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Math Help - Derifferentiating a "lemniscate"

  1. #1
    Member Chokfull's Avatar
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    Derifferentiating a "lemniscate"

    "Find the points on the lemniscate (x^2+y^2)=25(x^2-y^2)"

    I found the derivative of the equation, and solved for y', getting (x^2+y^2)=\frac {25} {4}, but i don't know where to go from here. I was given a couple hints, one of which being "Is the observation x^2-y^2=(x^2+y^2)-2y^2 helpful? And I don't think it is, but it's obvious where they got it--from x^2+y^2=x^2+y^2
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  2. #2
    Flow Master
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    Quote Originally Posted by Chokfull View Post
    "Find the points on the lemniscate (x^2+y^2)=25(x^2-y^2)" Mr F says: Points that satisfy what condition?

    I found the derivative of the equation, and solved for y', getting (x^2+y^2)=\frac {25} {4}, but i don't know where to go from here. I was given a couple hints, one of which being "Is the observation x^2-y^2=(x^2+y^2)-2y^2 helpful? And I don't think it is, but it's obvious where they got it--from x^2+y^2=x^2+y^2
    (x^2+y^2)=25(x^2-y^2) .... (1)

    dy/dx = whatever .... (2)

    Solve equations (1) and (2) simultaneously.
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  3. #3
    Member Chokfull's Avatar
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    Right, sorry, i never finished my sentence :P
    I meant the points at which the tangent line has a slope of 0
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  4. #4
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    Hello Chokfull

    I think you also missed out a power of 2. The equation should be
    (x^2+y^2)^2 = 25(x^2-y^2)
    So, when we differentiate:
    2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=50x -50y\frac{dy}{dx}
    and putting \frac{dy}{dx}=0 gives:
    x^2+y^2 = \frac{25}{2} (I think - not \frac{25}{4})
    Now we put this back into the original equation, using the hint that x^2-y^2 = (x^2+y^2)-2y^2:
    \left(\frac{25}{2}\right)^2=25\left(\frac{25}{2}-2y^2\right)
    which you can now solve for y, and hence for (x,y).

    Grandad
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  5. #5
    Member Chokfull's Avatar
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    I see, thanks a lot! Now i didn't see before that you could replace the x^2-y^2 with x^2+y^2 and use that to get a contant in place of every x--thus allowing me to solve for y!
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