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The functiong(x,y)=x2-y2is graphed in the figure as follows. use the definition of the partial derivatives to show thatfx(0,0)=fy(0,0)=0 what are the equations of the lines tangent to the surface on the x and y directions at (0,0,0). what is the equation of the plane containing these two lines.

Short Answer

Expert verified

The value of the partial derivative is fx(0,0)=0fy(0,0)=0

The equation of the lines tangent to the surface in the x and y direction at (0,0,0) is 0.

Equation of the plane containing these two line is z=0.

Step by step solution

01

Given information

We are given an equation g(x,y)=x2-y2.

02

Use the definition of partial derivatives to compute the partial derivatives 

Using the definition we have

limh→0g(x+h,y)-g(x,y)hlimh→0(x+h)2-y2-x2+y2hlimh→0x2+2xh+h2-y2-x2+y2hlimh→02x+hlimh→02xAtpoint(0,0)wehavethisvalueequaltozero

Similarly we have,

limk→0f(x,y+k)-f(x,y)klimk→0x2-(y+k)2-x2+y2klimk→0-2y+k=-2kAtpoint(0,0)wehavethisvalueequaltozero

03

Compute equations of the tangent line 

The equation of tangent line in x direction is given as

z=f(0,0)+fx(0,0)z=0

The equation of tangent line in y direction is given as

z=f(0,0)+fy(0,0)tz=0

Now we find the equation of the plane containing these lines to do this we find the cross product of

N=(1,0,fy(0,0))×(0,1,fx(0,0))N=(0,0,-1)

And now we take dot product of N and vector (x,y,z)

We get,

(0,0,-1)(x,y,z)-z=0

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