91Ó°ÊÓ

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

(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 $C∈□$

So, Rewrite the equation as follows

$ax+by=C$

$by=C-ax$

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

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

role="math" localid="1650496388272" $$\begin{gathered} =\frac{a}{b}x+DSinceC∈□ \\ so\frac{C}{b}=D∈□ \end{gathered}$$

As a result, the line represents the level curves of the supplied function$y=-\frac{a}{b}x+DwhereD∈□$

(b)

The goal is to show that any $∇f(x,y)$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,

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

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

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

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

$=ai+bj$

$=⟨a,bâŸ\copyright$

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

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

$=\frac{1}{b}\{⟨a,bâŸ\copyright ·⟨b,-aâŸ\copyright \}$

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

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

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

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