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1. Need help with this Q9 in exercise 4E, it just aint happening for me.
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 V m^3. The rate at which the fluid flows in m^3/min is proportional to thr square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation dh/dt = -ksqroot(h) where k is a positive constant.

Help would greatly be appreciated.
2. You'll need the chain rule, I think. How could you represent the flow in differential form?
3. All I can get is dh/dt = dh/dv * dV/dt and also I know that dV/dt = -ksqroot(V)
4. You know that , but you also know that, because it's a cylinder, . From this, you can find an equation for and then flip it to find ... combine these to get your answer.

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Since the radius and are constant, you can say that
5. (Original post by Junker)
Need help with this Q9 in exercise 4E, it just aint happening for me.
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 V m^3. The rate at which the fluid flows in m^3/min is proportional to thr square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation dh/dt = -ksqroot(h) where k is a positive constant.

Help would greatly be appreciated.
Can you see where I get the following from:

(A a positive constant), and ?

If so, use this and teh fact that:

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When you plug into the RHS of this equation you'll get which you can write as since everything but V is a constant. Now plug in V=pi r^2 h into this again...
6. AHh. thanks both of you, i think what I struggled with was differentiaing V = pi r^2 h
I was doing with respect to r, when I should have been with respect to h. But still why is dV/dh of V = pi r^2 h
just pi r^2? Sorry for my latex ignorance
7. Because pi r^2 is just a constant, since it is a cylinder.

So it's dV/dh of, a constant times h.
8. (Original post by Junker)
AHh. thanks both of you, i think what I struggled with was differentiaing V = pi r^2 h
I was doing with respect to r, when I should have been with respect to h. But still why is dV/dh of V = pi r^2 h
just pi r^2? Sorry for my latex ignorance
Because this is a cylinder we're considering r is just a fixed number (say, I dunno, 2.5 metres), so it acts like a constant when you're differentiating. Therefore, you're differentiating V with respect to h you're effectively just differentiating:

Ah

with respect to h, where A is a constant.

Spoiler:
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If you're still struggling, try answering the question again but adding in: "Water is flowing from a cylincer with radius 2m ...". You'll obviously then use the formula V=pi x 2^2 x h, which will give dV/dh = pi x 2^2.

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Updated: March 19, 2009
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