James Bennett
Posted on Friday, 23 May, 2003 - 10:47 pm:
The surface of a circular pond of radius a is being covered by weeds. The weeds are growing in a circular region whose centre is at the centre of the pond. At time t the region covered by the weeds has radius r and area A. An ecologist models the growth of the weeds by assuming that the rate of increase of the area covered is proportional to the area of the pond not yet covered.
dA/dt = 2pr(dr/dt)
The ecologist's model therefore leads to the differential equation:
2r(dr/dt) = k(a2 -r2 ),
where k is a constant.

a) By solving the differential equation, express r in terms of t, a and k, given that r=0 when t=0.

b) Will the weeds ever cover the whole pond? Justify your answer.

Philip Ellison
Posted on Friday, 23 May, 2003 - 10:49 pm:
How far have you got?
Marcos
Posted on Friday, 23 May, 2003 - 10:57 pm:
You need to 'separate the variables'...
Try dividing by (a2 - r2 ). Then notice that 2r is the derivative of r2 .
Can you finish the first part off now? (Ask if you need more help)

Once you get the first part ask if you can't see how to do the second part (Basically, the question is really asking if r = a for some value of t)

Hope that helps (again, ask if you need more help)
Marcos
James Bennett
Posted on Saturday, 24 May, 2003 - 06:12 pm:
I got to the point of separating the variables to:
ó
õ 		r

0 
 2r/(a2 - r2)  dr = k ó
õ 		t

0 
  dt

~Is this right?!

I'm really not sure about how to do this question from here tho, and not sure if I'm using the correct integral limits?
Please go through BOTH parts to this question, showing your own working!
Thank you!
Vicky Neale
Posted on Sunday, 25 May, 2003 - 10:14 am:
Excellent - you're on the right lines. Can you see of what the left hand side is the derivative? You should be able to do by inspection integrals where the integrand (the thing you are integrating) is of the form f'(x)/f(x). Do you see how this helps?

(In general, we'd much rather help you through the question than just give you the answer, since you'll learn more by doing the question for yourself - but if you need more hints, just ask!)

Vicky
James Bennett
Posted on Sunday, 25 May, 2003 - 11:13 am:
I think I need more hints! ;)
Marcos
Posted on Sunday, 25 May, 2003 - 12:13 pm:
Hi James,
The limits you specified are a bit strange as you're using the same variable in the integrand and in the limits. If this is for an A-Level question or similar examination, then 'indefinite' integration will suffice (and be simpler to manipulate)

 If you didn't understand what Vicky was trying to say you can always use a substitution. If we let v = a2 - r2 then 2r = -dv/dt.

Subsituting these results:
ó
õ 		2r/(a2 - r2) = ó
õ 		(-dv/dt)(1/v)  dt
Can you finish this part off now?

Basically, Vicky was hinting at the fact that òf¢(x) / f(x)  dx = log|f(x)| + C. Can you see how this helps? (Basically the same thing as before.)

Marcos

James Bennett
Posted on Sunday, 25 May, 2003 - 12:32 pm:
So
ó
õ 		2r/(a2 - r2)  dr = k ó
õ 		dt


=> ln |a2 -r2 | = kt + C

r=0 when t=0: ln |a2 -0| = 0k + C
=> C = ln |a2 |

~I'm thinking that this bit is wrong....?
"-By solving the differential equation, express r in terms of t, a & k, given that r=0 when t=0.
-Will the weeds ever cover the whole pond? Justify your answer."

-Please just go through this now as I'm wasting too much time on this question!
Chris Tynan
Posted on Sunday, 25 May, 2003 - 12:48 pm:
I think you need -log|a2 -r2 | since when you differentiate log|a2 -r2 | you get a negative...

Chris
Marcos
Posted on Sunday, 25 May, 2003 - 01:01 pm:
James,
Make sure you can see what Chris means.

We now have:
-log|a2 - r2 | = kt + C
a2 - r2 = Ae-kt
By putting in t=r=0, we get A = a2
Rearranging,
r2 = a2 (1 - e-kt )

Now, will e-kt ever be < 0? Can you now answer whether or not r can ever equal (or exceed) a and hence whether the weeds will cover the pond?

Marcos
James Bennett
Posted on Sunday, 25 May, 2003 - 02:13 pm:
OH I SEE!
I've just found some v.similar q's that I have done in class previously....

I understand why
ó
õ 		2r/(a2 - r2)  dr = - log|a2 - r2|.
For the final part, I hope my conclusion is correct - please let me know: Weeds will eventually cover the pond since as t ® ¥, e-kt ® 0, and r2 = a2 at some point - i.e. area covered by weeds will equal total area of pond.

What do you think?

Kerwin Hui
Posted on Sunday, 25 May, 2003 - 02:59 pm:
Very good except the last part - when will the weeds cover the pond? Today, tomorrow, this year, or next century?

The point about the last bit is that the weeds will never cover the pond, since e-kt > 0 for every real t meaning that a2 -r2 can never be zero.

Kerwin
James Bennett
Posted on Sunday, 25 May, 2003 - 03:43 pm:
OK - I understand.
Cheers all!
James Bennett
Posted on Sunday, 25 May, 2003 - 04:04 pm:
Is Kerwin's answer to last part of the question definitely right? Just wanted to make sure cos if you plug in to your calculator some power of e - i.e. e^(-10000) you get an answer of 0? is this due to rounding by the calc?! Or is it just the fact that the unit of t is undefined making it impossible for e-kt =0
Philip Ellison
Posted on Sunday, 25 May, 2003 - 04:12 pm:
This would be due to the calculator rounding down. I'm not sure what you mean when you say "the unit of t is undefined", however, ex > 0 for all x.
James Bennett
Posted on Sunday, 25 May, 2003 - 04:25 pm:
Yes - thought so.
Thanks