Question

Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for the plane is in simplified form.

Answer 1

Answer:

Required equation of tangent plane is $z=\frac{6}{5}(x-5y-11)$.

Step-by-step explanation:

Given surface function is,

$f(x,y)=6-\frac{6}{5}(x-y)$

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)$

So that,

$f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}$

$f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}$

$f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6$

Substitute all these values in (1) we get,

$z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}$

$\therefore z=\frac{6}{5}(x-5y-11)$

Which is the required euation of tangent plane.