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?
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.
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
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
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 course dv is just a thin circle on top of the cone (pi r^2) times the thickness (dh)!