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Finding equation of tangent line

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    The following point (x0,y0), is on the curve sqrtx +sqrty = 1


    Show that line equation of the tangent line in the point. (x0,y0)
    Is x/sqrtx0 + y/sqrty0 = 1

    I've found the slope which is
    -sqrty/sqrtx.
    So slope of the point is -sqrty0/sqrtx0



    2. Relevant equations


    3. The attempt at a solution
    y-y0=-sqrty0/sqrtx0(x-x0)

    Should I get the solution from the equation above?
     
  2. jcsd
  3. Dec 10, 2016 #2
    Maybe it's easier if you model the problem as one of "one" variable. If you write it as a function y of x, then it is [itex]y=x-2\sqrt{x}+1[/itex], do you know what the tangent if this is at an arbitrary point? Try to look at the equation for the tangentplane of a [itex]\textbf{R}^3[/itex]-function. The tangent-plane for such an equation at an arbitrary point [itex](x_0,y_0,z_0)[/itex] is [itex]f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0[/itex]
     
  4. Dec 10, 2016 #3
    Haven't det with R3 yet. All we have learned is finding the slope and then using the formula y-y0 = m(x-x0)
     
  5. Dec 10, 2016 #4
    SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)
     
  6. Dec 10, 2016 #5
    I think need a hint,
     
  7. Dec 10, 2016 #6
    Is it like rw
    is it like really easy to show from the equation I've written? I'm not sure which of the 4 variables should I substitute , I can't see how to get even 1 there
     
  8. Dec 10, 2016 #7

    Ray Vickson

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    You do not have 4 variables; you have only 2--namely x and y. These are the two objects that vary along the curve and along the tangent line. The other two objects (##x_0,y_0##) are just some constant input parameters. They do NOT vary along the curve or along the tangent line.
     
  9. Dec 10, 2016 #8

    Mark44

    Staff: Mentor

    Thread moved. @Dank2, questions involving derivatives should NOT be posted in the Precalc section.
     
  10. Dec 10, 2016 #9
    I meant I do not know which constant or variable should I subtitude in order to get the right form, from the main equation y= (1-sqrtx)^2 , y0=(1-sqrtx0)^2
     
    Last edited by a moderator: Dec 10, 2016
  11. Dec 10, 2016 #10

    Mark44

    Staff: Mentor

    IMO, and this is not a helpful hint.
     
  12. Dec 10, 2016 #11

    Mark44

    Staff: Mentor

    The slope at the point ##(x_0, y_0)## is ##\frac{-\sqrt{y_0}}{\sqrt{x_0}}##.
    You know the point -- ##(x_0, y_0)## -- and you know the slope of the tangent line there -- ##\frac{-\sqrt{y_0}}{\sqrt{x_0}}##. Finding the equation of a line through a given point and with a known slope should be straightforward.
     
  13. Dec 11, 2016 #12
    Yes it is (y-y_0)=−√y0/√x0*(x-x_0). but i need to show it equals :
    x/√x_0 + y/√y_0 = 1
     
    Last edited: Dec 11, 2016
  14. Dec 11, 2016 #13
    (y-y_0)=−√y0/√x0*(x-x_0) ==> y-y_0 = −√y_0*x/√x0 −√y_0*x_0/√x0
    i think that somehow i need to use y=x−2√x+1 from the main equation, i just struggle on this point to do the algebra
     
  15. Dec 11, 2016 #14

    Ray Vickson

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    You have ##\sqrt{y_0}## in your first equation and ##1/\sqrt{y_0}## in your second equation. How, then, can you get from the first to the second?
     
  16. Dec 11, 2016 #15
    1/√y_0 ? can't see it
     
    Last edited: Dec 11, 2016
  17. Dec 11, 2016 #16
    i need a hint
     
  18. Dec 11, 2016 #17

    Ray Vickson

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    You said you wanted to verify that ##x/\sqrt{x_0} + y/\sqrt{y_0} = 1##.
     
  19. Dec 11, 2016 #18
    that the tangent line equation at the point (x_0, y_0) is equal to it yes.
     
  20. Dec 11, 2016 #19
    i need to proof it
     
  21. Dec 11, 2016 #20
    i tried coming from the general equation of the tangent line by slope and point, and it didn't come out good.
     
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