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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
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    Such a ghastly question.
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    (Original post by 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!
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    tell me about it, hope i dont run into anything like this tommorow.
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    i'm ok for parts 1 and 2. It's 3 i need help with.
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    (Original post by Sang)
    i'm ok for parts 1 and 2. It's 3 i need help with.
    ∫2r/(a^2-r^2) = ∫k dt + c
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    Is 'a' a constant?
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    (Original post by Bhaal85)
    Is 'a' a constant?
    yes, it's the radius of the pond.
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    (Original post by Mysticmin)
    yes, it's the radius of the pond.
    brb
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    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?
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    (Original post by 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)
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    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
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    (Original post by 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).
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    Is it lnl(a+r)/(a-r)l=kt.
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    (Original post by 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 *
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    .. and it will never cover the whole pond by the way.
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    Ok, thanks guys, i think i get it now.
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    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.
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    In the heinemann P3 book, I have found ONE practice question in the revision exercise, on newton's laws of cooling :mad:
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    (Original post by Mysticmin)
    In the heinemann P3 book, I have found ONE practice question in the revision exercise, on newton's laws of cooling :mad:
    The wordy questions are just plain gay, I signed up for Maths not English Literature,.
 
 
 
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