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Hi I need help understanding this problem posed by my Edexcel maths textbook.
'Fluid flows out of a cylindrical tank with constant cross section. At time t minutes t>0, the volume of fluid remaining in the tank is

Vcm^3

. The rate at which the fluid flows in m cubed per/minute is proportional to the square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation.

\frac{dh}{dt} = -k \sqrt h

, where k is a positive constant'.

Okay feel free to correct if I'm wrong but here's what I understand so far.
The volume of a cylinder is given by the formula

V = \pi r^2 h

Since the rate at which volume is decreasing is proportional to the volume of the cylinder, I can state my known derivative to be

\frac{dV}{dt} = -k \sqrt (\pi r^2 h)

I'm looking for

\frac{dh}{dt}

so I can rewrite by the chain rule as

\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}

Ok

\frac{dh}{dV} = \frac{1}{\pi r^2}

As you can see my attempt is just an amateur recipe for disaster. :frown:

Reply 1

boner in jeans

Original post by sertobe

Hi I need help understanding this problem posed by my Edexcel maths textbook.
'Fluid flows out of a cylindrical tank with constant cross section. At time t minutes t>0, the volume of fluid remaining in the tank is

Vcm^3

. The rate at which the fluid flows in m cubed per/minute is proportional to the square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation.

\frac{dh}{dt} = -k \sqrt h

, where k is a positive constant'.

Okay feel free to correct if I'm wrong but here's what I understand so far.
The volume of a cylinder is given by the formula

V = \pi r^2 h

Since the rate at which volume is decreasing is proportional to the volume of the cylinder, I can state my known derivative to be

\frac{dV}{dt} = -k \sqrt (\pi r^2 h)

I'm looking for

\frac{dh}{dt}

so I can rewrite by the chain rule as

\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}

Ok

\frac{dh}{dV} = \frac{1}{\pi r^2}

As you can see my attempt is just an amateur recipe for disaster. :frown:

right

so

\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}

\frac{dV}{dt} = -kV = -k\sqrt{\pi r^{2} h}

\frac{dh}{dV} = \frac{1}{\pi r^{2}}

what do you get when you multiply them together?

Reply 2

sertobe

OP

Original post by boner in jeans

right

so

\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}

\frac{dV}{dt} = -kV = -k\sqrt{\pi r^{2} h}

\frac{dh}{dV} = \frac{1}{\pi r^{2}}

what do you get when you multiply them together?

urm I get

\frac{-k\sqrt{\pi r^{2} h}}{\pi r^{2}}

which looks nothing like the answer. :s-smilie:

Reply 3

boner in jeans

Original post by sertobe

urm I get

\frac{-k\sqrt{\pi r^{2} h}}{\pi r^{2}}

which looks nothing like the answer. :s-smilie:

good, now you can simplify that

-k \frac{\sqrt{\pi r^{2}} \sqrt{h}}{\sqrt{\pi r^{2}} \sqrt{\pi r^{2}}}

and you also know that the cross section is a constant

(edited 11 years ago)

Reply 4

ezioaudi77

Original post by sertobe

x

The question itself says that the cross-sectional area is constant. Therefore, V= A*h can be used (where A is a constant are), instead of V=(pi)*r^2*h ("r" is a variable).

So dV/dh = A, and carry on to find dh/dt through chain-rule as you did.

One more thing, when finding dV/dt, use a different letter instead of "k" for the constant term, such as "c", because the question asks you to prove that k=something.

Hope that helps, but if you are still stuck, I would be happy to post the method :smile:

Reply 5

sertobe

OP

Original post by ezioaudi77

The question itself says that the cross-sectional area is constant. Therefore, V= A*h can be used (where A is a constant are), instead of V=(pi)*r^2*h ("r" is a variable).

So dV/dh = A, and carry on to find dh/dt through chain-rule as you did.

One more thing, when finding dV/dt, use a different letter instead of "k" for the constant term, such as "c", because the question asks you to prove that k=something.

Hope that helps, but if you are still stuck, I would be happy to post the method :smile:

Sorry to be persistent, but I don't know what you mean. I didn't treat r as a variable because the radius of a cylinder is a constant. I looked up the definition of cross section but the explanations were too complicated to relate to my problem, I'd really appreciate a simplified explanation. (Based on this problem, and in general)

Reply 6

ezioaudi77

Original post by sertobe

Sorry to be persistent, but I don't know what you mean. I didn't treat r as a variable because the radius of a cylinder is a constant. I looked up the definition of cross section but the explanations were too complicated to relate to my problem, I'd really appreciate a simplified explanation. (Based on this problem, and in general)

First, read this if you are not sure about cross-sections of cylinders.
Usually

r

is a variable in

\pi{r^2}{h}

, which means as Volume or Area increases,

r

increases as well.

But, in this case, the cross-section of the cylinder (

\pi{r^2}

) is constant, which means

\pi{r^2}

is the same for any value of

V

that the cylinder takes, so

h

is the only variable for

V = \pi{r^2}{h}

.

I asked you to substitute

A

for

\pi{r^2}

in

V = \pi{r^2}{h}

, because it helps you to remember that

A

is a constant here. Also, you wouldn't accidentally differentiate

r

when finding the

\frac{dV}{dh}

of

V

.

Now, you have to find

\frac{dh}{dt}

. You can use either

A

or

\pi{r^2}

, but because you seem to get confused when using

A

, I will use

\pi{r^2}

.

Remember,

r

is a constant here.

V = \pi{r^2}{h}

\frac{dV}{dh}= \pi{r^2}

When finding

\frac{dV}{dt}

, you have used

k

as the constant term. You can use this usually, but you cannot use it here, because at the end of the question you have to use

k

again. So it is suggested that you use any letter other than

k

or other letters used here. I will use

c

.

\frac{dV}{dt}= -c \sqrt V = -c \sqrt {\pi{r^2}{h}} = -c (\sqrt {\pi} r) \sqrt {h}

Therefore,

\frac{dh}{dt}= \frac{dV}{dt} \times \frac{dh}{dV} = -c (\sqrt {\pi} r) \sqrt {h} \times \frac {1}{\pi{r^2}} = \frac {-c \sqrt h}{\sqrt {\pi} r} = -k \sqrt {h}

Now you have found the constant term

k

, where

k = \frac{-c}{\sqrt {\pi} r}

This way seems complicated, which is why

A

should be used for

\pi{r^2}

.

Sorry if the explanation is too simple :colone:

. Anyway, hope that helps.

(edited 11 years ago)

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