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Math Help - [SOLVED] Implicit diff question.

  1. #1
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    Question [SOLVED] Implicit diff question.

    I'm struggling with this question - I think I'm ok with the calculus its just the maniplutaion. Anyway, your help and guidence would be very much appreciated.

    The equ. of a curve is (6y^2) = (e^x)(y^3) - 2e^4x.

    Given that (a,b) is the point on the curve at which dy/dx = 0,
    show that b = 2e^a

    By substituting into the equ. obtain a further relationship between a and b, and hence find the values of a and b.

    OK

    So by implicit diff. and rearranging

    (e^x(y^3 - (8e^3x)) / 3y(4 - (e^x)y)) = 0

    I'm not really sure how to show the rest though..

    can anyone offer some help?

    D
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  2. #2
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    Quote Originally Posted by dojo View Post
    I'm struggling with this question - I think I'm ok with the calculus its just the maniplutaion. Anyway, your help and guidence would be very much appreciated.

    The equ. of a curve is (6y^2) = (e^x)(y^3) - 2e^4x.

    Given that (a,b) is the point on the curve at which dy/dx = 0,
    show that b = 2e^a

    By substituting into the equ. obtain a further relationship between a and b, and hence find the values of a and b.

    OK

    So by implicit diff. and rearranging

    (e^x(y^3 - (8e^3x)) / 3y(4 - (e^x)y)) = 0
    It looks like you solved for y' and then set it equal to 0. It would have been much easier to just put y'= 0 before solving for it.

    But, in any case, you should know that, in order that a fraction be equal to 0, the numerator must be 0- that equation leads immediately to [tex](e^x(y^3- 8e^{3x})= 0[tex] and, since e^x is never 0, y^3= 8e^{3x}.

    Since this is at (a, b), let x= a, y= b and solve for b.

    I'm not really sure how to show the rest though..

    can anyone offer some help?

    D
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  3. #3
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    Hi dojo,

    you differentiated nicely,

    good suggestion from HallsofIvy,
    notationally quite compact...

    6y^2=e^xy^3-2e^{4x}

    6y'2y=e^xy^3+e^xy'3y^2-8e^{4x}

    set y'=0

    0=e^xy^3-8e^{4x}\ \Rightarrow\ e^x\left(y^3-8e^{3x}\right)=0\ \Rightarrow\ y^3=8e^{3x}=2^3\left(e^x\right)^3=\left(2e^x\right  )^3

    \Rightarrow\ y=2e^x
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  4. #4
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    Question

    Thanks for the tips on diff - I see its much easier that way!

    Regarding the subbing in of y = 2e^x into the original equ.
    I get 24e^(2x) = 6e^(4x)


    i cant see a relationship between them at all?
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  5. #5
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    Quote Originally Posted by dojo View Post
    Thanks for the tips on diff - I see its much easier that way!

    Regarding the subbing in of y = 2e^x into the original equ.
    I get 24e^(2x) = 6e^(4x)


    i cant see a relationship between them at all?
    24e^{2x} - 6e^{4x} = 0

    6e^{2x}(4 - e^{2x}) = 0

    finish
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  6. #6
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    Quote Originally Posted by dojo View Post
    Thanks for the tips on diff - I see its much easier that way!

    Regarding the subbing in of y = 2e^x into the original equ.
    I get 24e^(2x) = 6e^(4x)


    i cant see a relationship between them at all?
    Just a few more lines and you have it, dojo,

    24e^{2x}=6e^{4x}

    \frac{24}{6}=\frac{e^{4x}}{e^{2x}}={e^{4x-2x}}=e^{2x}

    4=e^{2x}=\left(e^x\right)^2

    e^x=2

    x=ln2

    y=2e^{ln2}=4
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