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Differentiation - finding equation of normal to curve

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the normal to the curve y=(6x+3)^(1/2) at the point for which x=13.


    2. Relevant equations

    dy/dx=dy/du*du/dx

    y-y1=m(x-x1)


    3. The attempt at a solution

    y=(6x+3)^(1/2)
    dy/dx=1/3(6x+3)^2

    gradient of normal = -3(6x+3)^2

    at x=13, dy/dx=-3(6x13+3)^2=19683.

    obviously a silly answer and the textbook disagrees.

    So at this point I could simplify y=(6x+3)^(1/2) to y=(2x+1)^2 maybe? Even after doing this, I end up with an even slightly smaller silly answer and the textbook still disagrees, unless the textbook is wrong, although this is highly unlikely.

    HELP!
     
  2. jcsd
  3. Oct 13, 2011 #2

    Hootenanny

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    You will want to start by checking this:
     
  4. Oct 13, 2011 #3
    I inputted y=(6x+3)^(1/2) into wolfram alpha and the answer came out to be:

    √3/(2x+1)^(1/2)

    I'm confused. I used the chain rule to get:

    subu=6x+3 so that y=u^1/2
    dy/du=1/2u^-(1/2)=1/2(6x+3)^-(1/2) and du/dx=6

    => dy/dx=dy/du*du/dx=3(6x+3)^-(1/2).

    ???
     
    Last edited: Oct 13, 2011
  5. Oct 13, 2011 #4
    Work out the derivative by hand. It's straightforward.
     
  6. Oct 13, 2011 #5

    HallsofIvy

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    I put that into WolframAlpha and got the correct answer.
     
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