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Homework Help: Finding points on curve where tangent line given

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

    Find all points on the curve x^2 * y^2 + xy = 2 where the slope of the tangent line is -1.

    2. Relevant equations

    y' = -1

    3. The attempt at a solution

    I got the y' = [ -2xy^2 - y ] / [ 2x^(2)y + x ] to be the gradient which I am sure is right.
    Then I subbed in y' = -1 and tried to solve for y, where I reached the line -2x^(2)y + 2xy^(2) + y = x. Is there a way to factor out the y?
     
  2. jcsd
  3. Apr 21, 2010 #2

    Mark44

    Staff: Mentor

    I'll take your word that your derivative is correct.

    Since the slope is -1, you have
    -1 = [ -2xy^2 - y ] / [ 2x^(2)y + x ], which implies that 2xy^2 + y = 2yx^2 + x, or equivalently, 2xy^2 + y - 2yx^2 - x = 0.

    This can be factored.
     
  4. Apr 21, 2010 #3
    Oh I see. I got to:

    ( 2xy + 1)( y - x ) = 0
    y = - 1 / 2x OR y = x

    subbed it back into the original equation and for the y = -1 / 2x I got x^2(-1 / 2x)^2 + x(-1 / 2x) = 2 which simplifies to (1 / 4) - 2 = 2 and for y = x I got x^(2)(x)^2 + x(x) = 2 which simplified to x^(2)(x^(2) + 1) = 2 and I cant solve for any of the x values :(
     
  5. Apr 21, 2010 #4

    Mark44

    Staff: Mentor

    If y = x, then the original equation can be written as x^4 + x^2 - 2 = 0. This is quadratic in form and can be factored into two real solutions and two imaginary solutions.

    If y = -1/(2x), the original equation can be written as x^2/(4x^2) - x/(2x) = 0, or 1/4 - 1 = 0, which is impossible.
     
  6. Apr 22, 2010 #5
    For it x^(2) (x^2 + 1) =2 dont you get x = +/- (2)^1/2 OR x = +/- (2-1)^1/2?

    I got 1/4 - 1/2 = 2 as there is a 2 or the RHS of x^2/(4x^2) - x/(2x) = 2
     
  7. Apr 22, 2010 #6

    Mark44

    Staff: Mentor

    This isn't how factoring works. If a*b = 2, there are an infinite number of possibilities for a and b. However, if a*b = 0, then you can say for sure that a = 0 or b = 0.
    Factor the left side of x^4 + x^2 - 2 = 0 into two binomials.
    Yes, you get 1/4 - 1/2 = 2, not 1/4 - 1/2 = 0 as I said. Either way, there are no solutions when y = -1/(2x).
     
  8. Apr 22, 2010 #7
    Oh ok I got that. Now my final answer was (1, -2) and (1, 1) and also (-1, 2) and (-1, 1), yet the answer at the back of the book only said (1, 1) and (-1, 1). :confused:
     
  9. Apr 22, 2010 #8

    Mark44

    Staff: Mentor

    It goes back to what I said in post 4.
    This assumption here is that y = x. From that you get x^4 + x^2 - 2 = 0, so x = 1 or x = -1. Since by assumption y = x, then the points where y' = -1 are at (1, 1) and (-1, -1). I checked both of these against your derivative function and got -1 at both. Are you sure you book has (-1, +1) as one of the two points?
     
  10. Apr 23, 2010 #9
    Sorry yeah you right it (-1, -1) made a typing error. I looked at your 4th post but still not sure as I used the two real solutions (x = -1 and x = 1) and subbed them back into the original equation to get a quadratic for y on both occasions, which is why I have to 4 different points as opposed to the correct 2. It seems defined to me when I subbed the (1, -2) and the (1, 1) back into the equation which gave 2 = 2, which is true.
     
  11. Apr 23, 2010 #10

    Mark44

    Staff: Mentor

    When you set the derivative equal to -1, there were two cases: y = x and y = -1/(2x), the latter of which can't happen.

    Assuming y = x, you solved the equation x^4 + x^2 - 2 = 0, and found that x = 1 or x = -1.

    If x = 1, y = 1. If x = -1, y = -1. Period. These are the only two points on the graph of (xy)^2 + xy = 2 at which the derivative equals -1.

    Your mistake seems to be that you substituted (1, -2) and whatever the other point happens to be, into the original equation and got a true statement. All that says is that (1, -2) is a point on that curve. If you substitute this point into the derivative expression, you won't get -1.
     
  12. Apr 23, 2010 #11
    Oh ok I understand. Thanks
     
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