Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6 cm and a height of 12 cm, at the rate of 3 cm^3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep?

Question

anonymous · Answer

@CausticSyndicalist @emily0824 @Loser66 @satellite73 @ShadowLegendX @TheSmartOne

anonymous · Answer

Please help!

anonymous · Answer

@satellite73 @amistre64 @aaronq @Callisto @Compassionate @Zarkon @TheSmartOne @sammixboo @pooja195 @nincompoop @sleepyjess @bohotness @Data_LG2 @dtan5457 @Somy

anonymous · Answer

@tkhunny

tkhunny · Answer

What is the volume of water at a given depth "h"?

ConeVolume = V(h,r) = $\dfrac{1}{3}\pi r^{2}h$

Can you eliminate the 'r' by using the triangular, cross-sectional area of the upside down cone?  We need V(h).

anonymous · Answer

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6 cm and a height of 12 cm, at the rate of 3 cm^3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep?

tkhunny · Answer

Why are we repeating the question.  Answer my question, please.

We need V(h), instead of V(h,r).  Can you do it?  It is a geometry problem.

anonymous · Answer

@tkhunny i fugured it out

tkhunny · Answer

What did you do?  Share!