Question

Find an equation of the tangent plane to the surface given parametrically by x = u^2, y = v^2, z = u+4v at the point (1, 4, 9).

Answer #1

Find an equation of the tangent plane to the given parametric
surface at the specified point. x = u + v, y = 6u^2, z = u − v; (2,
6, 0)

Find the equation of the tangent plane (in terms of x, y and z)
to the surface given by x = u, y = v and z = uv at the point (3, 2,
6).

Find an equation of the tangent plane to the surface x y 2 + 3 x
− z 2 = 4 at the point ( 2 , 1 , − 2 ) An equation of the tangent
plane is

Find an equation of the tangent plane to the given surface at
the specified point.
z = 2(x − 1)2 + 4(y + 3)2 +
9, (2, −2, 15)

Find an equation of the tangent plane (in
variables x, y, z) to parametric surface
r(u,v)=〈3u,−5v-5u^2,−5v^2〉 at the point (3,0,−5)

Find an equation of the tangent plane to the given surface at
the specified point.
z = 2(x − 1)2 + 4(y + 3)2 +
1, (3, −1, 25)
Answer as z=

Problem 1. Find an equation of the tangent plane to the given
surface at the specified point. i) z = 2x 2 + y 2 − 5y, (1, 2, −4).
ii) z = e x−y , (2, 2, 1). iii) z = x sin(x + y), (−1, 1, 0)

Find an equation of the tangent plane to the given surface at
the specified point. z = 8x^2 + y^2 − 7y, (1, 3, −4)

Find the equation of the tangent plane for the surface
represented by x = u2, y = u - v2, and z =
v2 at the following point (1,0,1).

8).
a) Find an equation of the tangent plane to the surface z = x at
(−4, 2, −1).
b) Explain why f(x, y) = x2ey is differentiable at (1, 0). Then
find the linearization L(x, y) of the function at that point.

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