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

[chwala]

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?

 

[Orodruin]

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]

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

 

[Mark44]

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.

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.

given that

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

Where did this come from?

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.

 

[Mark44]

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.

 

[chwala]

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.

 

[chwala]

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.

 

[Lnewqban]

 

[Mark44]

$r= \dfrac{\sqrt{3}}{60}$ 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.

 

[chwala]

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.

 

[WWGD]

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]

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...

 

[erobz]

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]

$$ V(h) = \frac{\pi }{9} h^3$$

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

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?

 

[erobz]

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.

 

[chwala]

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?

 

[erobz]

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.

 

[chwala]

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.

$\dfrac{3Qt}{π} +C = \dfrac{h^3}{3}$

$h(t)=\sqrt[3]{\dfrac{9Qt}{π} +3C}$

 

[erobz]

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 $\frac{dh}{dt}$.

 

[erobz]

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 $h(t)$?

 

[Mark44]

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.

 

[erobz]

Or written in another way, $\left.\frac{dh}{dt}\right|_{t = 0.05}$

I think you mean: $\left.\frac{dh}{dt}\right|_{h = 0.05 \rm{m}}$

 

[Mark44]

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).

 

[erobz]

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 $\dot h(t)$.

 

[SammyS]

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

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.
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