Related rates

[Goldenboko]

Mid-term coming up, and I seem to be really rusty on my conic related rates.

Found a good example online, "A conical water tank with vertex down has a radius of 8 feet at the top and is 16 feet high. If water flows into the tank at a rate of 12ft^3 per minute, how fast is the depth of the water increasing when the water is 10 feet deep."

I'm sure I can remember how to do this, but seeing one out would help for my midterm as I won't be getting help from my teacher.

[Endless]

Ooo, I'm a bit rusty at that, so let's see.

Volume of a cone: 1/3 pi*r^2*h

Your cone fills up at a speed of 12ft^3/min, so in one minute, the height goes from h to H as follows:

1/3*pi*R^2*H-1/3*pi*r^2*h = 12
We know that at max height, r_max = 8 and h_max = 16, and since the radius is proportional to the height (it's a cone), we have r = h/2

Therefore 1/3*pi*(H/2)^2*H-1/3*pi*(h/2)^2*h = 12
H^3 = (12*3*4/pi)+h^3
H = (144/pi + h^3)^(1/3)

With h=10, your new height after one minute is H ~ 10.150512 ft, so the speed in ft/min is the difference in height:
S = H-h ~ 0.150512ft/min

If you want to generalize for any given height h, what is the speed in ft/min:

S = H-h = (144/pi + h^3)^(1/3)-h

Generalized more for any cone of radius alpha*h:

S = {[(36/pi)+((alpha)^2*(h^3))]/(alpha^2)}^(1/3) - h

And furthermore, with a speed v in ft^3/min:

S = {[(3*v/pi)+((alpha)^2*(h^3))]/(alpha^2)}^(1/3) - h

[Goldenboko]

Thanks! It was putting this part

We know that at max height, r_max = 8 and h_max = 16, and since the radius is proportional to the height (it's a cone), we have r = h/2

Together that impeded me from finishing the problem. Took the Exam 2 hours after reading, sadly after all that, there was no conic question!