Could somebody PLEASE help me with calc?

Question

You have a Flask. The bottom part of the flask is in the shape of a truncated cone with bottom radius of 10 cm, top radius 5 cm, height 10 cm. Sitting on top of the truncated cone is a tall cylindar whose radius is 5cm. Liquid is being poured into the flask at a rate of 2cm^3/s. Note that the volume of a cone whose bottom radius is r and whose vertical hieght is h is given by v=(1/3)π(r^2)h.

I also know- bottom iare is V=(1/3)π(r^2)h-(1/3)π((10-.5h)^2)•(20h).

I have had a lot of problems with this, and if somebody could walk me through this, it would be much appriciated.

A) When the liquid is 15cmhigh in the flask, how fast is the height of the liquid increasing? (I don't think I need to worry about the cone for this one, becasue the water is already out of the cone)

B)If the flask is initially empty, how long does it take the liquid to reach a height of 10 cm?

C How fast is the height of the liquid rising when the liquid is 5cm deep.

This problem deals with Rate of change at a point in a and c, and therefore, derivitives, unless I'm wrong here.

Answer 1

[Edit: I just read Paul's answer above, and I note that, especially for part C, you need dh/dt, the instantaneous rate of change of height. To get that, you need to differentiate a cubic equation as I do below. So it is properly a calculus question.

By the way, your "additional details" remarks are entirely correct. End edit.]

Okay, I know how to do this. Let's start by getting the radius of the flask at height h, for h<10. We have two given points for (h,r) -- (0,10) and (10,5) so the slope is m = (5-10)/10 = -1/2, and, using point-slope, r-10 = (-1/2)h, or r = 10 - h/2. Check: For h=0, r=10; and for h=10, r=5, so that's okay.

Also note from that equation that for r=0, h=20. (That's the height of the cone without truncation.)

Next we want the volume of liquid at height h, for h<10:

V = (pi/3)(100)(20) - (pi/3)(10 - h/2)^2 (20-h)
= (pi/3) {2000 - (20-h)[100 - 10h + (1/4)h^2]}
= (pi/3) {2000 - [2000 - 300h + 15h^2 - (1/4)h^3]
= (pi/3) [300h - 15h^2 + (1/4)h^3]
= pi [100h - 5h^2 + (1/12)h^3]

To check this, the cone without truncation has height 20, base radius 10, and volume (pi/3)(100)(20) = 2000 pi/3. The volume formula we just worked out is, for h=20:

V = pi [2000 - 5(400) + (1/12)(8000)] = 2000 pi/3

so our V formula is correct. The hard work is done, so now we just have to get your answers.

Let's do (B) first. The volume of liquid for h=10 is

V(10) = = pi [100(10) - 5(100) + (1/12)(1000)]
= pi (1000 - 500 + 250/3) = (pi/3) (1500 + 250) = 1750 pi/3

It fills up at 2 cm^3/s, so it takes 875 pi/3 seconds to fill it, or 15.27 minutes. (ANSWER B)

For answer C, we need dh/dt when h=5 and dV/dt=2. Use our volume formula:

V = pi [100h - 5h^2 + (1/12)h^3]
(1/pi) dV/dt = [100 - 10h + (1/4)h^2] dh/dt
dh/dt = (2/pi) / [100 - 10(5) + (1/4)(25)]
= (8/pi) / (200 + 25) = 8/(225 pi) = 0.0113 cm/sec (ANSWER C)

Finally, on answer A, you're right; you don't have to worry about much of anything. The volume of a cylinder is V = pi r^2 h, so dV/dt = pi r^2 dh/dt. For your problem, dV/dt=2 and r=5, so

dh/dt = 2/(25 pi) = 0.0255 cm/sec (ANSWER A)

That just about does it. Parts A and B were easy, but part C was sort of tricky. (That's the reason I decided to work on this one.) Interesting problem, and fun to do.

Answer 2

If by "calc" you mean calculus you are overthinking this problem. You don't need calculus to solve any of these questions. If you just mean calculations, then disregard the above.

A) Volumetric flow rate (volume/time) is equal to velocity of the liquid multiplied by cross sectional area. For example, if liquid moving at 10 cm/sec passes through an area 2 cm^2, then the flowrate is 20 cm^3/sec. We know the neck of the flask has a 5cm radius. Area is equal to (pi)(r^2). Velocity (or how fast the liquid is rising) is equal to flowrate divided by area:

V = (2 cm^3/sec)/((pi)(5 cm)^2) = 0.025 cm/sec

B) Volume = (flow rate) x (time)

or

Time = Volume / flow rate

You know the flow rate (2 cm^3/sec). You can calculate volume using the formula you gave above. (You can do the math)

C) The approach to this problem is similar to (A), but first you need to figure out the radius at a height of 5cm. The easiest way to do this is to notice that that 5 cm is halfway up the conic section. Therefore the radius is halfway between the radius at the bottom and top. Halfway between 10 and 5 is 7.5. To solve this problem use a radius of 7.5 instead of 5 and solve exactly like problem (A).

Paul