Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms

Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 8 ft high? (Round your answer to two decimal places.)

If height = x, then base radius = ½x (x is in feet)

So V = ⅓ π (½x)² x

. . . . = (π /12) x³ (ft³)

But also

V = 10t (t is in minutes)

So

10t = π x³ /12

x³ = 120 t/π

x = ³√(120 t/π)

So we have x as a function of t.

When x = 8 feet,

8³ = 120 t/π

t = 512 π / 120 = 64 π/15

So the diameter of the pile is 8 feet at time 64 π/15 minutes

and we want to find dx/dt at t = 64 π/15


x = (120 t/π)^⅓ which is better expressed now as

x = (120 / π)^⅓ ∙ (t^⅓)

dx / dt = (120 / π)^⅓ ∙ ⅓ t^ˉ⅔

. . . . . = {(120^⅓ ) / 3 (π^⅓ )} (t^⅔)

And putting t = 64 π/15 gives

. . . . . = (120^⅓ ) / 3 (π^⅓ ) {(64 π/15)^⅔}


I'll try to include sufficient detail here, but I haven't the time to fill all the details.

This expression simplifies to


. . . . . . . . . . (120^⅓ ) (15^ ⅔) . . . . (8^⅓) (15^⅓) (15^ ⅔) . . . . . . . . . 2 ∙ 15
. . . . . = . . ----------------------- . . = . . ---------------------------- . . = . . ---------------------------
. . . . . . . . . . 3 π^⅓ (64 π)^⅔ . . . . . 3 π^⅓ (64 ⅔) (π^⅔) . . . . . 3 (4²) (π^⅓ ) (π^ ⅔)


. . . . . . . . . .5
. . . . . = . . ----- . . . . . (do the cancelling and multiplications. To multiply powers, you)
. . . . . . . . . 8π . . . . . .( add the indices, remember.)


So the answer is (5/8π) feet per second. You can do the arithmetic yourself.

Do your own homework.

This is against the rules of YA!