# Rates of change horrible question

• Apr 13th 2009, 02:58 AM
Big_Joe
Rates of change horrible question
This is another question i came across which i really cant get my head around. Questions like this really really stop me in my tracks i just dont know how to think about them. If somebody could give me some guidelines on answering questions like these i would be really grateful because at the moment if one of these comes up in my exam i will be in trouble! (Crying)

Liquid is pouring into a large vertical circular cylinder at a constant rate of $1600cm^3s^{-1}$ and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is $4000cm^2$.

a)show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential equation: $\frac{dh}{dt} = 0.4-k\sqrt h$ where k is a positive constant. (3 marks)

When h = 25, water is leaking out of the hole at $400cm^3s^{-1}$

b) show that k = 0.02 (1 mark)

c) Separate the variables of the differential equation: $\frac{dh}{dt} = 0.4-0.02\sqrt h$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by: $\int_0^{100}\frac{50}{20-\sqrt h}\,dh$ (2 marks)

Using the substitution $h=(20-x)^2$, or otherwise,

d) find the exact value of $\int_0^{100}\frac{50}{20-\sqrt h}\,dh$ (6 marks)

e) Hence find the time taken to fill the cylinder from empty to a height of 100 cm, giving your answer in minutes and seconds to the nearest second. (1 mark)
• Apr 13th 2009, 09:10 AM
Stonehambey
The hardest thing about rates of change questions is deriving the equation. So let's have a think about it.

$\frac{dh}{dt}$ is the rate of change of the height of the liquid with respect to time. So what is causing the height to change? Well the liquid being poured in and the liquid leaking out. So let's write

Rate of change of height = rate of increase in height - rate of decrease in height.

Or another way, the combined rates of change, but remembering that the liquid leaking out gives a negative rate of change.

Well we are told that the rate of decrease is proportional to the square root of the height at that given time, so this is simply $k\sqrt{h}$

Oh I forgot to add that if k<0 then we would be adding the rates of change and the liquid leaking out would cause the height to increase faster, which is obviously false, so we must have that k>0

Well for the rate of increase in height it's a little harder, but not much. The volume is increasing by 1600cm^3/s, the area of the cross section of the cylinder is 4000cm^2. Well the height is the volume over the cross section, so after a time t, this would be (remember we're assuming there is no liquid leaking out for this bit)

$h = \frac{1600}{4000}t = 0.4t$.

So the rate of change of height is this expression, differentiated wrt t, which is 0.4

So combining these two values gives us the overall expression for rate of change of height, that is

$\frac{dh}{dt} = 0.4 - k\sqrt{h}$

as required. Well hope that gets you up and running, let us know if you're having any trouble with the rest of the quesion :)
• Apr 19th 2009, 07:15 AM
Big_Joe
I have attempted the question and although it seems a whole lot simpler now the only thing i don't understand is why is there a k constant in from on the root h and not just root h?
• Apr 26th 2009, 01:08 AM
Stonehambey
Quote:

Originally Posted by Big_Joe
I have attempted the question and although it seems a whole lot simpler now the only thing i don't understand is why is there a k constant in from on the root h and not just root h?

Remember proportionality from GCSE?

If a is directly proportional to b, then a=kb for some k. Usually k is to be found :)