grain is pouring down chute at 8 cu ft/min, forming conical

[Opus89]

Grain is being poured through a chute at a rate of 8 cubic feet per minute and forms a conical pile. The radius of the bottom of the pile is always half of the height of the pile. How fast is the circumference of the base increasing when the pile is 10 feet high?

for two weeks our Calc. teacher has been going over related rates problems with an object that is filling up with water. We have had one or two problems where a cone is being filled up with water and were asked to find the rate of change of the height of water. Now he has given us a problem where the cone is inverted and he wants us to find the rate of change of the circumference. I would think you would use the standard formula for the volume of a cone on this problem too, but then when i think about it, wouldn't the circumference of the base get bigger FASTER than the change in height of the cone? but then again the problem says that the base is always half the height of the pile so i don't know if he took that into a factor or not.

if anyone could help guide me on the right path to solving this problem, that would be greatly appreciated.

thank you for your time!

 

[galactus]

Re: conical rate change problem

Let r=radius of cone and h=height of cone.

If C is the circumference of the base, then we want dC/dt when h=10 given that dV/dt=8.

\(\displaystyle r=\frac{h}{2}\),

so \(\displaystyle C=2{\pi}r={\pi}h\)

\(\displaystyle \frac{dC}{dt}={\pi}\cdot \frac{dh}{dt}\).........[1]

Now use the volume of a cone formula, \(\displaystyle V=\frac{\pi}{3}r^{2}h\), and sub in the r we just found in the beginning.

V will now be entirely in terms of h. Differentiate V and solve for dh/dt. Then, sub dh/dt into [1] to find dC/dt.

Follow what I mean?. Let me know what you get.