rates of change

Half the stuff he writes doesn't make sense, or is hard to follow where it comes from, or is unclear as to what he's trying to do with it so I give up after a while. I don't really need to explain anything.

$\dfrac{dV}{dh} = 4000$ since $V=\pi R^2 h$ hence $\dfrac{dV}{dh} = \pi R^2 = A$, so that's fine.

But I don't understand where you pull the last line from. The RHS of it is basically $\dfrac{dV}{dt} = \pi R^2 \dfrac{dh}{dt} = \dfrac{dV}{dh} \dfrac{dh}{dt}$. So the LHS needs to reflect that it is change in volume over time, and dV/dt is most certainly not 4000 as you have it.

$dh/dv = 1/\pi r^2$, $dh/dv = 1/4000$

I get that.

and you do: $1/\pi r^2 = 1/4000$

where do you go from here?

thats what you wrote earlier.

either way, we have: $\dfrac{dV}{dh} = 4000$

$\dfrac{dV}{dh} = \pi R^2 = A$

where do I go from here?

The question tells us that liquid flows into the container at a rate of 1600, but is leaving at a rate proportional to the square root of the height.

Therefore, the rate of change of volume can be expressed as:
$dv/dt = 1600 - C\sqrt h$
So now we can just substitute our differentials into our original equation for $dh/dt = dh/dv\times dv/dt$ to get closer to our final answer.

ah, ok.

when we're dealing with proportions, do we always have a constant (in this case: c)?.

ok, I said that earlier, but I write: $dv/dt = 1600 - \sqrt h$

all help is welcome.

I thought: $dV/dh = A$ or $dV/dr = A$

but I'm unfamiliar with his approach although its probably the best. however, I still get stuck in the next part.


$4000 = A$

so, $4000 =\pi r^2$

or in the case of RDK: $4000 = \pi r^2 (dh/dt)$

Yes, a constant is usually used. If they are proportional, it is very different from being equal.

Go back to the beginning.

You said dh/dt = (dh/dV) x (dV/dt) which is correct

then you seemd OK with the poster who said:




Then you had:

,

Now put that altogether:

dh/dt = (1/4000) x (1600 - C h^1/2)

Then expand.

what happens to:

We are told pi r^2 = 4000 in the question.

ah ok, so we didn't need to solve $dV/dh$ but just needed to know that $dV/dh = A = 4000$ ??

thanks

In this question we were given the area of cross section which is constant in a cylinder. Quite often we are given a specific radius [when it changes e.g. a circle] or time to get the rate of change - this questiom is just a bit different.

Yes, a constant is usually used. If they are proportional, it is very different from being equal.

....

....

in the next part of the question we're solving the the change in volume or height?

can we use the [text]dh/dt

in the next part of the question we're solving the the change in volume or height?

can we use the [text]dh/dt

What do you think? Try it first

No because it tells us nothing about the rate at which the height changes.

From part (i) we know it leaks out at a range of $C\sqrt{h}$ which at $h=25$ is $5C$. We know now that this is 400. We also know that $k = \frac{C}{4000}$ from previous work, so you can show the result.

because its the water leaving, I did:

$-400 = - C\sqrt h$

a negative 400 because its water leaving.

$-400 = - C\sqrt 25$
$-400 = - C(5)$
$80 = C$
from the first part:
$k = C/4000$
$k = 80/4000$
$k = 1/50$