# 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
4
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