# P3 Differential Equations Questions :(

Edit: Appreciate all help, given rep or will do as soon as my giving rep limit clears in the next 24 hours hehe Thank you I wish you all could replace my maths teachers cos your all so mathtastic

p.s If anyone wanted to try these questions (they are from OCR MEI Ex 6C) for revision their quoted with the solutions below now....thank youuuuu

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15. A curve C is given by parametric equations x=t^2 , y=2t
ii) Find dy/dx in terms of t (dy/dx = 1/t) . Hence show that dy/dx = y/2x at any point on the curve...how? All i can think of is sub in t=y/2 which doesn't give that answer!

y/2x = 2t/2t^2 = 1/t = dy/dx as required.
10. The square horizontal cross-section of a container has side 2m. Water is poured in at the constant rate of 0.08m^3/s and, at the same time, leaks out of a hole in the base at the rate of 0.12x m^3/s , where x m is the depth of the water in the container at time s. So the voluem, V m^3 of the water in the container at time t is given by V=4x and the rate of change of volume is given by

dV/dt = 0.08 - 0.12x

Use these results to find an equation for dx/dt in terms of x.
Got 0.02 - 0.03x for this which is correct.

Solve this to find x in terms of t if the container is empty initially.
Couldn't get what they got which is x = 1/3[2-2e(-3t/100)]

I got as far as deducing that x = 2t/3 if that helps.
ZJuwelH
y/2x = 2t/2t^2 = 1/t = dy/dx as required.

Oh, working backwards! Why did that not occur to me Thanks!
Maisy
13. An industrial process creates a chemical C. At time t hours after the start of the process the amount of C produced is x kg. The rate at which C is produced is given by differential equation

dx/dt = k(2-x)(1+x)e(-t),
where k is a constant

v) Show that there is a finite limit to the amount of C which this process can produce, however long it runs, and determine the value of this limit
- Can't you do dx/dt = 0 ? But then I get x =2 and the answer is x=0.728kg

1/[(2 - x)(1 + x)] dx/dt = k e^(-t)
1/(2 - x) + 1/(1 + x) dx/dt = 3k e^(-t) . . . using partial fractions
ln[(1 + x)/(2 - x)] = constant - 3k e^(-t) . . . integrating wrt t
ln[(1 + x)/(2 - x)] = 3k + ln(1/2) - 3k e^(-t) . . . using x = 0 at t = 0
(1 + x)/(2 - x) = (1/2) e^[3k (1 - e^(-t))] . . . applying e^ to both sides
-1 + 3/(2 - x) = (1/2) e^[3k (1 - e^(-t))]
3/(2 - x) = 1 + (1/2) e^[3k (1 - e^(-t))]
2 - x = 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] }
x = 2 - 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] }

[Check:]

As t becomes very large,

(1) e^(-t) decreases, tending to 0
(2) 1 - e^(-t) increases, tending to 1
(3) e^[3k (1 - e^(-t))] increases, tending to e^(3k)
(4) 1 + (1/2) e^[3k (1 - e^(-t))] increases, tending to 1 + (1/2) e^(3k)
(5) x = 2 - 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] } increases, tending to 2 - 3/(1 + (1/2) e^(3k)) = 2 - 6 / (2 + e^(3k)). This final thing is 1.72833 if you assume k = 1.
Jonny W
1/[(2 - x)(1 + x)] dx/dt = k e^(-t)
1/(2 - x) + 1/(1 + x) dx/dt = 3k e^(-t) . . . using partial fractions
ln[(1 + x)/(2 - x)] = constant - 3k e^(-t) . . . integrating wrt t
ln[(1 + x)/(2 - x)] = 3k + ln(1/2) - 3k e^(-t) . . . using x = 0 at t = 0
(1 + x)/(2 - x) = (1/2) e^[3k (1 - e^(-t))] . . . applying e^ to both sides
-1 + 3/(2 - x) = (1/2) e^[3k (1 - e^(-t))] ..where did the minus come from as the other side is still the same..
3/(2 - x) = 1 + (1/2) e^[3k (1 - e^(-t))] ..how did you get +1 from above?

2 - x = 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] }
x = 2 - 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] }

[Check:]

As t becomes very large,

(1) e^(-t) decreases, tending to 0
(2) 1 - e^(-t) increases, tending to 1
(3) e^[3k (1 - e^(-t))] increases, tending to e^(3k)
(4) 1 + (1/2) e^[3k (1 - e^(-t))] increases, tending to 1 + (1/2) e^(3k)
(5) x = 2 - 3 / { 1 + (1/2) e^[3k (1 - e^(-t))] } increases, tending to 2 - 3/(1 + (1/2) e^(3k)) = 2 - 6 / (2 + e^(3k)). This final thing is 1.72833 if you assume k = 1.

