Q. 17. - Calculus

Question

Sketch level curves$z=-1,0$, and 1 for the function

$z=x^2-y^2$. Include the graphs of three gradient vectors

on each level curve. What do you observe?

Step-by-Step Solution

Answer

The graph of level curves $z=x^2-y^2$ for  is shown below:

The gradient and tangent vectors are orthogonal.

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

The function $z=x^2-y^2.$

The graph of the function $z=x^2-y^2$is a concentric circle, each of whose level curves is a circle centered at the origin.

The gradient is,
$$\begin{gathered} z=x^2-y^2 \\ \nabla z=2xi-2yj \end{gathered}$$

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.

Step 2: Draw a graph of level curves z = x 2 - y 2   for z = - 1 , 0 , 1   and write the observation

The level of curves  $z=x^2-y^2$for $z=-1,0,1$is

$$\begin{gathered} -1=x^2-y^2 \\ 0=x^2-y^2 \\ 1=x^2-y^2 \end{gathered}$$

The graph of level curves $z=x^2-y^2$for $z=-1,0,1$is shown below:

Therefore, the gradient and tangent vectors are orthogonal.