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    Hi, I've been working on some C4 differential equation questions and I have a few questions. I have attached a scan of my workings and a copy of the questions.

    For Q12, I am able to solve most of the question, but I'm unsure how you would "show that this model satisfies the differential equation dh/dt = -k(2h-h^2)^1/2. Do you just insert h=0 and dh/dt=0?

    For Q14, I'm not sure how to answer b), surely if you only use N=100 and t=0, then you would get a straight line graph? Also, for part c, my answer is slightly different to what they have. Mine is 1/500ln(n/[4(500-n)]) = 1/5000t whereas in the textbook it says 1/500ln(4n/500-n)=1/5000t. I think their answer is different as they placed the constant on the left rather than on the right (like I did), so they got a different value for c. But as I have a different value of c, it's completely wrong?

    For Q15, for b) I'm not sure where I substitute -1 into. And I don't really understand what c/d are saying.



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    (Original post by Bazinga?)
    .
    For q12, I've arrived at an inelegant solution. First consider \ h = 1- \sin(kt) differentiate and then manipulate the original equation to attain the given result.
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    (Original post by joostan)
    For q12, I've arrived at an inelegant solution. First consider \ h = 1- \sin(kt) differentiate and then manipulate the original equation to attain the given result.
    What do you mean by an inelegant solution?

    So when you integrate h=1-sin(kt) then you get dh/dt = -cos(kt). This is probably wrong, but do you equate the dh/dt's? So you get -cos(kt) = -h^1/2, so cos(kt) = h^1/2. Then can you substitute h=(1-1/2t)^2? However if I do this, I lose the h's and everything is in terms of t?
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    (Original post by Bazinga?)
    What do you mean by an inelegant solution?

    So when you integrate h=1-sin(kt) then you get dh/dt = -cos(kt). This is probably wrong, but do you equate the dh/dt's? So you get -cos(kt) = -h^1/2, so cos(kt) = h^1/2. Then can you substitute h=(1-1/2t)^2? However if I do this, I lose the h's and everything is in terms of t?
    Inelegant as in there's probably an easier way.
    \frac{dh}{dt} = -k\cos(kt)
    Spoiler:
    Show
    Chain rule.

    When I said original equation I meant rewrite \ h = 1-\sin(kt)
    Spoiler:
    Show
    Consider how sin and cos are related.
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    (Original post by Bazinga?)
    Hi, I've been working on some C4 differential equation questions and I have a few questions. I have attached a scan of my workings and a copy of the questions.

    For Q12, I am able to solve most of the question, but I'm unsure how you would "show that this model satisfies the differential equation dh/dt = -k(2h-h^2)^1/2. Do you just insert h=0 and dh/dt=0?
    You need to insert h=1-sinkt in both side and to show the equation
    For Q14, I'm not sure how to answer b), surely if you only use N=100 and t=0, then you would get a straight line graph? Also, for part c, my answer is slightly different to what they have. Mine is 1/500ln(n/[4(500-n)]) = 1/5000t whereas in the textbook it says 1/500ln(4n/500-n)=1/5000t. I think their answer is different as they placed the constant on the left rather than on the right (like I did), so they got a different value for c. But as I have a different value of c, it's completely wrong?
    For a) sub N=100 in the RHS you will get 8 for dN/dt as rate
    for b) Yes rhis the gradient line of the curve for N=constat
    fe. dN/dt=8 (for N=100) -> N=8t+c -< this c is 100 for t=0 => N=8t+100
    for other N=const you will get other lines. These lines cover the curve
    for c)
    1/(N(500-N))=A/N+B/(500-N)->(B-A)=0 and 500A=1 ->A=B and A=1/500
    So 1/500 (1/N+1/(500-N))dN=1/5000 dt
    from this you will get ln(N*(500-N)=1/10*t+C
    For Q15, for b) I'm not sure where I substitute -1 into. And I don't really understand what c/d are saying.
    you need to substitute y=1/4(x+1)^2 into dy/dx=Vy and to show the equation
    y>=0 because of Vy -> x>-1 after differentiation
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    (Original post by joostan)
    Inelegant as in there's probably an easier way.
    \frac{dh}{dt} = -k\cos(kt)
    Spoiler:
    Show
    Chain rule.

