# Related Rates: inverted cone

• Nov 11th 2010, 11:55 AM
Henderson
Related Rates: inverted cone
A cone (pointing downward) has a depth of 8m and a radius of 5m. Beginning at t=0, two things start happening: the radius begins expanding at 1 meter/sec, and water is poured in at a rate of 3 cubic meters/sec. The depth of the cone stays constant.

At what time does the depth of the water reach a maximum?

I've got an answer that I feel good about, but I'd like a more elegant method. Specifically, it'd be nice to know if this is possible without rewriting each part as a function in terms of time.

My answer in white text: t = 5.
• Nov 11th 2010, 01:47 PM
skeeter
Quote:

Originally Posted by Henderson
A cone (pointing downward) has a depth of 8m and a radius of 5m. Beginning at t=0, two things start happening: the radius begins expanding at 1 meter/sec, and water is poured in at a rate of 3 cubic meters/sec. The depth of the cone stays constant.

At what time does the depth of the water reach a maximum?

I've got an answer that I feel good about, but I'd like a more elegant method. Specifically, it'd be nice to know if this is possible without rewriting each part as a function in terms of time.

My answer in white text: t = 5.

I came up with depth as a function of time also ...

$\displaystyle h = \frac{9t}{\pi(5+t)^2}$

... and arrived at the same solution. Seems to me that getting depth as a function of time is about as "elegant" as any other method. I'd have to see an alternate approach to judge it "elegance" quotient.