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Answer 1 · 2017-04-30T22:26:20

(a)

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

(b)

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

$d f = f_{y} d y + f_{x} d x = 0$ $color(red)(df) = f_y dy +f_x dx color(red)(= 0)$

Then:

$\frac{d y}{d x} = - \frac{f_{x}}{f_{y}}$ $dy/dx = - (f_x)/(f_y)$ , the implicit function theorem

And this means that:

$\frac{d y}{d x} = - \frac{2 x - y}{- x + 2 y} = \frac{4}{5}$ $dy/dx = - (2x-y)/(- x + 2y) = (4)/(5)$

So we say that:

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

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

(c)

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