202210191436 Solution to 1969-CE-AMATH-II-XX

Calculate the gradient of the curve $x=3y+y^2$ at the point $(0,-3)$ .

Roughwork.

Differentiating $x(y)$ wrt to $y$ and taking reciprocal,

$\begin{aligned} x(y) & = 3y+y^2 \\ \frac{\mathrm{d}x}{\mathrm{d}y} & = 3 + 2y \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} \\ & = \frac{1}{3+2y} \\ \end{aligned}$

the gradient $\nabla y(x)$ at point $(x,y) = (0,-3)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(0,-3)} = \frac{1}{3+2(-3)} = -\frac{1}{3}}$ .

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