Question

Compressed air is escaping from a container. The pressure of the air in the container at time t is P, and the constant atmospheric pressure of the air outside the container is A. The rate of decrease of P is proportional to the square root of the pressure difference (P-A). Thus the differential equation connecting P and t is dP/dt = -k sqrt(P-A), where k is a positive constant.

Answer 1

i) Find, in any form, the general solution of this differential equation.
(ii) Given that P = 5A when t = 0, and that P = 2A when t = 2, show that k = sqrtA.
(iii) Find the value of t when P = A.
(iv) Obtain an expression for P in terms of A and t.

Answer 2

I'm not sure, but @Kainui is pretty awesome

Answer 3

Yeah I can help you with this, let's start with whichever part you want first, show me what you can do or what you know already.

Answer 4

i)

Answer 5

I dont get what do they mean by the general solution

Answer 6

solve the DE; its variables are seperable

Answer 7

General solution just means solve the differential equation but without solving for the constants of integration. For example, if I had the differential equation \[\frac{dy}{dx} = 2x+5\] then the general solution would be \[\Large y = x^2+5x+C\]

Answer 8

i dont know how to work that dP/dt out. D:

Answer 9

\[\quad\qquad \frac{\mathrm dP}{\mathrm dt} = -k \sqrt{P-A}\\
\quad\frac{\mathrm dP}{\sqrt{P-A}} = -k\,\mathrm dt\\
\int\frac{\mathrm dP}{\sqrt{P-A}} = \int-k\,\mathrm dt\\
\qquad\qquad\quad =\]

Answer 10

perform the integrations,

the one on the left can be found using a table of standard integrals,
the one on the right is simple

Answer 11

I got 1/ sqrt (P-A), i got the rhs. but dont really know how to continue lhs. D:

Answer 12

@UnkleRhaukus

Answer 13

For the left hand side use:\[\int\frac{\mathrm dx}{\sqrt{x\pm a}}=2\sqrt{x\pm a}+c_1\]
For the right hand side use:\[\int a\,\mathrm dx = a\int\mathrm dx = ax+c_2\]

Answer 14

I dont get how the 2 appears there

Answer 15

Anyways help me with ii? @UnkleRhaukus

Answer 16

I got 2 sqrtA = -k + 2 sqrt4A What should i do next

Answer 17

$$
\large \int\frac{\mathrm dx}{\sqrt{x\pm a}} \\
\large = \int (x \pm a)^{-1/2}~dx
\\ \large = \frac{(x \pm a)^{1/2}}{\frac{1}{2}} = 2(x \pm a)^{1/2}

$$

Answer 18

Nice @perl thanks.

Answer 19

don't forget the constant of integration

Answer 20

yes :) sorry

Answer 21

I'm done with part i. What about the ii

Answer 22

What did you get for part i) ?

Answer 23

[(it should have a \(t\) in it]

Answer 24

2 SQRT (P-A) = -kt+c

Answer 25

that's good

Answer 26

i work it out until 2 srqt A = -2k + 2sqrt4A then stuck

Answer 27

Your general solution is:
\[2 \sqrt{P-A} = -kt+c\]

The initial condition is that: \[P = 5A\quad\text{ when }\quad t = 0\]sub this into the general solution, to solve for \(c\)

Answer 28

i got 2 sqrt 4A for c

Answer 29

yes
\[2\sqrt{4A} = c\]
but you can simplify this

Answer 30

simplify to?

Answer 31

4sqrtA?

Answer 32

yep, so \(c=4\sqrt A\)

plugging this into the general solution,
we have a particular solution:
\[2 \sqrt{P-A} = -kt+4\sqrt A\]

Answer 33

I did alrd. then?

Answer 34

We also know that \(P = 2A\) when \(t = 2\),

plug these into the particular solution to find the constant of proportionality \(k\),

Answer 35

yeah yeah i did lol.

Answer 36

2 srqt A = -2k + 4sqrrt A

Answer 37

now solve for \(k\) . . .

Answer 38

THANKSS.

Answer 39

@UnkleRhaukus I need help with iv!

Answer 40

what did you get for part iii)

Answer 41

t = 4

Answer 42

and did you get k=sqrt A for ii) ?

Answer 43

I substitute them in, i got 2 sqrt(P-A) = -sqrt A t+4sqrt A

Answer 44

\[2 \sqrt{P-A} = -\sqrt A t+4\sqrt A\]
good,

now, can you solve this for \(P\)?

Answer 45

No. D:

Answer 46

factor the right hand side first,

Answer 47

then divide by 2,
and square both sides

Answer 48

i can't get the answer. D:

Answer 49

what do you get when you factor \(-\sqrt At+4\sqrt A\) ?

Answer 50

Sqrt A (4- t)

Answer 51

ooh got it!! thanks.

Answer 52

so what is \(P\) in terms of \(A\) and \(t\) ?

Answer 53

1/4 A (4+(4-t)^2)

Answer 54

good work!

Answer 55

:)