*2.8 Related rate from one cone to another cone

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In summary: The surface area of the solution is increasing at a rate of $.014\cdot cm^3\text{ sec}$ when the volume of the upper cone is equal to the volume of the lower cone.
  • #1
karush
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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.:cool:
 
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  • #2
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).
 
  • #3
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
 
  • #4
MarkFL said:
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:confused:
 
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  • #5
karush said:
I got \(\displaystyle h_L\approx 11.48\text{ cm}/s\) which seem to large:confused:

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)?
 
  • #6
MarkFL said:
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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  • #7
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.
 
  • #8
So just the area of a circle of surface of the solution at h ...
 
  • #9
karush said:
So just the area of a circle of surface of the solution at h ...

Yes, that's right. :D

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.
 
  • #10
MarkFL said:
\(\displaystyle h_L=h_T=\left(\frac{1}{2} \right)^{\frac{1}{3}}12=6\sqrt[3]{4}\approx9.52\text{ cm}\)

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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  • #11
karush said:
$S=\pi r^2$ or $\displaystyle\pi\left(\frac{h}{3}\right)^2$

then

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

Correct.
karush said:
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.
 
  • #12
$A_{circle} = \pi r^2$ 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...
 
  • #13
karush said:
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.:cool:

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:

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

This will answer part c). :D
 
  • #14
thanks for help... just have to move on to another topic
learned much on this one tho.
 

FAQ: *2.8 Related rate from one cone to another cone

What is a related rate in the context of cones?

In mathematics, a related rate is a type of problem that involves finding the rate of change of one variable with respect to another variable. In the context of cones, related rate problems typically involve finding the rate of change of the volume, surface area, or other properties of one cone while another cone is changing.

How do you set up a related rate problem with cones?

To set up a related rate problem with cones, you first identify the variables involved and the relationships between them. Then, you use the chain rule to find the derivative of the variable you are interested in with respect to time. Finally, you plug in the given values and solve for the unknown rate of change.

Can you give an example of a related rate problem with cones?

Sure, an example of a related rate problem with cones would be: "Two cones with radii of 6 inches and 9 inches, respectively, are placed base to base. If the height of the smaller cone is increasing at a rate of 2 inches per minute, at what rate is the height of the larger cone changing when the height of the smaller cone is 4 inches?"

What is the importance of related rates in the study of cones?

Related rates are important in the study of cones because they allow us to analyze how the properties of one cone changes in relation to another cone. This can help us understand real-world scenarios, such as how the melting rate of an ice cream cone affects its volume and surface area.

Are there any tricks or tips for solving related rate problems with cones?

One helpful tip for solving related rate problems with cones is to draw a diagram and label all the given and unknown variables. This can help you visualize the problem and identify the relationships between the variables. Another tip is to always use the chain rule and carefully plug in the given values to avoid making errors in calculations.

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