# Find rate at which the liquid level is rising in the problem

• chwala
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Rate of change

I was able to solve it using,

##\dfrac{dV}{dt} = \dfrac{dV}{dh}⋅\dfrac{dh}{dt}##

With, ##r = \dfrac{h\sqrt{3}}{3}##, we shall have

##\dfrac{dV}{dh} = \dfrac{πh^2}{3}##

Then,

##\dfrac{dh}{dt}= \dfrac{2×3 ×10^{-5}}{π×0.05^2}= 0.00764##m/s

My question is can one use the ##\dfrac{dV}{dt} = \dfrac{dV}{dr}⋅\dfrac{dr}{dt}## approach?

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That would tell you how fast the radius of the surface changes and would mean you would have to multiply with a constant in the end. Technically you can do that but already using h to describe the volume rather than r is the more direct path.

chwala
Orodruin said:
That would tell you how fast the radius of the surface changes and would mean you would have to multiply with a constant in the end. Technically you can do that but already using h to describe the volume rather than r is the more direct path.
Using,

##\dfrac{dV}{dt} = \dfrac{π}{3}\left[2rh \dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right]## and

given that

##r= \dfrac{\sqrt{3}}{60}##

and ##\dfrac{dV}{dr}= 0.003##

i then have,

##\dfrac{dr}{dt} = \dfrac{2×10^{-5}}{0.003}= 0.0066##

Now using:

##\dfrac{dV}{dt} = \dfrac{π}{3}\left[2rh \dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right]##

and substituting gives me,

##6 × 10^{-5}= 0.00006+\dfrac{π}{3600}⋅\dfrac{dh}{dt}##

##\dfrac{dh}{dt}=0 ##

which seems to be correct

The problem statement is somewhat unclear (i.e., "A conical vessel has a vertical angle of 60°.") My first take on this was that the central axis of the cone wasn't vertical. I later interpreted this to refer to the angle that the lateral side makes with respect to the axis of the cone.

Does the text include a drawing of the cone? The author's description in words leaves a lot to be desired.
chwala said:
Using,
##\dfrac{dV}{dt} = \dfrac{π}{3}\left[2rh \dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right]## and
Whatever the cone's actual shape is, there is no need to use partial derivatives (which you have done above but do not show). Instead, use the given information about the 60° to write the radius in terms of the height so that the volume can be written as a function of height alone.
chwala said:
given that

##r= \dfrac{\sqrt{3}}{60}##
Where did this come from?
chwala said:
and ##\dfrac{dV}{dr}= 0.003##

i then have,

##\dfrac{dr}{dt} = \dfrac{2×10^{-5}}{0.003}= 0.0066##

Now using:

##\dfrac{dV}{dt} = \dfrac{π}{3}\left[2rh \dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right]##

and substituting gives me,

##6 × 10^{-5}= 0.00006+\dfrac{π}{3600}⋅\dfrac{dh}{dt}##

##\dfrac{dh}{dt}=0 ##

which seems to be correct
Possibly you're being ironic in your conclusion. Otherwise, it doesn't make sense that the height of liquid in the cone doesn't change as more liquid is poured into it.

chwala said:
My question is are we able to use the ##\dfrac{dV}{dt} = \dfrac{dV}{dr}⋅\dfrac{dr}{dt}## approach?
No, for two reasons.
1) Your goal is to find ##\frac{dh}{dt}## not ##\frac{dr}{dt}##.
2) You're being fast and loose with your derivatives. Since V is a function of both r and h, the derivatives of V with respect to r and V with respect to h are both partial derivatives. This is almost certainly not the approach that the author intended you to use. In my previous post I explained that you could (and should) write V in terms of h alone.

Mark44 said:
The problem statement is somewhat unclear (i.e., "A conical vessel has a vertical angle of 60°.") My first take on this was that the central axis of the cone wasn't vertical. I later interpreted this to refer to the angle that the lateral side makes with respect to the axis of the cone.

