# [SOLVED]Related rate from one cone to another cone

#### karush

##### Well-known member
A solution is draining through a conical filter into an identical conical container
(both are $h=12$ and $r=4$ at top of cone)
The solution drips from the upper filter into the lower container at a rate of
$\displaystyle\frac{\pi\ cm^3}{\text{ sec}}$ and $\displaystyle ⁡V_{cone}=\frac{\pi}{3}\cdot r^2\cdot h$

since $\displaystyle r=\frac{h}{3}$ then $\displaystyle ⁡V_{cone}=\frac{\pi}{27}\cdot h^3$

a. How fast is the level in the upper filter dropping when the solution level in the upper filter is at $6 cm$?

When $h=6$ cm then $\displaystyle\frac{dV}{dt}=\frac{\pi}{3} 6^3 \frac{dh}{dt}=72\pi \frac{dh}{dt}$

So $\displaystyle\frac{dh}{dt}=\frac{1}{72\pi}\cdot \frac{\pi\ cm^3}{sec}=.014 \frac{cm^3}{sec}$

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 a 6cm
and now fast is the level in the lower filter rising?

c. How fast is the surface area of the solution in the lower filter increasing
when the volume in the upper filter equals the volume in the lower container?

not sure that a. is correct, but if it is, need some hints for b. and c.

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#### MarkFL

Staff member
For part a) I would state:

$$\displaystyle V=\frac{\pi}{3}hr^2$$

By similarity, we know:

$$\displaystyle r=\frac{h}{3}$$

Hence:

$$\displaystyle V=\frac{\pi}{27}h^3$$

Differentiating with respect to time $t$, we find:

$$\displaystyle \frac{dV}{dt}=\frac{\pi}{9}h^2\frac{dh}{dt}$$

Solving for $$\displaystyle \frac{dh}{dt}$$ we obtain:

$$\displaystyle \frac{dh}{dt}=\frac{9}{\pi h^2}\frac{dV}{dt}$$

Now, we are given:

$$\displaystyle \frac{dV}{dt}=-\pi\frac{\text{cm}^3}{\text{s}},\,h=6\text{ cm}$$

And so:

$$\displaystyle \left. \frac{dh}{dt} \right|_{h=6\text{ cm}}=\frac{9}{\pi \left(6\text{ cm} \right)^2}\left(-\pi\frac{\text{cm}^3}{\text{s}} \right)=-\frac{1}{4}\frac{\text{cm}}{\text{s}}$$

Do you see what you did wrong?

For part b) I would let $$\displaystyle h_L$$ be the depth of solution in the lower container, and use the fact the the amount of solution in both containers is equal to one full container:

$$\displaystyle \frac{\pi}{27}h_L^3+\frac{\pi}{27}6^3=\frac{\pi}{27}12^3$$

Now solve for $h_L$. Then to find $$\displaystyle \frac{dh_L}{dt}$$, use the same method as in part a).

Once you get parts a) and b) understood and worked, we can move on to part c).

#### karush

##### Well-known member
Yes I see where I went astray. Thnx for the insight I am on cell pH now so wiil do more on this tomorrow I don't see thanks button on the mobile version but will hit it tommorro

#### karush

##### Well-known member
For part b) I would let $$\displaystyle h_L$$ be the depth of solution in the lower container, and use the fact the the amount of solution in both containers is equal to one full container:

$$\displaystyle \frac{\pi}{27}h_L^3+\frac{\pi}{27}6^3=\frac{\pi}{27}12^3$$

Now solve for $h_L$.
I got $$\displaystyle h_L\approx 11.48\text{ cm}/s$$ which seems to large

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#### MarkFL

Staff member
I got $$\displaystyle h_L\approx 11.48\text{ cm}/s$$ which seem to large
I get:

$$\displaystyle \frac{\pi}{27}h_L^3+\frac{\pi}{27}6^3=\frac{\pi}{27}12^3$$

$$\displaystyle h_L^3+6^3=12^3$$

$$\displaystyle h_L=\sqrt[]{12^3-6^3}=\sqrt[3]{1512}=6\sqrt[3]{7}\approx11.48$$

This would just be cm, not cm/s, as it is a linear measure, not a rate. This may seem counter-intuitive, but the top cone has only $$\displaystyle \left(\frac{1}{2} \right)^3=\frac{1}{8}$$ of the total volume, and so the bottom cone would have the remaining $$\displaystyle \frac{7}{8}$$, hence:

$$\displaystyle h_L=12\sqrt[3]{\frac{7}{8}}=6\sqrt[3]{7}\approx11.48$$

Now, can you proceed with the remainder of part b)?

