Q. 14 - Calculus

Question

14. Sketch level curves $z=1,4,9$, and $16$ 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 graphs of three gradient vectors on each level curve are included and it is observed that the gradient and the tangent vector are orthogonal.

Introduction

The given is the function $z=x^2+y^2$with level curves $z=1,4,9$

The objective is to observe the gradient vectors on each level curve

Step 1

Let the function be $z=x^2+y^2$.

The goal is to draw the level curves for the function $z=1,4,9,16$.

The graph of the function $z=x^2+y^2$ is a paraboloid, with each of its level curves centred on the origin.

The gradient is,

$$\begin{gathered} z=x^2+y^2 \\ \nabla z=2 xi+2 y j \end{gathered}$$

As a result, every gradient vector extends directly from the origin.

Step 2

For the point $\left(x_0,y_0\right)$ the gradient is $\nabla z\left(x_0,y_0\right)=2x_0\mathbf{i}+2y_0\mathbf{j}$, and a tangent vector to the level curve containing the point $\left(x_0,y_0\right)$ is $y_0\mathbf{i}-x_0\mathbf{j}$.

The level curves $z=x^2+y^2$ for $z=1,4,9$, and $16$ is,

$x^2+y^2=1$

$x^2+y^2=4$

$x^2+y^2=9$

$x^2+y^2=16$

The graph of level curves $c=x^2+y^2$for $c=1,4,9$, and $16$ is as shown below.

As a result, the vectors of gradient and tangent are orthogonal.