Q. 9 - Calculus

Question

9. Continue with the function $f(x, y)=2x+3y$ from Exercise 8 .

(a) What are the level curves of $f$ ?

(b) Show that every gradient vector, $\nabla f(x,y)$, is orthogonal to every level curve of $f$.

Step-by-Step Solution

Answer

a, The level curves of the function is determined as the line $y=-\frac{2}{3}x+D$

b, The function is orthogonal to every level curves of $f$ as the function becomes $\nabla f(x,y)·r^{\text{'}}\left(t\right)=0$

Introduction

The given is the function $f(x,y)=2x+3y$

The objective is to find the level curves of the function and to prove that every gradient vector is orthogonal to the function

Step 1

(a)

Let the function be

$f(x, y)=2x+3y$

The goal is to locate the given function's level curves.

Let $f(x, y)=C$ be the case, with $C\in \square$.

Thus,

$$\begin{gathered} 2x+3y =C \\ 3y=C-2x \end{gathered}$$

$$\begin{gathered} y=\frac{C}{3}-\frac{2}{3} x \\ =\frac{2}{3} x+\frac{C}{3} \end{gathered}$$

$=-\frac{2}{3}x+D$

Use $C\in \square$, so $\frac{C}{3}=D\in \square$

As a result, the given function's level curve is the line $y=-\frac{2}{3} x+D$, where $D\in \square$.

Step 2

(b)

The goal is to show that any $\nabla f(x, y)$ gradient is orthogonal to every $f$ level curve. $x=t$ is the parameterization of the level curve $2 x+3 y=C$.

$$\begin{gathered} 3y=C-2 t \\ y =\frac{C}{3}-\frac{2}{3} t \end{gathered}$$

As a result, the parameterization is completed as .

The parameterization's tangent vector is

$=\frac{1}{3}⟨3,-2⟩$

The gradient of the function is

$\nabla f(x,y)=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}$

$=\left\{\frac{\partial }{\partial x}(2x+3y)\right\}\mathrm{i}+\left\{\frac{\partial }{\partial y}(2x+3y)\right\}\mathrm{j}$

$=\left(2\frac{\partial }{\partial x}\dot{x}+3\frac{\partial }{\partial x}y\right)\mathrm{i}+\left(2\frac{\partial }{\partial y}x+3\frac{\partial }{\partial y}y\right)\mathrm{j}$

$=(2·1+3·0)i+(2·0+3·1)\mathrm{j}$

$$\begin{gathered} =2 i+3 j \\ =<2,3> \end{gathered}$$

Step 3

So,

$\nabla f(x,y)·r^{\text{'}}\left(t\right)=⟨2,3⟩·\frac{1}{3}⟨3,-2⟩$

$=\frac{1}{3}\{⟨2,3⟩·⟨3,-2⟩\}$

$=\frac{1}{3}\{2·3+3(-2\left)\right\}$

$=\frac{1}{3}\left(0\right)$

$=0$

The gradient vector $\nabla f(x, y)$ is orthogonal to every level curve of $f$ because $\nabla f(x, y) ·r^{\text{'}}\left(t\right)=0$.

$\nabla f(x,y)·r^{\text{'}}\left(t\right)=⟨2,3⟩·\frac{1}{3}⟨3,-2⟩$