A puddle in the shape of a right circular cylinder is evaporating at a rate of 5 cubic inches and the height of the puddle is .4 inches, the radius is decreasing at the rate of .5 inches per hour. At the instant, what is the rate of change of the height of the puddle with respect to time.

Question

A puddle in the shape of a right circular cylinder is evaporating at a rate of 5 cubic inches and the height of the puddle is .4 inches, the radius is decreasing at the rate of .5 inches per hour. At the instant, what is the  rate of change of the height of the puddle with respect to time.

isaiah.feynman · Answer

Calculus huh?

anonymous · Answer

yep

isaiah.feynman · Answer

The volume of the cylinder relates the height and radius. So differentiate the formula implicitly to get and solve for the missing one.

anonymous · Answer

The thing is that I don't know what formula to use?

isaiah.feynman · Answer

|dw:1384362225275:dw| That's the formula for volume of a cylinder, as you see it relates the radius and height! Now can you differentiate it implicitly?

anonymous · Answer

Thank you!

isaiah.feynman · Answer

No problem

isaiah.feynman · Answer

No radius given.

anonymous · Answer

A puddle in the shape of a right circular cylinder is evaporating at a rate of 5 cubic inches per hour. At the instant when the radius is 40 inches and the height of the puddle is .4 inches, the he radius is decreasing at the rate of .5 inches per hour. At the instant, what is the  rate of change of the height of the puddle with respect to time. 

sorry i wrote it wrong

isaiah.feynman · Answer

Okay, lol. Cos I tried solving and I noticed there was no radius.

anonymous · Answer

$$V=Pi r ^{2}h$$
$$d/dt[\pi r^2 \times h] $$
$$dv/dt=2\pi r (dr/dt) (h)+ \pi r^2 dh/dt$$
Do you just plug in the numbers afterwards

isaiah.feynman · Answer

|dw:1384363416713:dw| Yep, you plug in numbers. I prefer writing mine in prime notation. :D