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Math Help - Finding equation of normal to a curve.

  1. #1
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    Finding equation of normal to a curve.

    Well I got this question:

    Differentiate from first principals (i did it)  f=x^3  f'(x) = 3x^2

    Find the equation of the normal to the curve at point where x = 3.

    So first I found y.  y=(3)^3 y=27 That means we got  (3,27)

    Then I subbed 3 into  3x^2 to get the gradient of the tangent. Answer is  27

    Since the normal and tangent are perpendicular the formula is  m_{2} = \frac {-1}{m_{1}} So that makes  \frac {-1}{27}

    Then I subbed it into y-y_{1} = m(x-x_{1}

     y-27=\frac{-1}{27}(x+3)

    Finally I get

     -27y+726=x+3 but for some reason the answers in the book says it's different

    Did I do anything wrong?


    Also, I need help with another question:

    If  y=x^3-3x^2 find the derivative and find the values of y when \frac{dy}{dx} = 0

    How would I go about starting this one?
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  2. #2
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    Quote Originally Posted by jgv115 View Post
    Well I got this question:

    Differentiate from first principals (i did it)  f=x^3  f'(x) = 3x^2

    Find the equation of the normal to the curve at point where x = 3.

    So first I found y.  y=(3)^3 y=27 That means we got  (3,27)

    Then I subbed 3 into  3x^2 to get the gradient of the tangent. Answer is  27

    Since the normal and tangent are perpendicular the formula is  m_{2} = \frac {-1}{m_{1}} So that makes  \frac {-1}{27}

    Then I subbed it into y-y_{1} = m(x-x_{1})

     y-27=\frac{-1}{27}(x-3) <=== you used the wrong sign in the bracket

    Finally I get

     -27y+726=x+3 but for some reason <=== maybe they believe that 27 = 729
    the answers in the book says it's different

    Did I do anything wrong?


    Also, I need help with another question:

    If  y=x^3-3x^2 find the derivative and find the values of y when \frac{dy}{dx} = 0

    How would I go about starting this one?
    To your last question: I know that you know how to find the drivatie: Differentiate y wrt x. Solve the equation for x:

    3x^2-6x = 0
    Last edited by mr fantastic; July 17th 2010 at 03:01 PM. Reason: Moved from orignal thread.
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  3. #3
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    Quote Originally Posted by earboth View Post
    To your last question: I know that you know how to find the drivatie: Differentiate y wrt x. Solve the equation for x:

    3x^2-6x = 0
    mm what do you mean i put the wrong sign? I'm pretty sure x is 3.

    What does y wrt x mean?
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  4. #4
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    It's actually the next step which is wrong, since when you multiply both sides by -27 you should get

    -27y + 729 = x - 3.


    "wrt" means "with respect to". So Earboth is saying find \frac{dy}{dx}.
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    Quote Originally Posted by Prove It View Post
    It's actually the next step which is wrong, since when you multiply both sides by -27 you should get

    -27y + 729 = x - 3.


    "wrt" means "with respect to". So Earboth is saying find \frac{dy}{dx}.
    Yea, I actually typed it wrong.

    I was meant to type:
    -27y + 726 = x. I got 729 then I had to subtract 3 and i get 726. The answer says it's 732 though :S

    And why do you have the equate the \frac {dy}{dx} to 0?

     3x^2-6x = 0

     3x(x-2) =0

     x = 0  x= 2?
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    Because that's what you were told to do in the question...

    You have found the x values of the stationary points, now substitute back into the original function to find the corresponding y values.
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    y=0 and y=4

    yay!
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  8. #8
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    Quote Originally Posted by jgv115 View Post
    Yea, I actually typed it wrong.

    I was meant to type:
    -27y + 726 = x. I got 729 then I had to subtract 3 and i get 726. The answer says it's 732 though :S

    And why do you have the equate the \frac {dy}{dx} to 0?

     3x^2-6x = 0

     3x(x-2) =0

     x = 0  x= 2?
    Like I said, you should have

    -27y + 729 = x - 3.

    Now you have to ADD 3 to both sides (to undo the -3 ) which gives

    -27y + 732 = x as required.
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  9. #9
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    yea i get it!! (stupid mistake)
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