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Differentiating x² + 2xy - 3y² + 16 = 0

  1. Jun 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A curce has equation:

    x² + 2xy - 3y² + 16 = 0

    find the co-ordinates of the points on the curve where dy/dx = 0

    2. Relevant equations



    3. The attempt at a solution

    I dont have a clue. Do I convert them into parametric equations?
     
  2. jcsd
  3. Jun 8, 2008 #2

    Air

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    Differentiate Implicitly.

    You should get [itex]\mathrm{d}y/ \mathrm{d}x[/itex] within the differentiated equation twice. Factorise [itex]\mathrm{d}y/ \mathrm{d}x[/itex] (by taking it to one side) and rearrange it so that it is only on one side. Then solve for co-ordinates.
     
    Last edited: Jun 8, 2008
  4. Jun 8, 2008 #3
    i tried looking at implicit differentiation here http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node49.html
    and my book but i dont understand it.

    dy/dx (x² + 2xy - 3y² + 16)
    = 2x + 2y - 3y² + 16

    Apparently i should use the chain rule somewhere?
    Thanks
     
  5. Jun 8, 2008 #4

    Redbelly98

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    We should write that first line as

    d/dx (x² + 2xy - 3y² + 16) = d/dx (0)

    i.e., it's "d/dx" and not "dy/dx" in front, since we want to take the derivative with respect to x.

    Then use the chain rule, and when needed the product rule. For example:

    [tex]
    \frac{d}{dx}(x^2 y^4)
    = 2x \cdot y^4 + x^2 \cdot 4 y^3 \frac{dy}{dx}
    [/tex]
     
  6. Jun 8, 2008 #5
    so if i do:

    d/dx (x² + 2xy - 3y² + 16)

    i get

    2x + 2y - 3y²

    how am i ment to use the chain rule? I dont see.

    dy/dx = dy/du x du/dx is the chain rule. What is u and y

    Thanks :)
     
  7. Jun 8, 2008 #6

    Redbelly98

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    Let's suppose u = y^4, and we want to differentiate that with respect to x.

    d/dx (y^4)
    = d/dx (u)
    = du/dx
    and by the chain rule
    = (du/dy) * (dy/dx)
    = 4 y^3 * (dy/dx)

    Moreover:

    d/dx (2xy) is not 2y. You need the product rule here, applied to the functions "x" and "y".
     
  8. Jun 8, 2008 #7

    tiny-tim

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    Hi thomas49th! :smile:

    Simple question:
    Simple answer:

    y is 3y², and u is y. :biggrin:

    oh yes they are …

    d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx. :smile:
     
  9. Jun 8, 2008 #8
    im still lost completely. I know the chain rule. I dont see how y is 3y² nor why u is y. Can you show me why. Why do you set u equal to y?
    Thanks :)
     
  10. Jun 8, 2008 #9

    tiny-tim

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    Hi thomas49th! :smile:

    Your original problem was implicit differentiation of x² + 2xy - 3y² + 16:

    that is, d/dx of x² + 2xy - 3y² + 16.

    Obviously , d/dx of x² + 16 is really easy, and d/dx of 2xy is fairly easy … I assume you can do that …

    So that leaves d/dx of 3y².

    You know how to do d/dy of 3y².

    So you use the chain rule: d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx.

    Does that make sense? :smile:
     
  11. Jun 8, 2008 #10
    is d/dx of 2xy
    = 2(x+y)
    = 2x + 2y ???

    i don't think it is. I've used the product rule (dy/dx = v du/dx + u dv/dx).

    I know about the chain rule, quotient and product rule. I just dont know how to differeriate parametric equations and equations with both x and y in (i can do y = ... but not y² + yx + x²).

    d/dx of x² + 16 is really easy, it's 2x. easy peasy. This implicit differentiation is really confusing me. I cant find it in my book!!

    Thanks :)
     
  12. Jun 8, 2008 #11
    thomas the idea is really simple.

    if you have y^2 then different as if it was x , you should have 2y. simple right?

    Now all you have to do is multiply the 2y by dy/dx. Just write 2y . dy/dx and that's the answer.

    So whenever you different with the letter y, do it as if it was x but just multiply it by dy/dx. This is because of the chain rule, since y is a function of x.
     
  13. Jun 8, 2008 #12

    tiny-tim

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    d/dx of 2xy

    = d(2xy)/dx = yd(2x)/dx + 2x dy/dx = 2y + 2x dy/dx.
     
  14. Jun 8, 2008 #13
    so the final answer is

    2x + 2y + 2x . (dy/dx) -6y . dy/dx

    NOTE is it 2y + 2x . (dy/dx) or 2y . (dy/dx) + 2x

    So now i've differntiated it i need to find these co-ordinates

    Well ill set 2x +2y + 2x . (dy/dx -6y . dy/dx = 0

    2x + (2x . dy/dx) = (6y . dy/dx) - 2y

    But now im stuck with how to treat these dy/dx...

    Thanks :)
     
  15. Jun 8, 2008 #14

    tiny-tim

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    You're right … it's definitely the first one.

    (2y . (dy/dx) would be the d/dx of y²)

    Then, as you said in your first post:
    So just put dy/dx = 0 in 2x + 2y + 2x . (dy/dx) - 6y.dy/dx

    that gives you … ? :smile:

    (going to bed now … :zzz:)
     
  16. Jun 8, 2008 #15
    ahhh YOU ACUTLALY SUBST FOR dy/dx = 0
    2x + 2y = 0
    y = x

    please say that is right...

    Thanks :)
    Night Night.
     
  17. Jun 8, 2008 #16

    Redbelly98

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    Not quite but you are really close.

    2x + 2y = 0
    Divide both sides of equation by 2:
    x + y = 0
    And then
    x = ???
     
  18. Jun 8, 2008 #17

    HallsofIvy

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    No. It is NOT right.

    2x+ 2y= 0 is right. But you solved that equation wrong!
     
  19. Jun 9, 2008 #18

    tiny-tim

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    … yawn … stretch … rubs eyes …

    It's right!! :biggrin:

    well … apart from the minus sign!! :rolleyes:

    Now put y = -x into the original curve equation x² + 2xy - 3y² + 16 = 0, to find the points where dy/dx = 0, as required. :smile:
     
  20. Jun 9, 2008 #19
    y = -x

    silly mistake about the minus



    right anyhow

    x² + 2xy - 3y² + 16 = 0

    = x² - 2(x)² - 3x² = -16
    -4x² = -16
    x = +-2


    SOOOOOOOO ill sub x = - y

    y² - 2y² - 3y² = 0

    well that's easy y = +-2

    so (2,-2),(-2,2)....how do i not know it's (-2,-2) (2,2)?
     
  21. Jun 9, 2008 #20

    tiny-tim

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    Hi thomas49th! :smile:
    erm … you're missing the obvious … you have x = ±2, and x = -y.

    So you don't need to put y back into the equation, do you? :wink:
     
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