#### karush

##### Well-known member
can you proceed with the remainder of part b)?
I will try a part of this at a time....

first it asks for what happens when the volumns are equal which doesn't mean h/2.

so a full cone is $\displaystyle V_{full} = \frac{\pi}{27}\cdot 12^{\ 3} = 64\pi cm^{\ 3}$

then $V/2$ is $\displaystyle 32\pi cm^{\ 3}= \frac{\pi}{27}\cdot h^{\ 3}$
so $h_{V/2}\approx 9.52\$ cm when $V_{\ T}= V_{\ L}$

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#### MarkFL

Staff member
I think you have jumped ahead to part c), and in part c) I am fairly certain they are referring to the surface area of the solution, i.e., that part which is exposed to the air.

#### karush

##### Well-known member
So just the area of a circle of surface of the solution at h ...

#### MarkFL

Staff member
So just the area of a circle of surface of the solution at h ...
Yes, that's right.

You have correctly found that when the two cones have equal volume the depth of the solution in each is:

$$\displaystyle h_L=h_T=\left(\frac{1}{2} \right)^{\frac{1}{3}}12=6\sqrt[3]{4}\approx9.52\text{ cm}$$

I would use the exact value. So, state the surface area as a function of $h$, then differentiate with respect to time $t$. You will then need to find the time rate of change of the depth to complete the question.

#### karush

##### Well-known member
$$\displaystyle h_L=h_T=\left(\frac{1}{2} \right)^{\frac{1}{3}}12=6\sqrt[3]{4}\approx9.52\text{ cm}$$

I would use the exact value. So, state the surface area as a function of $h$, then differentiate with respect to time $t$. You will then need to find the time rate of change of the depth to complete the question.
$S=\pi r^2$ or $\displaystyle\pi\left(\frac{h}{3}\right)^2$

then

$\displaystyle\frac{dS}{dt}=\frac{2\pi h}{9} \frac{dh}{dt}$

is $\displaystyle\frac{dh}{dt}$ still $\displaystyle -\pi\frac{cm^2}{s}$ ?

or use
 $\displaystyle \left. \frac{dh}{dt} \right|_{h=6\sqrt[3]{4}\text{ cm}}$ to get the new $\displaystyle\frac{dh}{dt}$

then we just plug in $h=6\sqrt[3]{4}$

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#### MarkFL

Staff member
$S=\pi r^2$ or $\displaystyle\pi\left(\frac{h}{3}\right)^2$

then

$\displaystyle\frac{dS}{dt}=\frac{2\pi h}{9} \frac{dh}{dt}$
Correct.

is $\displaystyle\frac{dh}{dt}$ still $\displaystyle -\pi\frac{cm^2}{s}$ ?

or use
 $\displaystyle \left. \frac{dh}{dt} \right|_{h=6\sqrt[3]{4}\text{ cm}}$ to get the new $\displaystyle\frac{dh}{dt}$

then we just plug in $h=6\sqrt[3]{4}$
You want to compute $$\displaystyle \frac{dh}{dt}$$ since it varies with time.

#### karush

##### Well-known member
$A_{circle} = \pi r^2$ or $\pi\left(\frac{h}{3}\right)^2$
$\frac{dA}{dt}=\frac{2h\pi}{9}$
$\displaystyle \left. \frac{dh}{dt} \right|_{h=6\sqrt[3]{4}\text{ cm}} =\frac{9}{\pi \left(6\sqrt[3]{4}\text{ cm} \right)^2} \left(-\pi\frac{\text{cm}^3}{\text{s}} \right) =-.5953 \frac{\text{cm}}{\text{s}}$
then
$\left. \frac{dA}{dt}\right|_{h=(6\sqrt[3]{4})} =\frac{2(6\sqrt[3]{4})\pi\text{ cm}}{9} \cdot \frac{ -.5953\text { cm}}{s} =\frac{11906\pi}{9}\frac{ \text { cm}^2}{s}$

another stab in the dark....

#### MarkFL

Staff member
A solution is draining through a conical filter into an identical conical container
(both are $h=12$ and $r=4$ at top of cone)
The solution drips from the upper filter into the lower container at a rate of
$\displaystyle\frac{\pi\ cm^3}{\text{ sec}}$ and $\displaystyle ⁡V_{cone}=\frac{\pi}{3}\cdot r^2\cdot h$

since $\displaystyle r=\frac{h}{3}$ then $\displaystyle ⁡V_{cone}=\frac{\pi}{27}\cdot h^3$

a. How fast is the level in the upper filter dropping when the solution level in the upper filter is at $6 cm$?

