Q. 15 - Calculus

Question

15. Sketch level curves $z=9,16,21$, and $24$ for the function $z=25-x^2-y^2$. 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 observed that the gradient and tangent vectors are orthogonal

Introduction

The given is the function $z=25-x^2-y^2$with level curves $z=9,16,21$

The objective is to include the gradient vectors and to observe the function

Step 1

Let the function be $z=25-x^2-y^2$.

The goal is to draw the level curves for the function $z=9,16,21$ and $24$.

The graph of the function $z=25-x^2-y^2$ is a concentric circle, with each of its level curves centred on the origin.

The gradient is,

$$\begin{gathered} z=25-x^2-y^2 \\ \nabla z =-2 xi-2 yj \end{gathered}$$

As a result, the magnitude of the gradient vectors increases.

Step 2

The level curves $z=25-x^2-y^2$ for $z=9,16,21$ and $24$ is,

$\begin{array}{ll}{x}^2+y^2=16& \left(\text{ Since, }9=25-x^2-y^2\right)\\ x^2+y^2=9& \left(\text{ Since, }16=25-x^2-y^2\right)\\ x^2+y^2=4& \left(\text{ Since, }21=25-x^2-y^2\right)\\ x^2+y^2=1& \left(\text{ Since, }24=25-x^2-y^2\right)\end{array}$

.The graph of level curves $z=25-x^2-y^2$ for $z=9,16,21$ $z=9,16,21$ and $24$ is as shown below.

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