Water is poured into a conical reservoir at a rate

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Reference no: EM13842568

Water is poured into a conical reservoir at a rate of pi cubic feet per second (about 3 ft/sec). The reservoir has a radius of 6 ft across the top and a height of 12 ft. At what rate is the level of the water inside the reservoir increasing when the depth is 6ft?

So we want dh/dt, when h=6ft. dv/dt is pi

For height/Radius I used 6/12=r/h with r=1/2h and volume=1/3(pi)(r^2)h

Plugged in 1/2h for r. v=1/3pi(h^2)/2)h dv/dt=pi=pi/12(3h^2)dh/dt

This is where i get stuck assuming I'm correct up to this point. I was given the answer of increasing rate of 1/9 ft/sec at 6ft. Not sure how to get it there.

Reference no: EM13842568

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