AP Calculus Question AB Please help!!!!?

Question

I need major help and description on how to do this so I dont fail one like this on my AP.

A solution is draining through a conical filter into an identical conical container as shown (cone with radius 4 and height of 12). The solution drips from the upper filter into the lower filter at a rate of Pi cm^3/sec

A) How fast is the leel in the upper filter dropping when the solution level in the upper fileter is at 6 cm?
B) If the conical filter is initially full, what is the level of the solution in the lower level when the solution level in the upper filter is at 6 cm and how fast is the level in the lower filter rising?
C) HOw fast is the surface area of the solution increasing when the volume in the upper filter ='s the volume in the lower container?

Thanks so much. I really just need to understand it .

Answer 1

The basic approach is to write down whhat you know about depth, area, volume, etc. and take derivatives as needed to get your solution.

You know the volume of a cone is V = (pi*h*r^2)/4

The surface area is a circle A = pi*r^2

The radius and height of a cone are linearly related: h = k*r where k is a constant. In your case, when h is 12, r is 4 so k=3.

You also know how fast the top cone is draining (pi cc/sec). That is the rate of volume change. So for the top cone, dv/dt = -pi (negative because it is getting smaller). The bottom cone is growing in volume dv/dt = pi.

Armed with all this, lok at your problem.

A) How fast is the lever dropping, you need to figure dh/dt for the top cone. The cone volume is V = (pi*h*r^2)/4 but you aren't concerned with radius here so combine it with h = k*r to eliminate r. You get: V = (pi*h*(h/k)^2)/4 = pi/(4k) h^3

To get dh/dt, differenitate on time: dV/dt = (pi/(4k)) (3 h^2) dh/dt
Rearranging: dh/dt = 4k/(3 pi h^2) dV/dt

You know everything on the right side so just substitute.

B) Here, by the cone volume formula, you know the total volume of liquid and how much is left in the top, so you can figure how much is in the bottom. Once you know that, you can figure the depth in the bottom. From there, you solve for the rate of height change just like part A.

C) Here, you know that half the volume is in each cone so you can figure how deep it is and what the radius is. The area is pi r^2 so the cone volume is: V = Ah/4. You need to get rid of h. You know h = k*r and A = pi*r^2. You can combine these to get h = k*sqrt(A/pi), so the volume becomes:

V = k/(4*sqrt(pi)) A^(3/2)

Now differentitate as above to get dA/dt as a function of dV/dt

Answer 2

I hate these questions, but I'll help XD.

First thing first is to develop a ratio between h and r and develop the ratio between their dervatives (with respect to time). This will help later.

So when h=12, then r=4
and so r=1/3h
when means dr/dt=1/3 dh/dt.

And we need the equation for the volume of a cone
V=1/3*pi*r^2*h

a)
When the height is 6cm, that means, through the equation we established before, than r is at 2cm.

And this question, is, in another sense, asking how much is the height decreasing. SO we find the derivate of the volume, because we know the change in volume and it just so happens that equation has height in it.

V=1/3*pi*r^2*h
dv/dt = 1/3*pi*2*r*h*dr/dt + 1/3*pi*r^2*dh/dt

Substitute for dr/dt with dh/dt as we found above and dv/dt.

-pi = 1/3*pi*2*r*1/3dh/dt+ 1/3*pi*r^2*dh/dt
(it's -pi because it's decreasing)

Now solve for dh/dt. This is just plug r with 2, h with 6, and solve for dh/dt.

b) the first part of b is just finding the intial volume and subtracting the second volume so...

V=1/3*pi*r^2*h

V(inital)=1/3*pi*4^2*12

V(second)=1/3*pi*2^2*6

and subtract those 2 volumes which is 25.13274123!

Now you find the height with the new volume equation

V= 1/3*pi*r^2*h

substitute r

25.13274123 = (1/3)*pi*(1/3h)^2 * h

Solve for h

h = 6

Second part is similar to part A, finding the dh/dt.

v=1/3*pi*r^2*h
dv/dt = 1/3*pi*2*r*dr/dt*h + 1/3*pi*r^2*dh/dt
pi = 1/3*pi*2*(2)*(1/3dhdt)*6+1/2*pi*2^2*dhdt

now solve for dh/dt.

c) Whew! One more to go!

Now we must find dr/dt and take the ratio of the two.
Now you'll be thinking "Oh no! Not that!"
But there's a shortcut :)

We already establed 1/3dhdt = dr/dt

But the dhdt in for each cone and you'll find the drdt for each cone.
And then just put it into ratio form..


And sorry in advance if I made any math mistakes. That's why I don't like to solve out equations.

Answer 3

Boy, your teacher is a real slave driver.
To start you need to know the volume of a cone, which is pi/3 x r^2 x h ( I hope). From the geometry of the cone, 3r=h, so we can get V= V(h) as V= pi/3 x h^3/9. We can find dV/dh as
pi/27 x 3h^2 = pi/9 x h^2.
We know dV/dt, which is constant. Then we can write, dV/dh x dh/dt = constant. The constant is pi cm^3/sec, and dV/dh at 6 cm is 36pi/9. So dh/dt at that location must be 9/36 cm/sec.

Part b stumps me, and I frankly don't have the time to tackle it. Apparently, while water is entering the lower cone at pi cm^3/sec, but not draining. Perhaps another answer can move you along.

Answer 4

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