Posted by **EA** on Friday, March 28, 2014 at 11:39pm.

Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h is the depth of the water )

- Calculus -
**Damon**, Friday, March 28, 2014 at 11:48pm
the rate of change of volume is the surface area times the rate of change of height

450 ft^3/min = surface area * dh/dt

surface area = pi r^2

450 = pi (30^2) dh/dt

dh/dt = .159 ft/min

You could do this by saying

V = pi r^2 h

dV/dh = pi r^2

dV/dh*dh/dt = pi r^2 dh/dt

chain rule

dV/dt = pi r^2 dh/dt

but most of us just look at the lake rising an inch and seeing that the added volume is the area times one inch

- Calculus -
**EA**, Sunday, March 30, 2014 at 10:17pm
Thanks Damon, that really clears it up for me

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