Question

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

Answer 1

Answer:

x - 8y - z = 1

Step-by-step explanation:

Data provided according to the question is as follows

f(x,y) = z = ln(x - 8y)

Now the equation for the tangent plane to the surface

For z = f (x,y) at the point P $(x_0,y_0,z_0)$ is

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

Now the partial derivatives of f are

$f_x(x,y) = \frac{1}{x-8y} \\\\f_y(x,y) = \frac{8}{x-8y} \\\\P(x_0,y_0,z_0) = (9,1,0)\\\\f_z(9,1,0) = (\frac{1}{x-8y})_^{(9,1,0)}$

$\\\\=\frac{1}{9-8}$

= 1

Now

$f_y(9,1,0)=(\frac{8}{x-8y})_{(9,1,0)}\\\\ = -\frac{8}{9 - 8}$

= -8

So, the tangent equation is

$z - 0 = 1\times (x - 9) -8\times (y - 1)$

Now after solving this, the following equation arise

z = x - 9 - 8y + 8

z = x - 8y - 1

Therefore

x - 8y - z = 1