Results 1 to 2 of 2

Thread: [SOLVED] Tangent plane

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    [SOLVED] Tangent plane

    At what angle does the tangent plane to the surface $\displaystyle S: z=2x^3-x^2y-y^2-2y+11$ in the point (2,3,0) meet the xy-plane?

    $\displaystyle F=2x^3-x^2y-y^2-2y+11 - z=0$

    $\displaystyle grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)$
    $\displaystyle grad F(2,3,0) = (12,-12,-1)$

    The normal vector to the xy-plane, n, is (0,0,1).


    $\displaystyle grad F \cdot n = |17||1|cos\theta$
    $\displaystyle grad F \cdot n = -1$

    So we have $\displaystyle cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}$

    The correct answer is $\displaystyle \theta = arccos \frac{1}{17}$ though, which is what you get if you select $\displaystyle (0,0,-1)$ as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Spec View Post
    At what angle does the tangent plane to the surface $\displaystyle S: z=2x^3-x^2y-y^2-2y+11$ in the point (2,3,0) meet the xy-plane?

    $\displaystyle F=2x^3-x^2y-y^2-2y+11 - z=0$

    $\displaystyle grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)$
    $\displaystyle grad F(2,3,0) = (12,-12,-1)$

    The normal vector to the xy-plane, n, is (0,0,1).


    $\displaystyle grad F \cdot n = |17||1|cos\theta$
    $\displaystyle grad F \cdot n = -1$

    So we have $\displaystyle cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}$

    The correct answer is $\displaystyle \theta = arccos \frac{1}{17}$ though, which is what you get if you select $\displaystyle (0,0,-1)$ as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?
    we usually take $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}.$ so if $\displaystyle n_1, n_2$ are the normal vectors of two planes, then $\displaystyle \theta,$ the angle between the planes, is definied to be $\displaystyle \theta = \arccos \left|\frac{n_1 \cdot n_2}{||n_1|| \cdot ||n_2||} \right|.$

    what you've found is the supplement of $\displaystyle \theta.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 10:59 AM
  2. Tangent plane parallel to another plane
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 22nd 2009, 03:19 PM
  3. tangent plane problem (almost solved)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 16th 2009, 10:06 PM
  4. Replies: 1
    Last Post: May 28th 2009, 05:26 PM
  5. Replies: 2
    Last Post: May 9th 2009, 10:35 AM

Search Tags


/mathhelpforum @mathhelpforum