C4 - Differential equations QUESTIONS

Hi,

I'm stuck on these two questions and would really appreciate it if someone could show me how to do them:

1) Water flows out of a tank such that the depth h metres of water in the tank falls at a rate which is proportional to the square root of h.

Show that the general solution of the differential equation may be written as h=(c-kt)² where c and k are constants.

I did dh/dt = -k√h

∫ h^(-1/2) dh = ∫ -k dt

2h^(1/2) = -kt+c

but I'm not sure how to get it in the form h=(c-kt)² as with my equation I'd have to divide by 2 and then square everything, which would give
h=(c/2-kt/2)² ?

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2) Liquid is poured into a container at a constant rate of 20cm³/s. After t seconds liquid is leaking from the container at a rate of v/10 cm³/s, where vcm³ is the volume of liquid in the container at the time.

Show that

-10 dv/dt = v - 200

Given that v=500 when t=0

I'm just not sure how to do this one.

Thank you very much! :smile:

Reply 1

meef cheese

10

sweet_gurl

Hi,

I'm stuck on these two questions and would really appreciate it if someone could show me how to do them:

1) Water flows out of a tank such that the depth h metres of water in the tank falls at a rate which is proportional to the square root of h.

Show that the general solution of the differential equation may be written as h=(c-kt)² where c and k are constants.

I did dh/dt = -k√h

∫ h^(-1/2) dh = ∫ -k dt

2h^(1/2) = -kt+c

but I'm not sure how to get it in the form h=(c-kt)² as with my equation I'd have to divide by 2 and then square everything, which would give
h=(c/2-kt/2)² ?

I think, given that k and c are arbitrary constants, if they are divided by 2, you just get another two arbitrary constants. These shouldn't really be called k and c, but there doesn't seem to be anything wrong with your solution.

Reply 2

sweet_gurl

OP

4

Thank you, can anyone please help me with my second question?

Reply 3

dvs

10

rate of increase of liquid = (rate at which its added) - (rate at which its leaking)
dv/dt = 20 - v/10

Reply 4

ImperceptibleNinja

13

No offence here, but have you checked you've typed the second question out correctly? It just seems a little disjointed. I would expect them to ask something about getting the rate, but then to ask you to use the v=500 when t=0 information to get an equation for v in terms of t, like this:

"2) Liquid is poured into a container at a constant rate of 20cm³/s. After t seconds liquid is leaking from the container at a rate of v/10 cm³/s, where vcm³ is the volume of liquid in the container at the time.
Show that
-10 dv/dt = v - 200
Given that v=500 when t=0"

overall rate, dv/dt, is the sum of the different rates:
dv/dt = 20 - v/10, since you have 20cm³ coming in every second and v/10 cm³ going out every second.

rearranging this, you get -10dv/dt = v-200

rearranging again:
dv/(v-200) = dt/(-10)

Integrating:
ln (v-200) = t/(-10) + c

I call this bit exponentiating. No idea what it's really called...
e^[ln(v-200)] = e^[t/(-10) + c]

simplify:
v-200 = e^c.e^(-t/10)

v = Ae^(-t/10) + 200 where A = e^c

Plug in given values:
at t=0, v=500

500 = Ae^(0) +200

A = 300

v = 300e^(-t/10) + 200

Then they often ask you to find out a certain volume at a given time, or vice-versa. If you check the question, it's probable that it asks you to get an equation for the volume in terms of the time. All the questions I've ever come across on this kind of thing tend to ask that sort of stuff - not just for the rate.

I hope that helped. I hope it was relevant too :smile:

(ARGH. I typed that all out last night and got disconnected... Sorry for the delay)

C4 - Differential equations QUESTIONS

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