Question

Find the planes tangent to the following surfaces at the indicated points.

a) z = x² + y², at the point (1, 2, 5)
b) z = 3x + 2y - 4, at the point (2, -1, 6)
c) z = e^(xy), at the point (0, 1, 1)
d) z = ln(x + y), at the point (1, 1, 0)

Answer 1

1. Tangent Plane for z = x² + y² is:
   z - 5 = 2(x - 1) + 4(y - 2).

2. Tangent Plane for z = 3x + 2y - 4 is simply:
   z = 3x + 2y - 4.

3. Tangent Plane for z = e^(xy) is:
   z - 1 = e(y - 1).

4. Tangent Plane for z = ln(x + y) is:
   z = 0.5(x - 1) + 0.5(y - 1).

Step-by-step explanation:

We want to find the planes tangent to the given surfaces at the specified points.

1. Tangent Plane for z = x² + y²

   To find the tangent plane to the surface z = x² + y² at the point (1, 2, 5), we first calculate the partial derivatives of z with respect to x and y, which are 2x and 2y respectively. At the point (1, 2, 5), these derivatives are 2 and 4. Hence, the equation of the tangent plane is:
   z - 5 = 2(x - 1) + 4(y - 2).

2. Tangent Plane for z = 3x + 2y - 4

   For the linear function z = 3x + 2y - 4, the plane itself is the tangent plane at every point, including (2, -1, 6). Thus, the equation of the tangent plane is simply:
   z = 3x + 2y - 4.

3. Tangent Plane for z = e^(xy)

   To get the tangent plane for z = e^(xy) at the point (0, 1, 1), we again calculate partial derivatives. This time they are ye^(xy) for x, and xe^(xy) for y. At (0, 1, 1) these are 0 and e. Therefore, the equation is:
   z - 1 = e(y - 1).

4. Tangent Plane for z = ln(x + y)

   Finally, for z = ln(x + y) at point (1, 1, 0), the partial derivatives are 1/(x + y) for both x and y. These evaluate to 1/2 at the point of interest. The equation is:
   z = 0.5(x - 1) + 0.5(y - 1).