness (dh)!

Answer 5

exactly. I am a Naval Architect. We have a hydrostatic characteristic called "tons per inch" which is how many tons of cargo you can add to increase the draft by one inch as a function of draft. It is basically the water plane area * density of seawater :)