91Ó°ÊÓ

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

(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 role="math" localid="1650491266937" $C∈□$.

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 $C∈□$, so $\frac{C}{3}=D∈□$

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

(b)

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

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

As a result, the parameterization is completed as .

The parameterization's tangent vector is

$=\frac{1}{3}⟨3,-2âŸ\copyright$

The gradient of the function is

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

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

$=(2·1+3·0)i+(2·0+3·1)\mathrm{j}$

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

So,

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

$=\frac{1}{3}\{⟨2,3âŸ\copyright ·⟨3,-2âŸ\copyright \}$

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

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

$=0$

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

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