C4: Differential Equations

8. Liquid is pouring into a container at a constant rate of 20 cm3 s–1 and is leaking out at a rate proportional to the volume of the liquid already in the container.
(a) Explain why, at time t seconds, the volume, V cm3, of liquid in the container satisfies the differential equation
dV/dt = 20 – kV,
where k is a positive constant.

The container is initially empty.
(b) By solving the differential equation, show that
V = A + Be^(–kt),
giving the values of A and B in terms of k.

Given also that
dV/dt = 10 when t = 5,
(c) find the volume of liquid in the container at 10 s after the start.

I can do part a) and part b), but I'm having trouble understanding how to do part c). The answer to part b) is V = 20/k - 20/k e^(-kt)
Could someone point me in the right direction?

Reply 1

vc94

dV/dt = 20 – kV
10 = 20 – kV
kV=10

and...with t=5

V = 20/k - 20/k e^(-5k)
kV = 20 - 20e^(-5k)

Solve to find k

Reply 2

coldplasma

vc94

dV/dt = 20 – kV
10 = 20 – kV
kV=10

and...with t=5

V = 20/k - 20/k e^(-5k)
kV = 20 - 20e^(-5k)

Solve to find k

Oh cool, thanks :smile:

The mark scheme says something completely different and more complicated, but I can see the logic in your method.

Reply 3

SunGod87

Slightly easier method:

Unparseable latex formula:

[br]\[[br]\begin{array}{l}[br] V = \frac{{20}}{k} - \frac{{20}}{k}e^{ - kt} \\ [br] \frac{{dV}}{{dt}} = 20e^{ - kt} \\ [br] \frac{{dV}}{{dt}} = 10\,\,\,when\,\,\,t = 5 \\ [br] 10 = 20e^{ - 5k} \\ [br] 0.5 = e^{ - 5k} \\ [br] \ln 0.5 = - 5k \\ [br] - \ln 2 = - 5k \\ [br] k = \ln 2 \\ [br] \end{array}[br]\][br]

Then just put the numbers in for V at t = 10.

Reply 4

coldplasma

SunGod87

Slightly easier method:

Unparseable latex formula:

Then just put the numbers in for V at t = 10.

This is what the mark scheme says, but I'm not sure how you jumped from the first step to the second...

Reply 5

SunGod87

I just differentiated V with respect to t.

Reply 6

Annabella33

could someone please show the method for part b please?

Reply 7

rlyneedtostudy

the key to part b is to recognise its a "separating the variables" question

if you separate the variables to get 1/(20-kV) dV = 1 dt and integrate you should get your answer.

Reply 8

Notnek

Original post by Annabella33

could someone please show the method for part b please?

Please start a new thread - this one is over 10 years old!

Reply 9

Annabella33

Original post by rlyneedtostudy

the key to part b is to recognise its a "separating the variables" question

if you separate the variables to get 1/(20-kV) dV = 1 dt and integrate you should get your answer.