-kx^2 = ? - Calculus relating to compound decay.

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Homework Statement

In a chemica reaction the rate of consumption of a compound x is proportional to the square of the concentration of x. That is,

dx/dt = -kx^2

where k is a constant show that

x = 1/k(t-c)

The Attempt at a Solution

from my understanding usually i would try and find the primative of dt/dx, but I am having a problem seeing how it gets to the final stage of x = 1/k(t-c)

Any help greatly appreciated!

 

[Mentallic]

Well you can't take the primitive just like that. Say you were asked for the primitive of dy/dx=x², well then that is simple, but what if you were asked for the primitive of dy/dx=y²?
This is what you have here. You need the variable -kx² to be in terms of t in order to take the primitive of dx/dt.

Remember this rule though: \frac{dy}{dx}=\frac{1}{dx/dy} so you can turn your problem into \frac{dt}{dx}=\frac{1}{-kx^2} and now you can take the primitive!

 

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Awesome :-) Managed to get the answer. Unfortunately i am having an issue now with the second question, where i am asked to find k and C, given that the initial concentration of x is 1.0 and it is half consumed in 2 seconds.

Im a little bit lost here, so really detailed explanations would be helpfull.

I have found a reference to problems with rate of change given in the form of k(N-P) having a general solution of N = P + Ae^kt

So in my case i have x = k(T-C) So i need to find somehow the values of k and c, so if i were to convert to the general solution i would then have:

x = C + Ae^kt

Subbing in my values for time 0, concentration 1 i am getting:

1 = C + A

And now i don't know where to go :-( (Thanks for the help previously, much appreciated)

 

[Mentallic]

No no, in your case, x=1/(k(T-C))

Don't worry about any general formulae at this point. Read what the question gives you and see what you can make of it:

find k and C, given that the initial concentration of x is 1.0 and it is half consumed in 2 seconds.

The initial concentration is when the time is at zero. So when t=0, x=1. Substitute this into your formula.
What can you make of the second info given?

 

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Ok :-)

With that,

1 = 1/k(0-c)

I get:

1 = 1/-kc

and, am unsure how to get one of the 2 variables i don't know. If i re-arrange i always have 2 unknowns stopping me from coming to an answer.

 

[Mentallic]

That's why you need to use the second info given in order to have two equations in 2 unknowns, which then you can solve simultaneously.

 

[Barchie]

Ok i think I am onto what you are saying, are you able to let me know if I am on the right track.

2 Equations:

1: 1 = 1/k(0-c)
2: .5 = 1/k(2-c)

Rearrange the first one to equal:

1 = 1/-c
-c = 1
c = -1

Substitute c into eq#2 to give:

.5 = 1/k(2+1)
.5 = 1/(2k + k)
.5(2k+k) = 1
k+k/2 = 1
2k + k = 2
k(2+1) = 2
k = 2/(2+1)

k = 2/3 ?
c = -1 ?

Thanks for ur assistance thus far, hope I am on the right track!

 

[Mentallic]

Ok i think I am onto what you are saying, are you able to let me know if I am on the right track.

2 Equations:

1: 1 = 1/k(0-c)
2: .5 = 1/k(2-c)

Rearrange the first one to equal:

1 = 1/-c

The equation was 1 = 1/k(0-c), so what happened to the k?

But you do have the right idea, you're nearly there.

 

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My bad, think i got a little excited.

ok... My attempt again:

Equation#1: 1 = 1/k(0-c)
Equation#2: .5 = 1/k(2-c)

Rearrange Equation 1 to get:

c = 1/-k

Sub into Equation#2 to get:

0.5 = 1/k(2-c)
0.5 = 1/2k-kc
0.5 = 1/2k-k(1/-k)
0.5 = 1/2k+1
k = 1

Therefor c = 1/-1 = -1

So i figured i would test it, by popping the concentration at 2 seconds into a re-arranged formula to see if it would give me the time of 2 seconds:

Re-arranged formula:

T = (1/k+xc)/x

x = .5 c = -1 k =1

t= (1+.5*-1)/.5
t = 1

This isn't right :-( Any pointers :(

 

[Mentallic]

Yep, recheck this
0.5 = 1/2k+1
k = 1

 

[Barchie]

aha...

.5 = 1/2k+1
0.5(2k+1) = 1
k+.5 = 1
k = 1-.5
k = 0.5

But if i sub this into the original formula

x = 1/k(t-c)
x = 1/.5(2--1)
x = 1/.5*3

unfortunately still not getting it :(

This still doesn't work

 

[Barchie]

So if K = to .5 instead of 1 then popping it back into

c= 1/-k
makes c = -2

if c = -2 k =.5

Then putting it back into the first equation to check still doesn't get me the right answer:

x = 1/k(t-c)
x = 1/.5(2--2)
x = 1/2

I must have missed something :-(

 

[Mentallic]

Why have you missed something? You were already given that at time t=2 there was half consumed. This has been satisfied by the values of c and k that you found, and if you try to plug it into the other equation you should also satisfy the criteria of t=0, x=1. Your values are correct.

 

[Barchie]

ima douche.. been looking at this one for way too long now! thanks a heap for your help matey :)

 

[Mentallic]

No worries