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Find an equation of the tangent plane to the surface z = ln(x - 8y) at the point (9, 1, 0).

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Final answer:

To find the equation of the tangent plane to the surface z = ln(x - 8y) at the point (9, 1, 0), calculate the partial derivatives, evaluate them at the given point, and apply the formula for the equation of a tangent plane. The final equation is z = x - 9 - 8y + 8.

Step-by-step explanation:

To find an equation of the tangent plane to the surface z = ln(x - 8y) at the point (9, 1, 0), we start by calculating the partial derivatives of the function with respect to x and y. For a function z = f(x, y), the tangent plane at a point (x_0, y_0, z_0) is given by:

z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)

where f_x and f_y are the partial derivatives of f with respect to x and y, respectively.

The partial derivatives are:

f_x = 1 / (x - 8y)

f_y = -8 / (x - 8y)

Evaluating these at the point (9, 1), we get:

f_x(9, 1) = 1

f_y(9, 1) = -8

Therefore, the equation of the tangent plane at (9, 1, 0) is:

z - 0 = 1 * (x - 9) + (-8) * (y - 1)

Simplifying, the final equation of the tangent plane is:

z = x - 9 - 8y + 8

The complete qestion is: Find an equation of the tangent plane to the surface z = ln(x - 8y) at the point (9, 1, 0). is:

User Spanky Quigman
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