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

• May 5th 2012, 09:40 PM
math254
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 :)
• May 5th 2012, 11:55 PM
Prove It
Re: At what rate is the volume of the pyramid changing (vector calculus)
Quote:

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 :)

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?
• May 6th 2012, 01:05 AM
biffboy
Re: At what rate is the volume of the pyramid changing (vector calculus)
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.