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Trouble with Tangent of a Line

  1. Oct 18, 2006 #1
    I tried many thigns with this one:

    Determine the equation of the line that is perpendicular to the tangent to [tex]y=5x^2 at (1,5)[/tex]

    What I did was use the formula for instantaneous rate of change finding the equation of the secant.

    [tex](5x^2-5)/(x-1)[/tex]

    from there I found that the slope of the tangent is about 10 and from there i'm lost
     
  2. jcsd
  3. Oct 18, 2006 #2

    radou

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    Maybe it would be easier to differentiate the function and plug in the x-coordinate to find the slope of the tangent.
     
  4. Oct 18, 2006 #3
    but I have the slope....
     
  5. Oct 18, 2006 #4

    radou

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    Ok then. In what relation are the slope of a line with the slope of a line perpendicular to it?
     
  6. Oct 18, 2006 #5
    duh negative reciprocal
     
  7. Oct 18, 2006 #6

    radou

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    So, all you have to do now is find the equation of the line which passes through the given point and whose slope is, as you stated, negative reciprocal to the slope of the tangent at that very same point.
     
  8. Oct 18, 2006 #7
    answer is x+10y-51=0 but whres the 51 from
     
  9. Oct 18, 2006 #8

    radou

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    The equation of the line is [tex]y - y_{1} = -\frac{1}{10}(x-x_{1})[/tex], where [tex](x_{1}, y_{1})=(1, 5)[/tex]. Plug in the values and you should get the equation. (And see where 51 comes from.)
     
  10. Oct 18, 2006 #9
    ur not suggesting y-y1=m(x-x1)

    ??
     
  11. Oct 18, 2006 #10
    thats what i did one sec
     
  12. Oct 18, 2006 #11
    omg I'm so smart because I got to that last step u said but didnt see that 1/10 +5 is 5.1.......

    but thats why i'm so stupid too i didnt see it
     
  13. Oct 18, 2006 #12

    radou

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    You're not stupid, you're just slumpy. That can be cured easily. :smile:
     
  14. Oct 18, 2006 #13
    Thank you Mr Radou for your help and compliment
     
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