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Homework Help: Equation of Tangent Line

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Give the equation of the line tangent to the curve at the given point.

    ytan^-1x = xy at (sqrt3,0)

    2. Relevant equations

    3. The attempt at a solution

    Would it be right to do an implicit differentiation or to isolate for y here?

    I isolated for y and got y=1/(tan^-1x-x)

    deriving: y' = x^2/((1+x^2)(tan^-1x-x))

    That's where I'm confused, at the initial differentiation.

    Any guidance would be greatly appreciated!
  2. jcsd
  3. May 24, 2013 #2


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    There is something funny about your equation. Are you sure it is copied correctly? If ##y\ne 0## you have ##\arctan(x) = x## and if ##y=0##, then ##x## can be anything.
  4. May 24, 2013 #3
    Yep, it's copied correctly.

    y*tan^-1(x) = x*y
  5. May 24, 2013 #4
    LCKurtz, when I plug this into Wolfram Alpha, I get a straight vertical line... There is no tangent line for a straight vertical line am I right?
  6. May 24, 2013 #5


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    Look at what I observed earlier. If ##y=0##, then ##x## can be anything. That is the ##x## axis, not the ##y## axis.
  7. May 24, 2013 #6
    Hmm, I am a little confused. To my uneducated eye, if y= 0, doesn't that just make x=0 as well?

    0*arctanx = x*0

    Is this wrong?
  8. May 24, 2013 #7


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    Please quote the post to which you are replying, like I just did. No it doesn't "make x = 0". When ##y=0## it doesn't tell you anything about x because it gives 0=0 no matter what x is. Every value of ##x## works if ##y=0##.
  9. May 24, 2013 #8
    Oh sorry about the quote thing!

    I see what you're saying. Do you think there is a typo? I haven't encountered a problem like this and I doubt it's a "thinker" question if it's throwing a retired math professor a curve ball hah.
  10. May 24, 2013 #9


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    Isn't that what I suggested in my first post? But given the question as posted, you can still answer it since the only function that equation represents is ##y=0##.
  11. May 24, 2013 #10
    How did you isolate y ? Let's try it

    Assume [tex]y \neq 0[/tex]

    We begin with

    [tex] y arctan(x) \, = \, xy [/tex]

    Moving all the terms to the left hand side (subtracting xy)

    [tex] yarctan(x) \, - \, xy \, = \, 0 [/tex]

    Factoring out y on the left hand side..

    [tex] y(arctan(x) \, - \, x) \, = \, 0 [/tex]

    Dividing both sides by (arctan(x) - x)

    [tex] y \, = \, \frac{0}{arctan(x) \, - \, x} [/tex]


    [tex] y \, = \, 0 [/tex]

    Of course you could have seen this by inspecting the original equation. If you divide both sides by y you get

    [tex]arctan(x) \, - \, x \, = \, 0 [/tex]
  12. May 24, 2013 #11
    Ah yes. In my original factoring process, I left a 1 on the right hand side for some reason. So now, thanks to the help of both you and LCKurtz, I have y = 0.

    Can the question even be answered? For y= 0 there is no tangent line. Unless, that is the answer they want...
  13. May 24, 2013 #12


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    Didn't I already answer that? Do straight lines have tangent lines?
  14. May 24, 2013 #13
    Well, y=0 is a linear equation.We could write it as y = 0x, it's slope being 0. Geometrically its the x-axis as for any x, y = 0. So it's tangent line is y=0, in essence the "tangent line" of a linear function is the function itself, although it is not as intuitive to visualize geometrically as with many nonlinear functions.

    I stumbled across this paper which is kind of interesting. If you like reading about mathematical stuff you may enjoy this.

  15. May 24, 2013 #14


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    Can you go back and double check, right back to the original source?
  16. May 24, 2013 #15
    Yeah, I quadruple checked the original source. I'll probably speak with my tutor (it's a distance course) about it.

    I really appreciate the input folks!
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