Question

Let f (x, y) = e⁽⁶ˣ⁻⁷ʸ⁾. Find the equation for the tangent plane to the graph of f at the point (1,2).

(Use symbolic notation and fractions where needed.)
z =

Answer 1

To find the equation of the tangent plane to f at (1,2), we calculate the partial derivatives of f, determine the gradient at (1,2) as the normal vector, and use the point-normal form of the plane equation.

Step-by-step explanation:

The goal is to find the equation of the tangent plane to the graph of f at the point (1,2). First, we calculate the partial derivatives of f with respect to x and y at the point (1,2). The gradient of f at (1,2) gives the normal vector to the tangent plane.

The function is f(x, y) = e⁶¹⁷⁵. The partial derivatives are:

• f_x(x, y) = 6e⁶¹⁷⁵
• f_y(x, y) = -7e⁶¹⁷⁵

At the point (1,2), these derivatives are f_x(1, 2) = 6e and f_y(1, 2) = -7e. Therefore, the normal vector is (6e, -7e, -1).

The point on the plane is f(1,2) = e⁶²×¹⁷ײ or z = e. Using the point-normal form of the plane equation: (x - 1)f_x(1, 2) + (y - 2)f_y(1, 2) + (z - e) = 0, the equation of the tangent plane is 6e(x - 1) - 7e(y - 2) + (z - e) = 0.

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