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Implicit differentation problem

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    find the lines that are (a)tangent and (b)normal to the curve at the given point:
    x2 - √(3)xy + 2y2 = 5, (√3, 2)

    2. Relevant equations

    dy/dx



    3. The attempt at a solution

    i am completely confused about implicit differentiation and the chain rule. ive learned about 3 ways to do the chain rule and apparently they are all the same thing which confuses me more.

    x2 - √(3)xy + 2y2 = 5
    2y2(d/dx) = 5(d/dx) + √(3)xy(d/dx) - x2(d/dx)
    4(dx/dy) = 0 + (1/2)3-1/2y' - 2x
    (dy/dy) = (((1/2)3-1/2y' - 2x) / 4

    and i am stuck.
     
  2. jcsd
  3. Oct 6, 2009 #2

    Mark44

    Staff: Mentor

    d/dx as you are using it above doesn't mean anything. d/dx means "take the derivative with respect to x of whatever is to the right. dy/dx represents the derivative of y with respect to x. In implicit differentiation you are assuming that y is some differentiable function of x, and so are differentiating implicitly.

    So let's start again from your second line, where you moved the terms around.
    2y2 = 5 + √(3)xy - x2
    d/dx(2y2) = d/dx(5) + d/dx(√(3)xy) - d/dx(x2)
    Two of the derivatives in the equation above are very straightforward, since you are differentiating a function of x, with respect to x. The other two require the chain rule. That is, the one on the left side and the 2nd one on the right side, which also requires the product rule (which has to be done first).

    Can you carry out these differentiations?


     
  4. Oct 6, 2009 #3
    oh so you can't take the derivative of 2y2, the same as your take 2x2?

    4y(y') = √3(√3)y - 2x
    4y(y') = 3y - 2x

    is that in the right direction?
     
  5. Oct 6, 2009 #4

    Mark44

    Staff: Mentor

    Partly. 4yy' is correct and -2x is correct, but the other one is not.
    d/dx(√3xy) = √3 d/dx(xy) = ?
    Hint: product rule
     
  6. Oct 6, 2009 #5
    = √3(y+xy')
    4yy' = √3y + √3xy'
    y' = (√3y/4y) + (√3xy'/4y)
    y' = (√3/4) + (√3xy'/4y)

    closer?
     
  7. Oct 6, 2009 #6

    Mark44

    Staff: Mentor

    Closer, but still no cigar.
    Start from here.
    d/dx(2y2) = d/dx(5) + d/dx(√(3)xy) - d/dx(x2)
     
  8. Oct 6, 2009 #7
    d/dx(2y2) = d/dx(5) + d/dx(√(3)xy) - d/dx(x2)
    4yy' = √(3)y' - 2x
    y' = (√(3)y')/4y - x/2y

    any cigar yet?
     
  9. Oct 6, 2009 #8

    Mark44

    Staff: Mentor

    Nope. You didn't get the derivative right for d/dx(√(3)xy) this time. You had it in post 5, but apparently forgot that you needed the product rule.
     
  10. Oct 6, 2009 #9
    oh i didn't know if i was doing it right.

    d/dx(2y2) = d/dx(5) + d/dx(√(3)xy) - d/dx(x2)


    4yy' = √3(y+xy') - 2x
    y' = ((√3(y+xy')) - 2x)/4y
     
  11. Oct 6, 2009 #10

    Mark44

    Staff: Mentor

    That's correct, but you haven't solved for y' (aka dy/dx).
    Start with this equation and get all the terms with y' on one side, and then divide both sides by whatever is the coefficient of y'.
    4yy' = √3(y+xy') - 2x
     
  12. Oct 6, 2009 #11
    4yy' = √3(y+xy') - 2x
    4yy' + √3xy' = √3y - 2x
    (4y + √3x)y' = √3y - 2x
    y' = (√3y - 2x)/(4y + √3x)

    right?
     
  13. Oct 6, 2009 #12

    Mark44

    Staff: Mentor

    No. What did you do to go from the first equation to the second?
     
  14. Oct 7, 2009 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Surely you do not believe that [itex]\sqrt{(a+ b)}= \sqrt{a}+ \sqrt{b}[/itex]?
     
  15. Oct 7, 2009 #14

    Mark44

    Staff: Mentor

    That's not applicable in this problem. physjeff12's expression is √3(y+xy'), which is the derivative of √3 xy. At least, this is how he presented it in his first post in this thread. I don't believe he intended the x and y factors to be in the radicand.
     
  16. Oct 7, 2009 #15

    Mark44

    Staff: Mentor

    The second equation doesn't follow from the first. Do you know what you did wrong?
     
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