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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 ).

User OscarVGG
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1 Answer

4 votes

Answer:


z-y+1=0

Explanation:

We are given that equation of surface


z=ln(x^4+y)

Point (0,1,0)


z_x=(1)/(x^4+y)(4x^3)


z_x(0,1,0)=0


z_y=(1)/(x^4+y)(1)


z_y(0,1,0)=1

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


z-z_0=z_x(x-x_0)+z_y(y-y_0)

Using the formula


z-0=0+1(y-1)=y-1


z=y-1


z-y+1=0

User Newmaniese
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