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P3 Exam Question

The surface area 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 regoin 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.

(i) Show that dA/dt = 2(pi)r dr/dt

(ii) Hence show that the ecologist's model leads to the differential equation, 2r (dr/dt) = k(a^2 - r^2), where k is a constant.

(iii) By solving the differential equation in part (ii), express 'r' in terms of 't', 'a' and 'k', given that r=0 when t=0.

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

*****************************************************
Thanks to you for reading this far. I've managed to do parts (i) and (ii) but am stuck on (iii). I've got the markscheme but even after looking at that I don't know what to do. A clear walkthrough would be appreciated.

Answers are (for (iii) and (iv)), r = a sqrt(1 - e^-kt);
No since r->a as t->infinity

Reply 1

Bhaal85

9

Such a ghastly question.

Reply 2

Bhaal85

9

Sang

The surface area 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. A time 't' the regoin 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.

(i) Show that dA/dt = 2(pi)r dr/dt

(ii) Hence show that the ecologist's model leads to the differential equation, 2r (dr/dt) = k(a^2 - r^2), where k is a constant.

(iii) By solving the differential equation in part (ii), express 'r' in terms of 't', 'a' and 'k', given that r=0 when t=0.

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

*****************************************************
Thanks to you for reading this far. I've managed to do parts (i) and (ii) but am stuck on (iii). I've got the markscheme but even after looking at that I don't know what to do. A clear walkthrough would be appreciated.

Answers are (for (iii) and (iv)), r = a sqrt(1 - e^-kt);
No since r->a as t->infinity

OK

't' = the region covered

Area (A) = Pi(R)^2 therefore dA/dr = 2pi(r)

dA/dt = dA/dr * dr/dt = 2pi(r) * dr/dt

The rest, I hate with a passion!

Reply 3

Sang

OP

tell me about it, hope i dont run into anything like this tommorow.

Reply 4

Sang

OP

i'm ok for parts 1 and 2. It's 3 i need help with.

Reply 5

Bhaal85

9

Sang

i'm ok for parts 1 and 2. It's 3 i need help with.

∫2r/(a^2-r^2) = ∫k dt + c

Reply 6

Bhaal85

9

Is 'a' a constant?

Reply 7

Mysticmin

10

Bhaal85

Is 'a' a constant?

yes, it's the radius of the pond.

Reply 8

Bhaal85

9

Mysticmin

yes, it's the radius of the pond.

brb

Reply 9

Sang

OP

Right, INT(2r/(a^2 - r^2)) = -ln(a^2 - r^2). Thats whats written in the markscheme. It resembles INT(f'(x)/f(x)) = ln|f(x)| + K. Could you just kindly explain to me how the minus sign gets there?

Reply 10

meepmeep

Sang

The surface area 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. A time 't' the regoin 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.

(i) Show that dA/dt = 2(pi)r dr/dt

(ii) Hence show that the ecologist's model leads to the differential equation, 2r (dr/dt) = k(a^2 - r^2), where k is a constant.

(iii) By solving the differential equation in part (ii), express 'r' in terms of 't', 'a' and 'k', given that r=0 when t=0.

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

*****************************************************
Thanks to you for reading this far. I've managed to do parts (i) and (ii) but am stuck on (iii). I've got the markscheme but even after looking at that I don't know what to do. A clear walkthrough would be appreciated.

Answers are (for (iii) and (iv)), r = a sqrt(1 - e^-kt);
No since r->a as t->infinity

For part (ii), we have 2r (dr/dt) = ka^2-kr^2 where k and a are constants.

dr/dt = k[(a^2)/r-r]/2
dr/[(a^2)/r-r]=kdt/2
rdr/[a^2-r^2]=kdt/2
Integrating both sides gives:
-0.5ln(a^2-r^2)=kt/2+lnc
ln(a^2-r^2)=-kt+lnC
a^2-r^2=Ce^-kt
r^2=a^2-Ce^-kt

Given that r=0 when t=0, 0=a^2-C

Thus, C=a^2
r^2=a^2(1-e^-kt)
r=asqrt(1-e^-kt)

Reply 11

m:)ckel

14

okay, i can do (iii):

∫2r/(a^2-r^2) dr = ∫k dt

use the substitution: u = a^2 - r^2

du/dr = -2r
dr = -(du/2r)

so the integral now becomes:

- ∫ (1/u)du = ∫ k dt

-ln u = kt + c

- ln(a^2 - r^2) = kt + c

t=0, r=0 => c = -ln(a^2)
just rearrange etc.....should get the right answer

Reply 12

meepmeep

Sang

Right, INT(2r/(a^2 - r^2)) = -ln(a^2 - r^2). Thats whats written in the markscheme. It resembles INT(f'(x)/f(x)) = ln|f(x)| + K. Could you just kindly explain to me how the minus sign gets there?

If you differentiate the bottom bit with respect to r, as a is a constant, you get -2r. So you need to times the lot by -1 after integrating as the top bit has the wrong sign (+ instead of -) so it ends up as -ln(a^2-r^2) instead of just ln(a^2-r^2).

As for the last bit, e^t>0 for all real t. The same is true of e^-t. As t tends to infinity, e^-t tends to 0 (but never reaches it), so 1-e^-kt never quite reaches 1, so r^2 tends to a^2 as t tends to infinity, but never reaches it (this is an overly long explanation. Just saying, as t tends to infinity, e^-kt tends to 0 and r^2 tends to a^2 so it never covers the whole pond would do).

Reply 13

Ralfskini

10

Is it lnl(a+r)/(a-r)l=kt.

Reply 14

MalaysianDude

4

Bhaal85

Such a ghastly question.

1) dA/dt = dr/dt x dA/dr
it is known that a circle, A = (pi)r²
therefore dA/dr = 2(pi)r
thus
dA/dt = 2(pi)r x dr/dt (proven)

2)
It is given that the rate of increase of the area covered is proportional to the area of the pond not yet covered.
therefore
dA/dt = k(pi)(a²-r²)

therefore from the equation dA/dt = dr/dt x dA/dr
put dA/dt = k(pi)(a²-r²) and
dA/dr = 2(pi)r

2r (dr/dt) = k(a^2 - r^2) (proven)

3) *give me some time to figure out *

Reply 15

Ralfskini

10

.. and it will never cover the whole pond by the way.

Reply 16

Sang

OP

Ok, thanks guys, i think i get it now.

Reply 17

Mysticmin

10

It's not any of the mathematical principles invovled in these rates of change questions, it's the actual processing written problems into mathematical symbols. I know I sound lame, but hey. Those questions are long winded.