Does the text include a drawing of the cone? The author's description in words leaves a lot to be desired.
Whatever the cone's actual shape is, there is no need to use partial derivatives (which you have done above but do not show). Instead, use the given information about the 60° to write the radius in terms of the height so that the volume can be written as a function of height alone.
Where did this come from?

Possibly you're being ironic in your conclusion. Otherwise, it doesn't make sense that the height of liquid in the cone doesn't change as more liquid is poured into it.
@Mark44 the text does not include a drawing. The problem is posted as it is directly from the textbook. Yes, I am quite conversant with the short route to destination. My interest was to alternatively use the long route. Thks.

Mark44 said:
The problem statement is somewhat unclear (i.e., "A conical vessel has a vertical angle of 60°.") My first take on this was that the central axis of the cone wasn't vertical. I later interpreted this to refer to the angle that the lateral side makes with respect to the axis of the cone.

Does the text include a drawing of the cone? The author's description in words leaves a lot to be desired.
Whatever the cone's actual shape is, there is no need to use partial derivatives (which you have done above but do not show). Instead, use the given information about the 60° to write the radius in terms of the height so that the volume can be written as a function of height alone.
Where did this come from?

Possibly you're being ironic in your conclusion. Otherwise, it doesn't make sense that the height of liquid in the cone doesn't change as more liquid is poured into it.
##r= \dfrac{\sqrt{3}}{60}## is from ##r = \dfrac{h\sqrt{3}}{3}## as posted in post 1.

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chwala
chwala said:
##r= \dfrac{\sqrt{3}}{60}## is from ##r = \dfrac{h\sqrt{3}}{3}## as posted in post 1.
You have ##r = \dfrac{h\sqrt{3}}{3}## in post #1 and ##r= \dfrac{\sqrt{3}}{60}## in post #2, but there's no explanation of how the first one is transformed to the second. Other members have posted in other threads that you sometimes omit steps, thus making it difficult to follow what you're trying to do.

SammyS
chwala said:
Homework Statement: See attached.
Relevant Equations: Rate of change

View attachment 334357

I was able to solve it using,

##\dfrac{dV}{dt} = \dfrac{dV}{dh}⋅\dfrac{dh}{dt}##

With, ##r = \dfrac{h\sqrt{3}}{3}##, we shall have

##\dfrac{dV}{dh} = \dfrac{πh^2}{3}##

Then,

##\dfrac{dh}{dt}= \dfrac{2×3 ×10^{-5}}{π×0.05^2}= 0.00764##m/s

My question is can one use the ##\dfrac{dV}{dt} = \dfrac{dV}{dr}⋅\dfrac{dr}{dt}## approach?
Just a question on this related rates. As the liquid is flowing into the cone...the radius is also increasing...is the rate in reference to increase in depth constant? I am thinking that since the radius ( in any case the diameter of the cone) is increasing then we should see a drop in the rising of the liquid depth...unless i am not getting it clearly. The rate at which depth is increasing is not constant in my take.

chwala said:
Just a question on this related rates. As the liquid is flowing into the cone...the radius is also increasing...is the rate in reference to increase in depth constant? I am thinking that since the radius ( in any case the diameter of the cone) is increasing then we should see a drop in the rising of the liquid depth...unless i am not getting it clearly. The rate at which depth is increasing is not constant in my take.
Is the rate you refer to ##\frac{dh}{dt} ##?
If so, the rate of change is constant at 0.00764 m/s

chwala
WWGD said:
Is the rate you refer to ##\frac{dh}{dt} ##?
If so, the rate of change is constant at 0.00764 m/s
Agreed, no problem with that.

I do not know whether you understood my question. Imagine a big cone having some liquid being poured in...my thoughts are that initially the rate (increase) in depth will be faster as the radius is small...but after some time there should be a reduction in the rate of increase of depth as the diameter is increasing gradually...

WWGD
chwala said:
Agreed, no problem with that.

