Q. 16. - Calculus

Question

Sketch level curves $z=1,4,9$ and 16 for the function $z=\frac{x^2}{4}+\frac{y^2}{9}$. 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 $z=\frac{x^2}{4}+\frac{y^2}{9}$for $z=1,4,9 and 16$is shown below.

The gradient and tangent vectors are orthogonal.

Step 1: The objective is to sketch the level curves z = 1 , 4 , 9   a n d   16 for the function.

The function $z=\frac{x^2}{4}+\frac{y^2}{9}$

The graph of the function $z=\frac{x^2}{4}+\frac{y^2}{9}$is an ellipse.

The gradient is,

$$\begin{gathered} z=\frac{x^2}{4}+\frac{y^2}{9} \\ \nabla z=\frac{x}{2}i+\frac{2y}{9}j \end{gathered}$$

Hence, the gradient vectors are perpendicular to the level curves and point to the $x$- and $y$-axes, respectively. The magnitude of the gradient vectors grows.

Step 2: Draw a graph of level curves z = x 2 4 + y 2 9   for z = 1 , 4 , 9   a n d   16 and write the observation

The level of curves $z=\frac{x^2}{4}+\frac{y^2}{9}$ for $z=1,4,9 and 16$ is

$$\begin{gathered} 1=\frac{x^2}{4}+\frac{y^2}{9} \\ 4=\frac{x^2}{4}+\frac{y^2}{9} \\ 9=\frac{x^2}{4}+\frac{y^2}{9} \\ 16=\frac{x^2}{4}+\frac{y^2}{9} \end{gathered}$$

The graph of level curves $z=\frac{x^2}{4}+\frac{y^2}{9}$ for $z=1,4,9 and 16$ is shown below:

Therefore, the gradient and tangent vectors are orthogonal.