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Finding the slope of the tangent

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the slope(s) of the tangent at x = 1 for 2y^2 - xy -x^2 = 0

    2. Relevant equations



    3. The attempt at a solution

    Ive figured the derivative to be 2x+4/4y+x

    But I dont know how to make this equation work to find a y value so I can find the slope. I tried singling out a y value and got y = 1 but im not sure if thats the correct way to go about doing this question.

    Thank You for your help!
     
  2. jcsd
  3. Dec 28, 2008 #2

    Dick

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    You have a sign error in the derivative, I'd check it. But as to your original question the original curve is quadratic in y. You can have two different values of y that are both on the curve at x=1. They will give you two different slopes. If the problem doesn't give you any clue as to which to use, you'd better work out both.
     
  4. Dec 28, 2008 #3
    Can you point out the sign error for me? I tried working it out again and I got the following

    4y dy/dx - y + x dy/dx - 2x = 0

    Which led me to 2x + y / 4y + x = dy/dx

    As for the slopes, Im not sure how to go about finding y for the quadratic. Its the xy thats throwing me off.

    Thank you for your help though!
     
  5. Dec 28, 2008 #4

    Dick

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    I get 4y dy/dx - y - x dy/dx - 2x = 0. How did you get a + sign on the x dy/dx term? Don't worry about the xy term. Substitute x=1 into the equation first, then solve for y.
     
  6. Dec 28, 2008 #5
    I used the product rule for derivatives to differentiate the xy.
     
  7. Dec 28, 2008 #6

    Dick

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    Sure, I'm with you so far. But then shouldn't both of the terms you get from the product rule have a minus sign in front of them, since xy is SUBTRACTED?
     
  8. Dec 28, 2008 #7
    okay i just thought of this and I get it now, my equation should have been

    4y dy/dx -(y + x dy/dx) - 2x = 0


    which leads me to the derivative that you got. Thank you. Now onto the slope. If I plug 1 into the equation to get a y value I get

    2y^2 - 1y - 1^2 = 0
    = 2y^2 -y - 1 = 0

    Do I combine the y in the equation before solving for y?
     
  9. Dec 28, 2008 #8

    Dick

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    You must have solved a quadratic equation before, right? You can either try to factor, complete the square or use the quadratic equation. Your choice.
     
  10. Dec 28, 2008 #9
    Thank you, i remember that now.

    For y I got 1 and -1/2

    What do you think?
     
  11. Dec 28, 2008 #10

    Dick

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    That's what I got.
     
  12. Dec 28, 2008 #11
    Thanks Dick! I apprecaite your help very much.

    Would you be able to look at the other question I posted up in a couple of threads below?
     
  13. Dec 28, 2008 #12
    Oh one more thing, for the slope for this question, I got 1 and 2/3. Is the 2/3 right or is it - 1/6?
     
  14. Dec 28, 2008 #13

    Dick

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    If I put y=(-1/2) into (2+y)/(4y-1) I don't get either 2/3 or -1/6. What are you doing?
     
  15. Dec 28, 2008 #14
    I got the 2/3 by putting x and y into the deriviative giving me, 2(1) - 1/2 / 4(-1/2) - 1. I cancelled out the -1/2

    I got the - 1/6 by mistake, I actually meant -2/3 but I got that by working out the equation. I got 3/2 / -3 which I turned into - 2/3
     
  16. Dec 28, 2008 #15

    Dick

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    (3/2)/(-3)=(-1/2). I think you might want to review arithmetic with fractions, it looks like you have similar problems in the other thread.
     
  17. Dec 28, 2008 #16
    arrgg. im not even thinking straight. I added the numerator while multiplying the denomenator. Its such a stupid mistake. Thanks for the help though!
     
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