Question

12. A surface is given parametrically by the equations -u2-u y lue, = (3u-20)". Find an equation of the plane tangent to the surface at the point where u u = 1

Answer 1

From the given question A surface is given parametrically by the equation x = u^2 - v^2 and also we have y = 4uv and z = (3u - 2v)^2 x = u^2 - v^2 here u = v = 1 x = (1)^2 - (1)^2 = -1 - 1 x = > 0 and slso y = 4uv = 4(1)(1) y = 4 and z = (3u - 2v)^2 = (3(1) - 2(1)^2 = (3 - 2)^2 implies (1)^2 z = 1 \r\n Now (x, y, z) = (0, 4, 1) r_uv = (u^2 - v^2, 4uv,(3u - 2v)^2) r_u = 2v - 0, u(1) v, 2(3u - 2v) (3(1) - 0) r_u = 2u, 4v, 6(3u - 2v) > so r_u = < 2u, 4u, 6(3u - 2v) > Now r_v = (- 2v, 4(1)u, 2(3u - 2v) (0 - 2(1)) > r_v = -2v 4u 6u - 2v (0 - 21)) r_v = < - 2v, 4u, - 4(3u - 2v) > Now we can do that the multiplication of matrix of r_u times r_v