Thanks that's right as k=1/3 Um do you think you could say how you got the lines in bold though? Much appreciated you are a genius! (Either that, or i'm incredibly thick!) Thanks again
Maisy
Thanks that's right as k=1/3 Um do you think you could say how you got the lines in bold though? Much appreciated you are a genius! (Either that, or i'm incredibly thick!) Thanks again

-1 + 3/(2 - x) = (1/2) e^[3k (1 - e^(-t))] ..where did the minus come from as the other side is still the same..
3/(2 - x) = 1 + (1/2) e^[3k (1 - e^(-t))] ..how did you get +1 from above?

First bit: division, he means -1 + [3/(2-x)] I think.

And the +1 is a result of bringing the -1 over to the RHS.
ZJuwelH
First bit: division, he means -1 + [3/(2-x)] I think.

And the +1 is a result of bringing the -1 over to the RHS.

ooh ok but what about that 3? suddenly that one x has gone and replaced by a 3.. sorry hehe
Maisy
ooh ok but what about that 3? suddenly that one x has gone and replaced by a 3.. sorry hehe

He's said that (1 + x)/(2 - x) = -1 + [3/(2-x)].

-1+[3/(2-x)] = [-(2-x)/(2-x)] + [3/(2-x)] = (-2+x+3)/(2-x) = (1+x)/(2-x)
ZJuwelH
He's said that (1 + x)/(2 - x) = -1 + [3/(2-x)].

-1+[3/(2-x)] = [-(2-x)/(2-x)] + [3/(2-x)] = (-2+x+3)/(2-x) = (1+x)/(2-x)

i see...working backwards but how on earth you'd think of that from (1 + x)/(2 - x) in the beginning and working it from that, I can't get
Maisy
i see...working backwards but how on earth you'd think of that from (1 + x)/(2 - x) in the beginning and working it from that, I can't get

Synthetic (or long) division, you must have been taught this?

...... -1__
x+1 |-x+2
....- (-x-1)
.......3

Hence you get -1 + 3(2-x).
Maisy
Quick One! - may be misprint error..? 9. A hemispherical bowl of radius a has its axis vertical and is full of water. At time t=0 water starts running out of a small hole in the bottom of the bowl so that the depth of water in the bowl at time t is x. The rate at which the volume of water is decreasing is proportional to x. The volume of water in the bowl when the depth is x is pi(ax^2 - 1/3x^3), so pi(2ax - x^2)dx/dt = -kx

Given the bowl is empty after a time T, show that
k=[3(pi)a^2]/2T - right I get this but with the missing 1/2 ??

pi (2ax - x^2) dx/dt = -kx
(2a - x) dx/dt = -k/pi . . . dividing by x*pi
2ax - (1/2) x^2 = constant - kt/pi . . . integrating wrt t
2ax - (1/2) x^2 = (3/2) a^2 - kt/pi . . . because x = a when t = 0
0 = (3/2) a^2 - kT/pi . . . because x = 0 when t = T
k = (3pi a^2) / (2T)
Maisy
i see...working backwards but how on earth you'd think of that from (1 + x)/(2 - x) in the beginning and working it from that, I can't get

It's like partial fractions. Suppose

(1 + x)/(2 - x) = A + B/(2 - x)

for some constants A and B. Then, multiplying by (2 - x),

1 + x = A(2 - x) + B . . . (*).

Put x = 2 into (*) to get B = 3.

Put x = 0 into (*) to get 1 = 2A + B = 2A + 3, and so A = -1.

So (1 + x)/(2 - x) = -1 + 3/(2 - x).
Jonny W
pi (2ax - x^2) dx/dt = -kx
(2a - x) dx/dt = -k/pi . . . dividing by x*pi
2ax - (1/2) x^2 = constant - kt/pi . . . integrating wrt t
2ax - (1/2) x^2 = (3/2) a^2 - kt/pi . . . because x = a when t = 0
0 = (3/2) a^2 - kT/pi . . . because x = 0 when t = T
k = (3pi a^2) / (2T)

thanks but um one question, x = a when t = 0? if a is radius, isn't the depth = diameter when t = 0 ie when it's full? sorry, thanks

what i did was
when bowl empty: v=0
so 0 = ax^2 - 1/3x^3
0 = x^2(a -1/3x)
a-1/3x = 0
x=3a (substitute)

pi[2a(3a)-(3a)^2] dx/dt = -k(3a)
pi[6a^2 - 9a^2]dx/dt = -3ak
k = pi(a) dx/dt (integrate)

kt = pi(a)x +c
kt = 3pi(a^2) + c
so k = (3pi.a^2)/t +c

Maisy
thanks but um one question, x = a when t = 0? if a is radius, isn't the depth = diameter when t = 0 ie when it's full? sorry, thanks

It is a hemisphere, not a sphere.
Maisy

what i did was
when bowl empty: v=0
so 0 = ax^2 - 1/3x^3
0 = x^2(a -1/3x)
a-1/3x = 0
x=3a (substitute)

x = 0 when the bowl is empty. x = 3a is impossible because the bowl has height a.
Maisy

pi[2a(3a)-(3a)^2] dx/dt = -k(3a)

If you put a particular value of x (in your case, 3a) into the differential equation, the expression you get for dx/dt only holds when the water is at the height you specify. So you can't then integrate that expression.
Jonny W
It is a hemisphere, not a sphere.

x = 0 when the bowl is empty. x = 3a is impossible because the bowl has height a.

