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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

(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

Scroll to see replies

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

(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!

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

(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)

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

∫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

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?

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).

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 *

- Messed up CIE Math P3 and Physics P4.
- last min history and english alevel help!!!!
- Leaving A level mathematics a week before the exam.
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- error
- Maths grade
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- written wrong numbers to answers in exam? HELP!
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- Getting ready for the exams.
- Physics
- P3 A level international maths exam edexcel Jan 08 2024
- Retaking Exam
- CAIE AL Maths Grade Boundaries

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