A2 maths question need help

[loggerhead]

Im stuck on this question can someone please walk me through it.

An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is
empty. At time t hours after filling begins, the volume of liquid is V m³ and the depth of liquid is h m.
It is given that V = 4/3h³
The liquid is poured in at a rate of 20m³ per hour, but owing to leakage, liquid is lost at a rate
proportional to h².

When h = 1, dh/dt = 4.95.

(i) Show that h satisfies the differential equation

dh/dt = 5/h2 - 1/20

(ii) Verify that 20h²/(100 - h²) = 20 + 2000/((10 - h)(10 + h))

(iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h.

This is CIE, 9709/03 (mathematics paper 3), October November 2008, question 8. I've attached the paper

Any help would be much appreciated.

[astarmathsandphysics]

This question has been answered. I will look for the link tomorrow.

[nid404]

I'm still not good with calculus...will try doing it though
And maybe I've seen the question before...so I'll look for the link

[astarmathsandphysics]

Here is the link

https://studentforums.biz/index.php/topic,740.0.html

[loggerhead]

I cant seem to follow the link, i just get this

"The topic or board you are looking for appears to be either missing or off limits to you."

what do i do??

[nid404]

8iii) dh/dt=  Separate variables to get dh=dt then use previous part and integrate


-20h-ln(100-h)+ln(100+h)+C=t so t=-20h+ln((100+h)/(100-h))+C

when h=0, t=0
then C=0 and t=t=-20h+ln((100+h)/(100-h))


This is what astar answered in that thread

[astarmathsandphysics]

Thanks nid

[nid404]

anytime

[loggerhead]

thanks guys much appreciated. I'm sure there will be more questions to come.