If you put a particular value of x (in your case, 3a) into the differential equation, the expression you get for dx/dt only holds when the water is at the height you specify. So you can't then integrate that expression.

Ok I see now! Thanks SO much!! that really helps Gave you rep
Anyone fancy having a stab at Q10 and 11? I appreciate all the help thanks
Maisy
10. The square horizontal cross-section of a container has side 2m. Water is poured in at the constant rate of 0.08m^3/s and, at the same time, leaks out of a hole in the base at the rate of 0.12x m^3/s , where x m is the depth of the water in the container at time s. So the voluem, V m^3 of the water in the container at time t is given by V=4x and the rate of change of volume is given by

dV/dt = 0.08 - 0.12x

Use these results to find an equation for dx/dt in terms of x.
Got 0.02 - 0.03x for this which is correct.

Solve this to find x in terms of t if the container is empty initially.
Couldn't get what they got which is x = 1/3[2-2e(-3t/100)]

x = (1/4) V
dx/dt = (1/4) dV/dt = 0.02 - 0.03x = (3/100)*(2/3 - x)
1 / (2/3 - x) dx/dt = 3/100
-ln(2/3 - x) = constant + 3t/100 . . . integrating wrt t
-ln(2/3 - x) = -ln(2/3) + 3t/100 . . . because x = 0 when t = 0
ln(2/3 - x) = ln(2/3) - 3t/100
2/3 - x = (2/3) e^(-3t/100) . . . applying e^ to both sides
x = (2/3) [1 - e^(-3t/100)]
11. A patch of oil pollution in the sea is approximately circular in shape. When first seen its radius was 100m and its radius was increasing at a rate of 0.5m/minute. At a time t minutes later, its radius is r metres. An expert believes that, if the patch is untreated, its radius will increase at a rate which is proportional to 1/(r^2)

ii) Using the initial conditions, find the value of k. (k = 5000). Hence calculate the expert's prediction of the radius of the oil patch after hours - am I missing something here? Because when i solve the differential equation, you can sub in k but you want to find r and you don't know c or anything else? I thought i could use t=0 and r=100 to find c, but doesn't work! (ans: r=141m)

The expert thinks that that if the oil patch is treated with chemicals then its radius will increase at a rate proportional to 1/[r^2(2+t)]

iv) Calculate the expert's prediction of the radius of the treated oil patch after 2 hours. Again, how do I find c if I don't know anything? (ans:104m)

ii)
at t=0, r=100, dr/dt = 0.5

we are given that,

dr/dt = k/r²

puting in the initial conditions,

0.5 = k/100²
k = 0.5*10,000
k = 5000
======

dr/dt = 5000/r²
dr = 5000 dt
r³/3 = 5000t + c

at t=0, r=100,
=> c=10^6/3

= 15000t + 10^6

at t= 120, (this time not actually given ! )

= 15000*120 + 10^6
= 2,800,000
r = 140.9
r = 141 m
======

iv)
dr/dt = k/[r²(2+t)]

at t=0, dr/dt = 0.5, r=100

0.5 = k/[100²*2]
k = 10,000
=======

dr/dt = 10,000/[r²(2+t)]
dr = 10,000/(2+t) dt
r³/3 = 10,000ln(2+t) + c

at t=0, r = 100,

10^6/3 = 10,000ln(2) + c
c = 10^6/3 - 10,000ln2
= 30,000ln(2+t) + 10^6 - 30,000ln2
= 30,000ln(1+t/2) + 10^6

at t=120,

= 30,000ln(1+60) + 10^6
= 123,326 + 10^6
= 1,123,326
r = 103.9
r = 104 m
======
Jonny W
x = (1/4) V
dx/dt = (1/4) dV/dt = 0.02 - 0.03x = (3/100)*(2/3 - x)
1 / (2/3 - x) dx/dt = 3/100
-ln(2/3 - x) = constant + 3t/100 . . . integrating wrt t
-ln(2/3 - x) = -ln(2/3) + 3t/100 . . . because x = 0 when t = 0
ln(2/3 - x) = ln(2/3) - 3t/100
2/3 - x = (2/3) e^(-3t/100) . . . applying e^ to both sides
x = (2/3) [1 - e^(-3t/100)]

Thanks again! hm I tried something similar though but I must've gone wrong somewhere unless that answer actually goes to x = 1/3[2-2e(-3t/100)] somehow