题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The curve $C$ has equation $y=f(x)$ , $x>0$ .

Given that

(a) find, in simplest form, $f(x)$ .

(5)

The line $l$ is the normal to $C$ at $P$ .

(b) Find the coordinates of the point at which $l$ crosses the $y$ -axis.

(4)

解答

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已知 $f'(x)$ ，要求 $f(x)$ ，所以积分一次。积分后要加常数 $c$ ，再用点 $P(9,8)$ 求出这个常数。

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Rewrite the derivative as

\begin{align*} f'(x)=12x^{-\frac12}+\frac13x-4. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int \left(12x^{-\frac12}+\frac13x-4\right)\,dx\\ =&\,24x^{\frac12}+\frac16x^2-4x+c. \end{align*}

Use $P(9,8)$ :

\begin{align*} 8 =&\,24(9)^{\frac12}+\frac16(9)^2-4(9)+c\\ =&\,24(3)+\frac{81}{6}-36+c\\ =&\,72+\frac{27}{2}-36+c\\ =&\,\frac{99}{2}+c. \end{align*}

Thus

\begin{align*} c =&\,8-\frac{99}{2}\\ =&\,-\frac{83}{2}. \end{align*}

Therefore

\begin{align*} f(x)=24x^{\frac12}+\frac16x^2-4x-\frac{83}{2}. \end{align*}

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先求切线斜率 $f'(9)$ ，法线斜率就是它的负倒数。然后用点 $P(9,8)$ 写出法线方程，令 $x=0$ 求 $y$ 轴截距。

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At $x=9$ ,

\begin{align*} f'(9) =&\,\frac{12}{\sqrt9}+\frac93-4\\ =&\,4+3-4\\ =&\,3. \end{align*}

So the gradient of the normal is

\begin{align*} -\frac13. \end{align*}

The normal passes through $P(9,8)$ :

\begin{align*} y-8=-\frac13(x-9). \end{align*}

At the $y$ -axis, $x=0$ , so

\begin{align*} y-8=&\,-\frac13(0-9)\\ y-8=&\,3\\ y=&\,11. \end{align*}

Therefore the point is

\begin{align*} (0,11). \end{align*}