Question

Find the tangent plane to the surface z = cos(xy) when (x, y) = (π, 0).

Answer #1

Compute equations of tangent plane and normal line to the
surface z = x cos (x+y) at point (π/2, π/3, -√3π/4).

Let S be the portion of the surface z=cos(y) with 0≤x≤4 and
-π≤y≤π. Find the flux of F=<e^-y,2z,xy> through S:
∫∫F*n dS

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

(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))

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

Find the equation of the tangent plane to the surface determined
by x2y4+z−35=0 at x=3, y=4.

Identify the surface with parametrization x = 3 cos θ sin φ, y =
3 sin θ sin φ, z = cos φ where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Hint: Find
an equation of the form F(x, y, z) = 0 for this surface by
eliminating θ and φ from the equations above. (b) Calculate a
parametrization for the tangent plane to the surface at (θ, φ) =
(π/3, π/4).

find volume lies below surface
z=2x+y and above the region in xy plane bounded by x=0 ,y=1
and x=y^1/2

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.

1. a) For the surface f(x, y, z) = xy + yz + xz = 3, find the
equation of the tangent plane at (1, 1, 1).
b) For the surface f(x, y, z) = xy + yz + xz = 3, find the
equation of the normal line to the surface at (1, 1, 1).

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