P3 Differential Equations Questions :(

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.

Reply 2

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.

Reply 3

OP

ZJuwelH

y/2x = 2t/2t^2 = 1/t = dy/dx as required.

Oh, working backwards! Why did that not occur to me :rolleyes:

Thanks!

Reply 4

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.

Reply 5

OP

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.. :confused:

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 :biggrin:

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 :biggrin:

Reply 6

Maisy

Thanks that's right as k=1/3 :biggrin:

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 :biggrin:

-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.

Reply 7

OP

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 :tongue:

Reply 8

Maisy

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

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)

Reply 9

OP

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 :frown:

Reply 10

makesomenoise

1

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 :frown:

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).

Reply 11

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)

Reply 12

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 :frown:

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).

Reply 13

OP

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 :smile:

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

:confused:

Reply 14

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 :smile:

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.

Reply 15

OP

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 :biggrin:

Gave you rep :smile:

Reply 16

OP

Anyone fancy having a stab at Q10 and 11? I appreciate all the help thanks :smile:

Reply 17

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)]

Reply 18

Fermat

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
r² dr = 5000 dt
r³/3 = 5000t + c

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

r³ = 15000t + 10^6

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

)

r³ = 15000*120 + 10^6
r³ = 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)]
r² 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
r³ = 30,000ln(2+t) + 10^6 - 30,000ln2
r³ = 30,000ln(1+t/2) + 10^6

at t=120,

r³ = 30,000ln(1+60) + 10^6
r³ = 123,326 + 10^6
r³ = 1,123,326
r = 103.9
r = 104 m
======

Reply 19