Question

3. Suppose you need to know an equation of the tangent plane to a surface S at the point P(3, 1,3). You don't have an equation for S but you know that the curves r(t) = (3 + 3t, 1-t2,3 - 5t + t2) rz(t) = (2+u, 2u3 – 1,2u + 1) both lie on S. Find an equation of the tangent plane at P.

Answer 1

8 (4) = (3 + 3t, 1-4, 3- 5 4 + +) 82 (4) = (2+ 4², 24²-1, 2u+1) P (3, 1, 3) We know two curves contained in the surface We know at least two vectors tangent to the surfan s, namely the velocity vectors (t) and (w) of these two curves. (+) = (3,-25,- 5+2+) CU 20, Gu? 2) since, we are interested the point P (3,1, 3), we need to find the value of the parameter t when A CH) passes through P. when to them (3+3 t, 1-t², 3- 5€ + t²) = £3, 1, 3) Similarly, the path & (u) passes through P a when u=l. (0) = ( 3,0,-5) B (1) = (2,6,2) We find the equation of the tangent plane

which means a plane containing the two vectors (30,-5) and <2,6,27. For the normal vector to this plane we take the cross - product of these two vectors. Now Cox V, U) = 1 130 - 5 1 2 6 2 - ?(30)- } (16) + 18 R. = 130,- 16, 18) we find the normal rector £30, -16, 187 to the plane and the point P (3, 1,3) on the plane. so, the equation for the plane is 301%-3) -1614-1) +38 (7-3) = 0. ² 30%-164 + 187 - 9o+ 16-54 = 0 30%-164 of 187 = 128. The required equation of tangent plane 30x-164 + 187 = 128 .