    When I said original equation I meant rewrite \ h = 1-\sin(kt)
    Spoiler:
    Show
    Consider how sin and cos are related.
    Ah yes, I thought I misinterpreted what you originally said. Thanks again for your help!

    (Original post by ztibor)
    You need to insert h=1-sinkt in both side and to show the equation

    For a) sub N=100 in the RHS you will get 8 for dN/dt as rate
    for b) Yes rhis the gradient line of the curve for N=constat
    fe. dN/dt=8 (for N=100) -> N=8t+c -< this c is 100 for t=0 => N=8t+100
    for other N=const you will get other lines. These lines cover the curve
    for c)
    1/(N(500-N))=A/N+B/(500-N)->(B-A)=0 and 500A=1 ->A=B and A=1/500
    So 1/500 (1/N+1/(500-N))dN=1/5000 dt
    from this you will get ln(N*(500-N)=1/10*t+C
    This is what I got, and you work out c to be 1/500ln0.25. However this is not true (well it's wrong according to the answers).
    you need to substitute y=1/4(x+1)^2 into dy/dx=Vy and to show the equation
    y>=0 because of Vy -> x>-1 after differentiation
    I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.
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    (Original post by Bazinga?)
    Ah yes, I thought I misinterpreted what you originally said. Thanks again for your help!



    This is what I got, and you work out c to be 1/500ln0.25. However this is not true (well it's wrong according to the answers).
    I wrote
    lnN\cdot (500-N)=\frac{t}{10}+C


    I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.[/QUOTE]
    \frac{dy}{dx}=\frac{d}{dx}(\frac  {1}{4}(x+1)^2)=\frac{1}{2}(x+1)
    \sqrt{y}=\sqrt{\frac{1}{4}(x+1)^  2}=\frac{1}{2}(x+1)&gt;=0 \rightarrow x&gt;=-1
    \sqrt{anything} is nonnegative by definition
    and anything \ge 0 for real set
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    (Original post by ztibor)
    I wrote
    lnN\cdot (500-N)=\frac{t}{10}+C


    I've substituted y=1/4(x+1)^2 into the dy/dx=y^1/2, but I don't really follow what I should do next.
    \frac{dy}{dx}=\frac{d}{dx}(\frac  {1}{4}(x+1)^2)=\frac{1}{2}(x+1)
    \sqrt{y}=\sqrt{\frac{1}{4}(x+1)^  2}=\frac{1}{2}(x+1)&gt;=0 \rightarrow x&gt;=-1
    \sqrt{anything} is nonnegative by definition
    and anything \ge 0 for real set[/QUOTE]

    For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

    I understand 15 now, thanks
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    (Original post by Bazinga?)
    \frac{dy}{dx}=\frac{d}{dx}(\frac  {1}{4}(x+1)^2)=\frac{1}{2}(x+1)
    \sqrt{y}=\sqrt{\frac{1}{4}(x+1)^  2}=\frac{1}{2}(x+1)&gt;=0 \rightarrow x&gt;=-1
    \sqrt{anything} is nonnegative by definition
    and anything \ge 0 for real set
    For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

    I understand 15 now, thanks [/QUOTE]

    I made a mistake so you are right
    SO
    ln\frac{N}{500-N}=\frac{1}{10}t+C
    \frac{N}{500-N}=c\cdot e^{\frac{t}{10}}
    so
    c=1/4 or C=ln(1/4)
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    (Original post by ztibor)
    For 14, I understand what you've done, but why can't I do what I did? I don't see where i hav gone wrong. And with what you're saying 1/500lnN(500-N) = 1/10t + c, when t=0 and N=100, then ln100(400) = c, so c = ln40000?

    I understand 15 now, thanks
    I made a mistake so you are right
    SO
    ln\frac{N}{500-N}=\frac{1}{10}t+C
    \frac{N}{500-N}=c\cdot e^{\frac{t}{10}}
    so
    c=1/4 or C=ln(1/4)[/QUOTE]

    Surely you'd sub in t=0 and N=100 to find c?
 
 
 
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