Question

Find an equation of the plane tangent to the following surface at the given points. z=ln(1+xy)​; (5,8,ln41) and (−5,−8,ln41) The tangent plane at (5,8,ln41) is z= The tangent plane at (-5,-8,

Answer 1

The equations for the tangent planes to the surface z = ln(1+xy) at the points (5,8,ln41) and (−5,−8,ln41) are z = ln41 + (8/41)(x - 5) + (5/41)(y - 8) and z = ln41 + (−8/41)(x + 5) + (−5/41)(y + 8), respectively.

Step-by-step explanation:

To find the equations of the tangent planes to the surface represented by z = ln(1+xy) at the given points (5,8,ln41) and (−5,−8,ln41), we first need to find the partial derivatives of the function with respect to x and y.

The partial derivative with respect to x is given by ∂z/∂x = y/(1+xy), and the partial derivative with respect to y is given by ∂z/∂y = x/(1+xy). At the point (5,8,ln41), the partial derivatives become ∂z/∂x = 8/41 and ∂z/∂y = 5/41.

The equation of the tangent plane at a point (x_0, y_0, z_0) is given by z = z_0 + (∂z/∂x)|_(x_0,y_0)(x - x_0) + (∂z/∂y)|_(x_0,y_0)(y - y_0). Plugging in the values for the point (5,8,ln41), we get:

z = ln41 + (8/41)(x - 5) + (5/41)(y - 8).

Similarly, the partial derivatives at the point (−5,−8,ln41) also turn out to be ∂z/∂x = −8/41 and ∂z/∂y = −5/41 since the expression involves the same constants. The tangent plane equation becomes:

z = ln41 + (−8/41)(x + 5) + (−5/41)(y + 8).