I do not know whether you understood my question. Imagine a big cone having some liquid being poured in...my thoughts are that initially the rate (increase) in depth will be faster as the radius is small...but after some time there should be a reduction in the rate of increase of depth as the diameter is increasing gradually...
$$V(h) = \frac{\pi }{9} h^3$$

$$\frac{dV}{dh} \frac{dh}{dt} = \frac{dV}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}$$

If ## \frac{dV}{dt} = \text{const.} = Q ## then;

$$\frac{dh}{dt} = \frac{3Q}{\pi h^2}$$

So initially ##h= 0##, and ##\frac{dh}{dt} \to \infty## and as ## h \to \infty, \frac{dh}{dt} \to 0 ##. Is that what you are asking?

chwala
erobz said:
$$V(h) = \frac{\pi }{9} h^3$$

$$\frac{dV}{dh} \frac{dh}{dt} = \frac{dV}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}$$

If ## \frac{dV}{dt} = \text{const.} = Q ## then;

$$\frac{dh}{dt} = \frac{3Q}{\pi h^2}$$

So initially ##h= 0##, and ##\frac{dh}{dt} \to \infty## and as ## h \to \infty, \frac{dh}{dt} \to 0 ##. Is that what you are asking?
Is the rate of change ##\dfrac{dh}{dt}## linear?

chwala said:
Is the rate of change ##\dfrac{dh}{dt}## linear?
It's certainly not linear in ##h##. When ##h## is small the rate of change is very large, and when ##h## is large the rate of change is small.

Lnewqban and chwala
erobz said:
It's certainly not linear in ##h##. when ##h## is small the rate of change is very large, and when ##h## is large the rate of change is small.
Then i think that answers my question. I guess i need to read more on the literature bit.
The ##0.00764## metres/second refers to change in depth per second for specifically ##0.05## metres correct?

Lnewqban
chwala said:
Then i think that answers my question. I guess i need to read more on the literature bit.
The ##0.00764## metres/second refers to change in depth per second for specifically ##0.05## m correct?
Find ##h## as an explicit function of time from:

$$\frac{3Q}{\pi}dt = h^2 dh$$

Then differentiate with respect to time and see if its a constant.

erobz said:
Find ##h## as a pure function of time from:

$$\frac{3Q}{\pi}dt = h^2 dh$$

Then differentiate with respect to time and see if its a constant.
separation of variables...integrate you suppose...differentiating will take me back to the start point.

##\dfrac{3Qt}{π} +C = \dfrac{h^3}{3}##

##h(t)=\sqrt[3]{\dfrac{9Qt}{π} +3C}##

chwala said:
The ##0.00764## metres/second refers to change in depth per second for specifically ##0.05## metres correct?
yeah, specifically at ##0.05 ~\rm{m} ##, that would be ##\frac{dh}{dt}##.

chwala
It seems like there was some confusion about whether or not ##\dot h## was constant in time. I think you can tell it's not given ##h(t)##?

erobz said:
yeah, specifically at ##0.05 ~\rm{m} ##, that would be ##\frac{dh}{dt}##
Or written in another way, ##\left.\frac{dh}{dt}\right|_{t = 0.05}##

In most related rates problems, the time derivatives of some quantity are relative to some specific time.

Lnewqban and chwala
Mark44 said:
Or written in another way, ##\left.\frac{dh}{dt}\right|_{t = 0.05}##
I think you mean: ##\left.\frac{dh}{dt}\right|_{h = 0.05 \rm{m}}##

chwala said:
Is the rate of change ##\dfrac{dh}{dt}## linear?
Not in this case. If the container happened to be a cylinder (circular or otherwise) then there would be a linear relationship between the flow rate (dV/dt) and the change in height (dh/dt).

chwala
chwala said:
differentiating will take me back to the start point.

