题目

Problem

A curve passes through the point $P(4,3)$ and is such that

$\frac{dy}{dx}=\frac{8}{x^2}-\frac{10}{(2x-3)^2}.$

(a) Find the equation of the normal to the curve at $P$ . Give your answer in the form $y=mx+c$ .

[3]

(b) Find the rate of change of the gradient of the curve when $x=4$ .

[3]

[5]

题目中文翻译

一条曲线经过点 $P(4,3)$ ，并满足

$\frac{dy}{dx}=\frac{8}{x^2}-\frac{10}{(2x-3)^2}.$

(a) 求曲线在 $P$ 处的法线方程。答案写成 $y=mx+c$ 的形式。

(b) 求当 $x=4$ 时，曲线梯度的变化率。

解答

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先由导数求切线斜率。法线与切线垂直，所以用负倒数作为斜率。

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At $x=4$ , the gradient of the curve is

\frac8{16}-\frac{10}{25}=\frac1{10}.

The normal has gradient $-10$ and passes through $(4,3)$ , so its equation is $y-3=-10(x-4)$ , or $y=-10x+43$ .

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斜率的变化率就是二阶导数。对第二项求导时，注意内层 $2x-3$ 的导数是 $2$ 。

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Differentiating the gradient gives

\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =-\frac{16}{x^3}+\frac{40}{(2x-3)^3}.

At $x=4$ , this is $-\frac{16}{64}+\frac{40}{125}=-\frac14+\frac8{25}=\frac7{100}$ .