Rate of water through a conical cone, in order to find constant k

[i_m_mimi]

Homework Statement

[10marks] A water tank has the shape of a vertex-down right circular cone. The depth
of the tank is 9 meters, and the top of the tank has radius 6 meters. Water
flows into the tank from a hose at a constant rate of 14 cubic metres per
hour, and leaks out of a hole at the bottom of the tank at a rate of kh cubic
metres per hour when the depth of water in the tank is h metres. Here k is a
constant. When the water is 3 metres deep in the tank, its surface is rising at
the instantaneous rate of 2 metres per hour. Find the value of the constant k.

Answer: k = 2pi

Homework Equations

V = 1/3 pi r^2 h

The Attempt at a Solution

made so many attempts on paper, including

r/h = 6/9

V = 1/3pi (6h/9)^2h = 4/27pi h^3

rate of change in volume = rate which water is poured in minus rate water leaks out

V' = 14pi m^3/hour - X

X = rate water leaks out

V' 12/27 (h^2)(h')
h' = 3
V' = 2/3 pi r r' h + h' pi r^2
V' = 2/3 pi (2)(2)h + (3) pi (2)^2
V' = 8/3 pi h + 4pi

X = 14pi - 8/3 pi h + 4pi
X = 10pi - 8/3 pi h

I'm not sure how to change the equation into rate water leaks out = kh form,
i keep on getting solutions where h is either not present or h^2
I think I've made mistakes somewhere in there and it is a complicated question worth 10 marks and with 20 minutes allocated to it on an exam.

thank you

 

[HallsofIvy]

"X= rate at which water leaks out"= kh, since you are told that the water leaks out at "at a rate of kh cubic
metres per hour". So the net "water in" is the rate at which it is coming in, 14 cubic meters per hour, minus the rate at which it is leaking out, kh. The net water in is 14- kh.

The rest you have done correctly. Yes, the ratio of radius of the top of the water to its height is that of the whole conical tank, r/h= 6/9= 2/3 so r= (2/3)h. The volume of a cone with height h and radius r is $(1/3)\pi r^2h$ which, with r= (2/3)h, is $V= (4/27)\pi h^3$.

Differentiating both sides with respect to time, t, $dV/dt= (4/9)\pi h^2 dh/dt$.

You kind of lose me with the rest since you have stopped writing full equations.
Putting together the fact that the volume is changing at $dV/dt= (4/9)\pi h^2 dh/dt$
with the fact that water is coming in at a net 14- kh, we have
$$\left(\frac{4}{9}\right)\pi h^2 \frac{dh}{dt}= 14- kh$$
which we can write as
$$\frac{h^2}{14- kh}= \frac{9 dt}{4\pi}$$

To integrate on the left, use the substitution u= 14- kh so that du= -k dh or dh= -du/k.
h= (u- 14)/k so that $h^2= (u^2- 28u+ 196)/k^2$ and the integral becomes
$-\frac{1}{k^3}\int \frac{u^2- 28u+ 196}{u} du$
$= -\frac{1}{k^3}\int \left(u- 28+ \frac{196}{u}\right)du$

 

[milesyoung]

Something seems off about the problem statement. You'd expect kh to be a positive number, but since:
dV/dt = 4/9*pi*h²*dh/dt = 4/9*pi*3²*2 m³/h ≈ 25.1 m³/h

that can't be the case, i.e. the solution k = 2*pi is wonky aswell.

 

[i_m_mimi]

"X= rate at which water leaks out"= kh, since you are told that the water leaks out at "at a rate of kh cubic
metres per hour". So the net "water in" is the rate at which it is coming in, 14 cubic meters per hour, minus the rate at which it is leaking out, kh. The net water in is 14- kh.

The rest you have done correctly. Yes, the ratio of radius of the top of the water to its height is that of the whole conical tank, r/h= 6/9= 2/3 so r= (2/3)h. The volume of a cone with height h and radius r is $(1/3)\pi r^2h$ which, with r= (2/3)h, is $V= (4/27)\pi h^3$.

Differentiating both sides with respect to time, t, $dV/dt= (4/9)\pi h^2 dh/dt$.

You kind of lose me with the rest since you have stopped writing full equations.
Putting together the fact that the volume is changing at $dV/dt= (4/9)\pi h^2 dh/dt$
with the fact that water is coming in at a net 14- kh, we have
$$\left(\frac{4}{9}\right)\pi h^2 \frac{dh}{dt}= 14- kh$$
which we can write as
$$\frac{h^2}{14- kh}= \frac{9 dt}{4\pi}$$

To integrate on the left, use the substitution u= 14- kh so that du= -k dh or dh= -du/k.
h= (u- 14)/k so that $h^2= (u^2- 28u+ 196)/k^2$ and the integral becomes
$-\frac{1}{k^3}\int \frac{u^2- 28u+ 196}{u} du$
$= -\frac{1}{k^3}\int \left(u- 28+ \frac{196}{u}\right)du$

I made a mistake in copying the question. Water from the hose pours in at a rate of 14pi cubic meters/hour instead of 14, if that makes things simpler. However, I'm still confused.
I solved to the point of getting the integral you got following what you did.
I got something like 98-14kh + (k^2)(h^2) - 392 + 28kh + 28kh + 196ln(14-kh) = 0

Not sure where to go from here.

 

[milesyoung]

If you got this far:
$$\left(\frac{4}{9}\right)\pi h^2 \frac{dh}{dt}= 14 \pi- kh$$
Then just solve for k and plug in the information given in the problem statement.

 

[i_m_mimi]

If you got this far:
$$\left(\frac{4}{9}\right)\pi h^2 \frac{dh}{dt}= 14 \pi- kh$$
Then just solve for k and plug in the information given in the problem statement.

I get something like

if h' = 3 since r' = 2

12/9 pi h + 14pi/h = k

Not sure how to isolate for so its 2pi h?

 

[milesyoung]

Have a look at the problem statement again, you're looking for information about h and dh/dt at some point in time.

Keep in mind that dh/dt means the instantaneous rate of change of h with respect to time.

 

[i_m_mimi]

Have a look at the problem statement again, you're looking for information about h and dh/dt at some point in time.

Keep in mind that dh/dt means the instantaneous rate of change of h with respect to time.

omg that's great i get it now :D

When the water is 3 metres deep in the tank, its surface is rising at
the instantaneous rate of 2 metres per hour.

That means h' = 2, not r' = 2

4/9 pi (3)^2 (2) = 14pi - 3k

k = 2pi

thanks so much

 

[milesyoung]

Good job