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Determining slope of the tangent line.

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    I've never seen such a problem before: Find the two straight lines that are perpendicular to [tex]y=0.25x[/tex] and tanget to the curve [tex]f(x)=1/x[/tex]


    2. Relevant equations

    [tex]y=0.25x[/tex] ; [tex]f(x)=1/x[/tex]

    3. The attempt at a solution

    Using the power rule, I found the derivatives of: [tex]f(x)=-x^{-2}[/tex] and [tex]y=0.25[/tex]

    But I am not sure what I do now? Do I find the slope of the equations? Do I equate y=f(x)?
     
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2
    If they are perpendicular to y=x/4 then the slope of the line is what?

    Now, to find a tangent to y=1/x.

    you need the equation for the straight line:

    [tex] y-y_0 = (x-x_0)f'(x_0)[/tex]

    what x0 is required to for the desired slope?
     
  4. Sep 20, 2009 #3
    Is it -4?

    I'm confused already :(.
     
  5. Sep 20, 2009 #4
    Any help anyone? :(
     
  6. Sep 20, 2009 #5
    All you have to do is not get mixed up, we need a line that is tangent to 1/x. It also needs to be perpendicular to that line you described.So, to get a line that is normal to x/4, you do need a line that has a gradient or slope of -4.

    [tex] y-y_0 = (x-x_0)m [/tex]

    First of all in order for the line to be tangent to 1/x it needs to parallel to it at a given point x0. The slope of f(x) is, as you say -1/x^2. The line is going to look like this:

    [tex] y-y_0 = (x-x_0)f'(x_0) [/tex]

    and

    [tex] f'(x_0)=-4 [/tex]

    Get the x0s then find y0 and plug them into the line equation.
     
  7. Sep 20, 2009 #6
    "the x0s then find y0"

    What are these? What are the equations that you come up with? :eek:
     
  8. Sep 20, 2009 #7
    Well the general equation for the line is going to take the form y=mx+c, in order to get that equation you need the slope (m) and a point on the line, which will be (x0), y0. For example if I knew my slope was 3 and that a point on my curve was (1,2) then I can work out that the equation for line is:

    [tex]y-2=3(x-1), y=3x-1[/tex]

    The same is with this problem with have here, we need to work out at what point the slope is -4 don't we. The derivative of the curve is -1/x^2.

    [tex] \frac{1}{x_0^2} = 4 \Rightarrow x_0 = \pm \frac{1}{2}[/tex]

    So you have two. Now all you do is work out the corresponding y values and use the equation for a straight line to work it out.
     
  9. Sep 20, 2009 #8
    I figred out that the functions cross at 2 and -2. I can't use these points?
     
  10. Sep 20, 2009 #9
    Just find out what value -1/x^2 = -4, that's x1. (there's two of them)

    Then plug it into y=1/x to get the y1. (there's two of them)

    Then plug it into

    y-y1=-4(x-x1)

    and you will get 2 lines.
     
  11. Sep 20, 2009 #10
    Okay, I got +/- 1/2 which gets me +/- 2

    Now I'm confused which is x and which is y :/
     
  12. Sep 20, 2009 #11
    You have two x values; plug them into f(x) = 1/x and you'll get the y values of the coordinates of the points tangent to 1/x. The plug them into the line equation y = mx + b, solve for b, then get the two equations of the tangent lines.
     
  13. Sep 20, 2009 #12
    x=-1/2
    so what is y at this point, 1/(-1/2) = -2
    So the corresponding y for -1/2 is -2. you can do the rest?
     
  14. Sep 20, 2009 #13
    Alright. I got y=-4x-4 and y=-4x+4

    Are these correct?
     
  15. Sep 20, 2009 #14
    They are correct. You should try graphing on a calculator or by hand, or look for an online graphing calculator. 1/x should be easy to graph by hand.
     
  16. Sep 20, 2009 #15
    graph2.jpg

    Red is -4x-4
    Yellow is -4x+4
    Green is x/4
    Blue is 1/x

    Nice!
     
  17. Sep 20, 2009 #16
    Thank you guys :)!
     
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