Rates of Change

• Feb 22nd 2010, 11:11 PM
xwrathbringerx
Rates of Change
Grain is ejected from a chute at the rate of 0.1 m3/min and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone is increasing, at the instant 3 minutes after the opening of the chute.

Setting out:
dV/dt = 0.1 m3/min
dh/dt
= dV/dt * dh/dV
= 0.1 * dh/dV
V = 1/3 pi r^2 h
= 1/3 pi h^3
dV/dh = pi h^2
dh/dV = 1/(pi h^2)
dh/dt = 1/(10 pi h^2)

BUT I am trying to substitute a value of t...HELP! where did i go wrong???
• Feb 23rd 2010, 03:24 AM
drumist
You are correct that you can't substitute t directly. However, you can use t to find out what V is, then you can use V to find out what h is, then finally you can use that h to plug into your expression.

Does this make sense?
• Feb 23rd 2010, 06:18 AM
xwrathbringerx
Ummm no (Crying) ... how on earth wuld i use t...
• Feb 23rd 2010, 03:15 PM
drumist
Since the volume is increasing at a rate of 0.1 m^3/min and we are looking at 3 minutes after the chute was opened, doesn't it make sense that the volume at that time is 0.3 m^3? I think this makes sense intuitively, but you can also write it in Calculus terms:

$V = \int_0^3 \frac{dV}{dt} \, dt = \int_0^3 0.1 \, dt = [0.1t]_0^3 = 0.3$

Then, to find the height, use your formula relating volume and height.

$0.3 = \frac{1}{3} \pi h^3 ~\implies~ h=\sqrt[3]{\frac{0.9}{\pi}}$