Rate of Change of a Cylindrical Volume?

Question

Suppose a cylindrical tank, whose base is a circle of radius 1.3 feet, is filling with water at the rate of 0.30 cubic feet per minute. How fast is the water level rising?

The correct answer I get is: .0565

V= PI (r^2) (height)
0.30/(PI(1.3)^2) = .0565
but, I don't understand how I reached that conclusion.
If I take the derivative of V, "h" (height) is cancelled out.

Answer 1

heres how it makes sense:

you take derivative of both sides with respect to time:

you get dV/dt = Pi (r^2) x dh/dt

Notice, Pi and the radius are a constant, only height is changing. So they are not affected by the derivative.

From there on, you know what you did. just plug those numbers in and you;ll have your answer for change in height per minute, that is feet/minute. and you have already solved that

Answer 2

What you are getting confused about is that you are taking the derivitive with respect to time not height.
You have the right fromula to start with:

V= PI (r^2) h

Take the derivitive with respect to time to get
dV/dt = Pi (r^2) dh/dt

dv/dt was given in the problem to be 0.3 feet^3/minute.
Substitute this on the left hand side. Solve for dh/dt. Since dh/dt represents how fast the height changes with time, this will be the answer you are looking for. The net result of doing it this way is to get the same thing you did to get 0.0565.

Answer 3

You don't need calculus to solve this problem, just some simple geometry and algebra!

The water in the cylindrical tank increases by .3 cubic feet per minute. If a cylinder of water has a radius of 1.3 feet and its volume is .3 cubic feet, then what is its height?

v = (pi)(r^2)(h) therefore h = v / ((pi)(r^2)); here r^2 = 1.69 square feet, so
h = (.3 cu.ft.) / ((3.14)(1.69 sq.ft)) = (.3 / 5.31) ft. = .0565 feet.

Thus the water in the cylinder rises .0565 feet per minute.

Answer 4

You have a volume rate of change with time

Let V=PI.r^2.l

dV/dl = PI.r^2 which is a constant

Now dV/dt the rate of change of volume with time can be written

dV/dl x dl/dt = PI.r^2 x dl/dt

You require dl/dt which from the last equation = dV/dt divided by the constant PI.r^2 (which is exactly the anwser you wrote with definitive units).

Answer 5

V = πr²h

dV/dt = πr²(dh/dt)
(dV/dt)/πr² = dh/dt

dh/dt = (dV/dt)/πr²

According to the information you provided,

dV/dt = 0.3 ft³/min
r = 1.3 ft

so plugging in those numbers we get

dh/dt = (0.3 ft³/min)/(π(1.3ft)²
dh/dt = 0.0565 ft/min