Question

Find the point on the surface z = x 2 + 2y 2 where the tangent plane is orthogonal to the line connecting the points (3, 0, 1) and (1, 4, 0).

Answer #1

Find equations of the tangent plane and normal line to the
surface x=2y^2+2z^2−159x at the point (1, -4, 8).
Tangent Plane: (make the coefficient of x equal to 1).
=0.
Normal line: 〈1,〈1, , 〉〉
+t〈1,+t〈1, ,

find the tangent plane to the surface x^2 + 2xy + z^3 = 4 at point
P (1,1,1)

Find equations of the tangent plane and normal line to the
surface z+2=xeycosz at the point (2, 0, 0).

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=

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 equations of the tangent plane and normal line to the
surface x=3y^2+1z^2−40x at the point (-9, 3, 2).
Tangent Plane: (make the coefficient of x equal to 1).
=0.
Normal line: 〈−9,〈−9, , 〉〉
+t〈1,+t〈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.

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 to the surface z = x^2 +
xy + 3y^2 at the point (1, 1, 5)

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

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