When $h=6$ cm then $\displaystyle\frac{dV}{dt}=\frac{\pi}{3} 6^3 \frac{dh}{dt}=72\pi \frac{dh}{dt}$

So $\displaystyle\frac{dh}{dt}=\frac{1}{72\pi}\cdot \frac{\pi\ cm^3}{sec}=.014 \frac{cm^3}{sec}$

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 a 6cm
and now fast is the level in the lower filter rising?

c. How fast is the surface area of the solution in the lower filter increasing
when the volume in the upper filter equals the volume in the lower container?

not sure that a. is correct, but if it is, need some hints for b. and c.
As a student, this is how I would have worked this problem. First, we have two identical conical containers. Let $R$ and $H$ represent the radii and heights respectively of the containers themselves and $r$ and $h$ represent the radius of the surface and depth of the solution respectively at time $t$.

Let $a$ be the ratio of their radii to their heights:

$$\displaystyle a=\frac{R}{H}\implies r=ah$$

Let $0<k$ be the rate at which the solution drains from the upper container into the lower container. Let all variables concerning the upper container have $T$ as subscripts and all variables concerning the lower container have $L$ as subscripts.

Since we are asked questions that involve the time rate of change of the levels of the solution in each container, we should express the volume $V$ as a function of the depth $h$:

$$\displaystyle V=\frac{\pi}{3}r^2h=\frac{\pi a^2}{3}h^3$$

Differentiating with respect to time $t$, we obtain:

$$\displaystyle \frac{dV}{dt}=\pi a^2h^2\frac{dh}{dt}$$

Hence:

$$\displaystyle \frac{dh}{dt}=\frac{1}{\pi (ah)^2}\frac{dV}{dt}$$

This will answer part a). Next, let's develop a formula relating the depths of the two containers. Let $A$ be the total amount of solution present, i.e.:

$$\displaystyle V_T+V_L=A$$

$$\displaystyle h_T^3+h_L^3=\frac{3A}{\pi}$$

Differentiating with respect to time $t$, we find after simplification:

$$\displaystyle h_T^2\frac{dh_T}{dt}+h_L^2\frac{dh_L}{dt}=0$$

From this, we may conclude:

$$\displaystyle \frac{dh_L}{dt}=-\left(\frac{h_T}{h_L} \right)^2\frac{dh_T}{dt}$$

Now, from the relation between the two depths, we obtain:

$$\displaystyle h_L=\sqrt[3]{\frac{3A}{\pi}-h_T^3}$$

Hence:

$$\displaystyle \frac{dh_L}{dt}=-\left(\frac{h_T}{\sqrt[3]{\dfrac{3A}{\pi}-h_T^3}} \right)^2\frac{dh_T}{dt}$$

And using the result $$\displaystyle \frac{dh}{dt}=\frac{1}{\pi (ah)^2}\frac{dV}{dt}$$, we may write:

$$\displaystyle \frac{dh_L}{dt}=-\left(\frac{h_T}{\sqrt[3]{\dfrac{3A}{\pi}-h_T^3}} \right)^2\frac{1}{\pi \left(ah_T \right)^2}\frac{dV_T}{dt}$$

And this can be written as:

$$\displaystyle \frac{dh_L}{dt}=\frac{k}{\pi}\left(\frac{1}{a\sqrt[3]{\dfrac{3A}{\pi}-h_T^3}} \right)^2$$

Now you have the formulas to answer part b). For part c), let $S$ be the surface area. we know:

$$\displaystyle V=\frac{Sh}{3}$$

And $$\displaystyle h=\frac{1}{a}\sqrt{\frac{S}{\pi}}$$

Hence:

$$\displaystyle V(S)=\frac{1}{3a\sqrt{\pi}}S^{\frac{3}{2}}$$

And so:

$$\displaystyle \frac{dV}{dt}=\frac{1}{2a\sqrt{\pi}}\sqrt{S}\frac{dS}{dt}$$

Hence:

$$\displaystyle \frac{dS}{dt}=2ak\sqrt{\frac{\pi}{S}}$$

Now, for part c) we are told:

$$\displaystyle V_T=V_L=\frac{A}{2}$$

Hence:

$$\displaystyle \frac{A}{2}=\frac{1}{3a\sqrt{\pi}}S^{\frac{3}{2}}$$

$$\displaystyle S=\left(\frac{3aA\sqrt{\pi}}{2} \right)^{\frac{2}{3}}$$

And thus we may write:

$$\displaystyle \frac{dS}{dt}=2ak\sqrt{\frac{\pi}{\left(\dfrac{3aA \sqrt{\pi}}{2} \right)^{ \frac{2}{3}}}}=2\sqrt[3]{\frac{2\sqrt{\pi}a^2}{3A}}$$