Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By earboth

Math Help - Find an equation of the tangent plane?

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    15
    Thanks
    3

    Find an equation of the tangent plane?

    Question: Find an equation of the tangent plane to the parametric surface x=1rcosθ, y=4rsinθ, z=r at the point (sqrt(2),4sqrt(2),2) when r=2, θ=pi/4.

    My attempt:

    Let m = <rcosθ, 4rsinθ, r>, then

    mr = <cosθ, 4sinθ, 1>
    m_θ = <-rsinθ, 4rcosθ, 0>

    Normal = (m_r x m_θ) = [4sinθ*0-4rcosθ*1]i - [cosθ*0 - (-rsinθ*1)]j + [4rcosθ*cosθ - 4sinθ*-rsinθ]k = <4rcosθ, -rsinθ, 4r>
    N at the point (sqrt(2),4sqrt(2),2) => <-4sqrt(2), sqrt(2), 8>

    Now, <(x-sqrt(2), y-4sqrt(2), z-2> dot <-4sqrt(2), sqrt(2), 8> = 0
    -4sqrt(2)((x-sqrt(2)) + (
    sqrt(2)(y-4sqrt(2))) + 8(z-2) = 0
    z = (4sqrt(2)x - sqrt(2)y + 16) / 8

    This is the wrong answer, could anyone point out where I went wrong or a better way to do this? Thanks!
    Last edited by MrCryptoPrime; April 17th 2012 at 07:40 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Find an equation of the tangent plane?

    Quote Originally Posted by MrCryptoPrime View Post
    Question: Find an equation of the tangent plane to the parametric surface x=1rcosθ, y=4rsinθ, z=r at the point (sqrt(2),4sqrt(2),2) when r=2, θ=pi/4.

    My attempt:

    Let m = <rcosθ, 4rsinθ, r>, then

    mr = <cosθ, 4sinθ, 1>
    m_θ = <-rsinθ, 4rcosθ, 0>

    Normal = (m_r x m_θ) = [4sinθ*0-4rcosθ*1]i - [cosθ*0 - (-rsinθ*1)]j + [4rcosθ*cosθ - 4sinθ*-rsinθ]k = <-4rcosθ, -rsinθ, 4r> <--- here is your mistake
    N at the point (sqrt(2),4sqrt(2),2) => <-4sqrt(2), sqrt(2), 8>

    Now, <(x-sqrt(2), y-4sqrt(2), z-2> dot <-4sqrt(2), sqrt(2), 8> = 0
    -4sqrt(2)((x-sqrt(2)) + (
    sqrt(2)(y-4sqrt(2))) + 8(z-2) = 0
    z = (4sqrt(2)x - sqrt(2)y + 16) / 8

    This is the wrong answer, could anyone point out where I went wrong or a better way to do this? Thanks!
    All your considerations are OK. You "only" made a tiny (but fatal) sign mistake.

    According to my results the tangent plane has the equation:

    z = \frac{x \cdot  \sqrt{2}}{2} - \frac{y \cdot  \sqrt{2}}{8}
    Attached Thumbnails Attached Thumbnails Find an equation of the tangent plane?-tangplanehyperboloid.png  
    Thanks from MrCryptoPrime
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    15
    Thanks
    3

    Re: Find an equation of the tangent plane?

    My assignment was actually due at midnight last night, but thank you for the explanation of where I went wrong and for the awesome picture (what program do you use to generate that?). Vielen Dank!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: October 20th 2010, 05:51 AM
  2. Find equation for tangent plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 16th 2010, 11:50 PM
  3. Replies: 2
    Last Post: May 9th 2009, 11:35 AM
  4. find tangent plane of this surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 30th 2008, 03:47 PM
  5. find a tangent plane on a sphere
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 26th 2008, 11:56 AM

Search Tags


/mathhelpforum @mathhelpforum