Results 1 to 2 of 2

Thread: gradient

  1. #1
    Junior Member
    Joined
    Aug 2008
    From
    England
    Posts
    40

    gradient

    find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
    then find the equations of the tangent and the normal to the curve at the point where x = 3


    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Henderson's Avatar
    Joined
    Dec 2007
    Posts
    127
    Thanks
    2
    $\displaystyle f(x) = x^3 - 3x^2$
    $\displaystyle f'(x) = 3x^2 - 6x$
    $\displaystyle f'(3) = 3(3)^2 - 6(3) = 9$

    So the slope of your tangent is $\displaystyle 9$, and the slope of your normal is $\displaystyle -\frac{1}{9}$.

    Also, since
    $\displaystyle f(3) = (3)^3 - 3(3)^2 = 0$,
    both equations will pass through the point $\displaystyle (3, 0)$.

    Let us know if you need more help from there.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Gradient
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jan 23rd 2010, 11:04 AM
  2. gradient
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 3rd 2008, 09:20 AM
  3. gradient
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 2nd 2008, 04:43 PM
  4. Gradient of f(x,y)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 30th 2008, 03:28 AM
  5. gradient
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 5th 2007, 08:46 AM

Search Tags


/mathhelpforum @mathhelpforum