Q. 11 - Calculus

Question

11. Continue with the function $f(x, y)=ax+by$ from Exercise 10 .

(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 the line $y=-\frac{a}{b}x+D$

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

Introduction

The given data is the function $f(x,y)=ax+by$

The objective is to find the level curves of the function and to show that the gradient vectors are orthogonal to every level curve of the function

Step 1

(a)

Let the function be

$f(x, y)=ax+by$

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

Let $f(x, y)=C$ where $C\in \square$

So, Rewrite the equation as follows

$ax+by=C$

$by=C-ax$

$y=\frac{C}{b}-\frac{a}{b}x$

$=-\frac{a}{b}x+\frac{C}{b}$

$$\begin{gathered} =\frac{a}{b} x+D Since C \in \square \\ so \frac{C}{b}=D \in \square \end{gathered}$$

As a result, the line represents the level curves of the supplied function$y=-\frac{a}{b} x+D where D\in \square$

Step 2

(b)

The goal is to show that any $\nabla f(x,y)$ gradient is orthogonal to every $f$ level curve.

The parameterization of the level curve $ax+by=C$ is $x=t$ and $by=C-at$

$y=\frac{C}{b}-\frac{a}{b}t$

So, the parameterization is .

The parameterization's tangent vector is

The function's gradient  is,

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

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

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

$=(a·1+b·0)\mathrm{i}+(a·0+b·1)\mathrm{j}$

$=ai+bj$

$=⟨a,b⟩$

Step 4

So, the function's gradient vector $f$ is as follows:

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

$=\frac{1}{b}\{⟨a,b⟩·⟨b,-a⟩\}$

$=\frac{1}{b}\{a·b+b(-a\left)\right\}$

$=\frac{1}{b}(ab-ab)$

$$\begin{gathered} =\frac{1}{b}\left(0\right) \\ =0 \end{gathered}$$

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