202210191506 Solution to 1970-CE-AMATH-II-XX

Calculate the gradient of the curve $x^2+xy+y^2=1$ at the point $(1,-1)$ .

Roughwork.

Let the function $f$ of variables $x$ and $y$ be

$f(x,y):=x^2+xy+y^2-1=0$

Then,

$\begin{aligned} \Delta f=2x\Delta x+x\Delta y+y\Delta x+2y\Delta y & =0 \\ 2x + x\frac{\Delta y}{\Delta x}+y+2y\frac{\Delta y}{\Delta x} & = 0 \\ (x+2y)\frac{\Delta y}{\Delta x} & = -2x-y \\ \frac{\Delta y}{\Delta x} & = - \frac{2x+y}{x+2y} \\ \end{aligned}$

The gradient $\nabla$ at point $(x=1,y=-1)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,-1)}=-\frac{2(1)+(-1)}{(1)+2(-1)}=1}$ .

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