tangent planes and linear approximations

Discussion in 'Math' started by wheetnee, Oct 7, 2009.

  1. wheetnee

    Thread Starter New Member

    Oct 1, 2009
    28
    0
    Suppose you need to know the equation of the tangent plane to a surface, S, at the point, P(2,1,3). You don't have have an equation for S, but you know the curves,

    r₁ (t) = < 2+3t , 1 – t² , 3 - 4t + t²>

    r₂ (u) = <1+u², 2u³ - 1, 2u + 1>

    both lie on S. Find an equation of the tangent plane at P.


    i already dervied r1(t) and r2(u)
    but im not sure what the next step is.
     
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    wheetnee,

    Well, let's see. It is easily seen that r1 and r2 pass through P(2,1,3) when t=0 and u=1 respectively.

    The direction vectors of r1 and r2 at P(2,1,3) will be the derivatives of r1 and r2 at t=0 and u=1 respectively. So, dr1/dt = 3,-2t,2t-4 = (3,0,-4) , and dr2/du = 2u,6u^2,2 = (2,6,2) .

    The cross product of (3,0,-4) and (2,6,2) is (24,-14,18), which reduces to a direction vector of (12,-7,9) which is perpendicular to (3,0,-4) and (2,6,2) at P(2,1,3) .

    So a plane perpendicular to the direction (12,-7,9) , and passing through P(2,1,3) is the dot product of (12,-7,9) and (x-2,y-1,z-3) = 0 , which is 12x-24 -7y+7+9z-27 =0 ====> 12x -7y + 9z = 44 .

    Ratch
     
    Last edited: Oct 9, 2009
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