Question

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)

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 an equation of the tangent plane to the parametric surface
r=(u,v)=ucosv I +usinv j +vk at u=1, v=pi/3
Find the surface area of the parametric surface r(u,v)=5sinucosv
I + 5sinusinv j+ 5cosu k, for 0 ,<= u <=pi and o<=v<=
2pi

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).

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 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).

Find the equation for the tangent plane to the surface
z=(xy)/(y+x) at the point P(1,1,1/2).

Find an equation of the tangent plane to the surface z = x^2 +
xy + 3y^2 at the point (1, 1, 5)

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.

(a) Find an equation of the plane tangent to the surface xy ln x
− y^2 + z^2 + 5 = 0 at the point (1, −3, 2)
(b) Find the directional derivative of f(x, y, z) = xy ln x −
y^2 + z^2 + 5 at the point (1, −3, 2) in the direction of the
vector < 1, 0, −1 >. (Hint: Use the results of partial
derivatives from part(a))

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