题目

Problem

The curve $C$ with equation $y=f(x)$ , $x>0$ , passes through the point $P(4,1)$ .

Given that

\begin{align*} f'(x)=4\sqrt{x}-2-\frac{8}{3x^2}, \end{align*}

(a) find the equation of the normal to $C$ at $P$ . Write your answer in the form $ax+by+c=0$ , where $a$ , $b$ and $c$ are integers to be found.

(4)

(b) Find $f(x)$ .

(5)

解答

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先求切线斜率 $f'(4)$ ，再取负倒数得到法线斜率。最后用点 $P(4,1)$ 写直线。

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\begin{align*} f'(4) =&\,4\sqrt4-2-\frac{8}{3(4)^2}\\ =&\,8-2-\frac{8}{48}\\ =&\,6-\frac16\\ =&\,\frac{35}{6}. \end{align*}

So the gradient of the normal is

\begin{align*} -\frac{6}{35}. \end{align*}

Using $P(4,1)$ ,

\begin{align*} y-1=&\,-\frac{6}{35}(x-4). \end{align*}

Multiply by $35$ :

\begin{align*} 35y-35=&\,-6x+24\\ 6x+35y-59=&\,0. \end{align*}

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把 $f'(x)$ 写成指数形式后积分。积分常数用点 $P(4,1)$ 求出。

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\begin{align*} f'(x)=4x^{1/2}-2-\frac83x^{-2}. \end{align*}

Integrating,

\begin{align*} f(x) =&\,4\cdot\frac{x^{3/2}}{3/2}-2x-\frac83\cdot\frac{x^{-1}}{-1}+c\\ =&\,\frac83x^{3/2}-2x+\frac{8}{3x}+c. \end{align*}

Use $f(4)=1$ :

\begin{align*} 1=&\,\frac83(4)^{3/2}-2(4)+\frac{8}{3(4)}+c\\ =&\,\frac{64}{3}-8+\frac23+c\\ =&\,14+c. \end{align*}

Hence

\begin{align*} c=-13. \end{align*}

Therefore

\begin{align*} f(x)=\frac83x^{3/2}-2x+\frac{8}{3x}-13. \end{align*}