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Find an equation for the plane that is tangent to the surface z equals ln (x Superscript 4 Baseline plus y )at the point Upper P (0 comma 1 comma 0 ).

Answer :

lublana

Answer:

[tex]z-y+1=0[/tex]

Step-by-step explanation:

We are given that  equation of surface

[tex]z=ln(x^4+y)[/tex]

Point (0,1,0)

[tex]z_x=\frac{1}{x^4+y}(4x^3)[/tex]

[tex]z_x(0,1,0)=0[/tex]

[tex]z_y=\frac{1}{x^4+y}(1)[/tex]

[tex]z_y(0,1,0)=1[/tex]

Now, the equation of tangent plane to the given surface at point (0,1,0) is given by

[tex]z-z_0=z_x(x-x_0)+z_y(y-y_0)[/tex]

Using the formula

[tex]z-0=0+1(y-1)=y-1[/tex]

[tex]z=y-1[/tex]

[tex]z-y+1=0[/tex]

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