At what rate is the volume of the pyramid changing (vector calculus)

The base of a square based pyramid is increasingat a rate of 1.5 meters/sec while its height is decreasing at a rate of 2.9meters/sec. At what rate is the volumeof the pyramid changing when the base is 5 meters and the height is 14meters.? The formula for the volume ofthe pyramid is
V=(b^2h)/3

thanks :)

Quote:

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Originally Posted by math254

The base of a square based pyramid is increasingat a rate of 1.5 meters/sec while its height is decreasing at a rate of 2.9meters/sec. At what rate is the volumeof the pyramid changing when the base is 5 meters and the height is 14meters.? The formula for the volume ofthe pyramid is
V=(b^2h)/3

thanks :)

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Am I correct in assuming that it's the SIDE LENGTH of the base of the pyramid which is increasing at 1.5 m/s?

Work out dV/dt (you will use implicit differentiation and the product rule). Then feed in db/dt=1.5, dh/dt = -2.9, b=5 and h=14.