# Question #7f620

Apr 30, 2017

(a)

$\nabla {f}_{\left(- 1 , 2\right)} = < 2 x - y , - x + 2 y {>}_{\left(- 1 , 2\right)} = < - 4 , 5 >$

(b)

We can do this many ways. But if we agree in differential terms that for a level curve:

$\textcolor{red}{\mathrm{df}} = {f}_{y} \mathrm{dy} + {f}_{x} \mathrm{dx} \textcolor{red}{= 0}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$, the implicit function theorem

And this means that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x - y}{- x + 2 y} = \frac{4}{5}$

So we say that:

$y = m x + c = \frac{4}{5} x + c$

And we plug in a value, $\left(- 1 , 2\right)$, so that: $y = \frac{2}{5} \left(2 x + 7\right)$

(c)