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Finding the equation of the normal line at

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1



    2. Relevant equations



    3. The attempt at a solution

    -I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
    -SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
    -Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

    Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if im wrong, point me in the right direction of being right!

    Cheers.
     
  2. jcsd
  3. Dec 15, 2008 #2

    berkeman

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    Staff: Mentor

    You are right that the derivative of [tex]3 {x^4}[/tex] is [tex]12{x^3}[/tex]

    But what is the derivative ot 1/x again?
     
  4. Dec 15, 2008 #3
    You made a minor mistake in your derivative, it should be 12x³ + 1/x², not 12x³ + 1. However, this doesn't affect the slope at x=1, and the rest of your derivation (including the answer) is correct.
     
  5. Dec 15, 2008 #4
    YUP i see where I made the mistake. Instead of subtracting the 1 from the x-1 I added. But your right it doesnt affect the outcome. But thanks guys!
     
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