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Chapter 12: Q. 17. (page 964)

Sketch level curvesz=-1,0, and 1 for the function

z=x2-y2. Include the graphs of three gradient vectors

on each level curve. What do you observe?

Short Answer

Expert verified

The graph of level curves z=x2-y2for is shown below:

The gradient and tangent vectors are orthogonal.

Step by step solution

01

The objective is to sketch the level curves z=-1,0,1 for the function

The function z=x2-y2.

The graph of the function z=x2-y2is a concentric circle, each of whose level curves is a circle centered at the origin.

The gradient is,
z=x2-y2鈭�z=2xi-2yj

Hence, the gradient vectors are perpendicular to the level curves and point toward the x-axis while avoiding the y-axis. The magnitude of the gradient vectors grows.

02

 Draw a graph of level curves z=x2-y2 for z=-1,0,1 and write the observation 

The level of curves z=x2-y2for z=-1,0,1is

-1=x2-y20=x2-y21=x2-y2

The graph of level curves z=x2-y2for z=-1,0,1is shown below:

Therefore, the gradient and tangent vectors are orthogonal.

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In Exercises, by considering the function f(x,y)=x2ysubject to the constraint x+y=0,you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.

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