##\dfrac{3Qt}{π} +C = \dfrac{h^3}{3}##

##h(t)=\sqrt[3]{\dfrac{9Qt}{π} +3C}##

Differentiating w.r.t. ##t## does not take you back to the start point. You started with ##\dot h (h)##, and you end with ##\dot h(t)##.

chwala
Let's go back to the question at the end of the OP.

chwala said:
Homework Statement: See attached.
Relevant Equations: Rate of change

I was able to solve it using ##\quad \dfrac{dV}{dt} = \dfrac{dV}{dh}⋅\dfrac{dh}{dt}##

With, ##\ \ r = \dfrac{h\sqrt{3}}{3}##, we shall have ##\quad \dfrac{dV}{dh} = \dfrac{πh^2}{3}##

Then, ##\dfrac{dh}{dt}= \dfrac{2×3 ×10^{-5}}{π×0.05^2}= 0.00764##m/s

My question is: Can one use the ##\dfrac{dV}{dt} = \dfrac{dV}{dr}⋅\dfrac{dr}{dt}## approach?
Well, of course it can be solved by finding ##dr/dt##, but as @Orodruin pointed out, you will then have to convert that to ##dh/dt## to complete the Exercise.

It seems that the place to start with this problem is to state that ## \displaystyle \quad V=
\dfrac{\pi r^2 h}{3}\ .##

You have ##r## and ##h## being related by ## \displaystyle \ r = \dfrac{h\sqrt{3}}{3}\ ##, which comes from ##\displaystyle \ h=r\tan 60^\circ ##.

As an aside:
You interpreted a vertical angle to be the angle that the lateral side of the cone makes w.r.t. the horizontal. I'll continue with that, but I agree with Mark that it makes more sense for it to be the angle that the lateral side of the cone makes w.r.t. the vertical axis of the cone, - also consistent with @Lnewqban's drawing.​

Substituting ## \displaystyle \ h =\sqrt{3}\, r\ ## into ## \displaystyle \ V= \dfrac{\pi r^2 h}{3}\ ## gives: ## \displaystyle \quad V= \dfrac{\pi r^3 }{\sqrt{3\,}}\ ##

so that ## \displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt} = \pi r^2\, \sqrt{3\,}\cdot\dfrac{dr}{dt}##

The relationship ## \displaystyle \ h =\sqrt{3}\, r\ ## also gives ## \displaystyle \ \dfrac {dh} {dt} =\sqrt{3}\, \dfrac {dr} {dt}\ ##

Do a little rearranging & substituting to get your desired result.
.

chwala

## How do I determine the rate at which the liquid level is rising in a tank?

To determine the rate at which the liquid level is rising in a tank, you need to know the inflow rate of the liquid and the cross-sectional area of the tank. The rate of rise of the liquid level (dh/dt) can be calculated using the formula: dh/dt = Q/A, where Q is the inflow rate (volume per unit time) and A is the cross-sectional area of the tank.

## What information do I need to solve a liquid level rise problem?

To solve a liquid level rise problem, you typically need the inflow rate of the liquid (often given in units like liters per minute or gallons per hour), the dimensions or shape of the tank (to determine the cross-sectional area), and any initial conditions such as the starting liquid level.

## How does the shape of the tank affect the rate at which the liquid level rises?

The shape of the tank affects the cross-sectional area, which in turn affects the rate at which the liquid level rises. For example, in a cylindrical tank with a constant radius, the cross-sectional area is constant, making the calculation straightforward. However, in a conical or irregularly shaped tank, the cross-sectional area changes with height, complicating the calculation.

## What is the role of calculus in finding the rate at which the liquid level is rising?

Calculus is often used to find the rate at which the liquid level is rising, especially when dealing with tanks of irregular shapes where the cross-sectional area changes with height. Techniques such as differentiation and integration can help determine the relationship between the volume of liquid and the height of the liquid level over time.

## Can I use software tools to find the rate at which the liquid level is rising?

Yes, there are various software tools and simulation programs that can help you find the rate at which the liquid level is rising. These tools can model the inflow of liquid, the shape of the tank, and other relevant factors to provide a detailed analysis. Common tools include MATLAB, Python with libraries like SciPy, and specialized engineering software.

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