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Horizontal tangent points (implicit differentiation)

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the points on the lemniscate: 2( x^2 + y^2 )^2 = 25( x^2 - y^2 ) where the tangent is horizontal

    2. Relevant equations

    Horizontal tangent: y' = 0

    3. The attempt at a solution

    I got the correct gradient of y' = [ 50x - 8x^3 - 8y^2 ] / [ (8x^2)y + 50y + 8y^3 ], and then solved for y = + ( -8x^3 + 50x)^1/2 ), after which I subbed this y into the original equation and tried to solve for x, whereby I reached a dead end. The last line I got to was complicated when solving for x: 128x^6 - 32x^5 + 2x^4 - 1800x^23 + 175x^2 + 1250x = 0.
     
  2. jcsd
  3. Apr 21, 2010 #2

    LCKurtz

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    To tell the truth, I didn't work through your stuff, but here's another idea that will work. First, notice that your equation has all symmetries so if you find an HTL in the first quadrant you will have three more by symmetry.

    Put your equation into polar form r = f(θ), then calculate y' by using the polar formula:

    [tex]\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

    = \frac{r(\theta)\cos\theta+r'(\theta)sin\theta}{-r(\theta)\sin\theta+r'(\theta)cos\theta}[/tex]

    You might want to get r' implicitly when you get there. To get a slope of 0 you just need to think about getting the numerator 0. With a bit of work you should be able to find a θ and then (x,y) that works.
     
  4. Apr 21, 2010 #3
    I havent learnt polar form, in a test we are expected to do it the hard way:frown:
     
  5. Apr 21, 2010 #4

    LCKurtz

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    Well, looking at your problem in rectangular coordinates, I get:

    [tex]\frac{dy}{dx}= \frac {8(x^2+y^2)x-50x} {8(x^2+y^2)y +50y}[/tex]

    You can set the numerator to zero, factor out x and solve the other factor for x in terms of y. Plug that back in to your original equation and you get a 4th degree equation in x, but no odd powers, so it is a quadratic equation in x2 which you can solve for x2, then for x. Seems a little long for an exam, but you can work through it.
     
  6. Apr 22, 2010 #5
    Sorry but I dont understand. I tried it again and found a mistake in my previous calculation and got:

    0 = [ 50x - 8x^3 - 8y^2 ] / [ (8x^2)y + 50y + 8y^3 ]
    0 = 2( 25x - 4x^3 - 4y^2 ) then solved for y^2
    y^2 = (25 / 4)x - x^3 subbed that back into my original equation:
    2[ x^(2) + (25 / 4)x - x^3]^2 = 25[ x^(2) - (25 / 4)x + x^3 ]

    which seems right to me, but now I am not sure how to simplify it
     
  7. Apr 22, 2010 #6

    LCKurtz

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    By my calculations, you are missing an x in the last term in the numerator. It should be -8y2x, so you get:

    x(50-8x2-8y2)

    in the numerator.
     
  8. Apr 22, 2010 #7
    Oh ok I got y^(2) = -2x^(2) + (25x / 2) then subbed it back into the equation to get

    2[x^(2) - 2x^(2) + (25x / 2)] = 25[x^(2) + 2x^(2) - (25x / 2)]
    2x^2 - 4x^(2) + (25x / 2) = 25x^(2) + 50x^(2) - (625x / 2)

    and I cant see that getting factorised anytime soon
     
  9. Apr 22, 2010 #8

    LCKurtz

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    Once again, the first thing you write down is wrong. You are apparently in a calculus class and you ought to be able to do simple algebra correctly. Once you correctly solve for y2 you should get a very easy equation to solve for x2 then x.

    Also, please learn to use the X2 button to make your equations more readable.
     
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