Respuesta :
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 [tex](x_0,y_0,z_0)[/tex] is
[tex]z - z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\\[/tex]
Now the partial derivatives of f are
[tex]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)}[/tex]
[tex]\\\\=\frac{1}{9-8}[/tex]
= 1
Now
[tex]f_y(9,1,0)=(\frac{8}{x-8y})_{(9,1,0)}\\\\ = -\frac{8}{9 - 8}[/tex]
= -8
So, the tangent equation is
[tex]z - 0 = 1\times (x - 9) -8\times (y - 1)[/tex]
Now after solving this, the following equation arise
z = x - 9 - 8y + 8
z = x - 8y - 1
Therefore
x - 8y - z = 1
The equation of the tangent plane is [tex]x-8y-z=1[/tex]
Tangent Plane:
An equation of the tangent plane to the given surface at the point [tex]P(x_0,y_0,z_0)[/tex] is,
[tex]z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)[/tex]
The function is,
[tex]z = ln(x-8y)[/tex]
And the point is (9,1,0)
Now, calculating [tex]f_x,f_y[/tex]
[tex]f_x(x,y)=\frac{1}{x-8y}\\ f_y(x,y)=\frac{x-8}{x-8y}[/tex]
Now, substituting the given points into the above functions we get,
[tex]f_x(9,1)=\frac{1}{9-8(1)}=1\\ f_y(x,y)=\frac{-8}{9-8(1)}=-8[/tex]
So, the equation of the tangent plane is,
[tex]z-0=1(x-9)-8(y-1)\\z=x-8y-1\\x-8y-z=1[/tex]
Learn more about the topic tangent plane:
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