题目

Problem

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

\begin{align*} f'(x)=3\sqrt{x}-\frac{9}{x\sqrt{x}}+\frac43 \end{align*}

Given that the point $P(9,20)$ lies on $C$ ,

(a) find $f(x)$ , simplifying the answer,

(5)

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

(4)

解答

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已知的是 $f'(x)$ ，所以先积分得到 $f(x)$ ，再用点 $P(9,20)$ 求积分常数。积分前先把根式和分式写成指数形式。

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Rewrite $f'(x)$ as

\begin{align*} f'(x)=3x^{1/2}-9x^{-3/2}+\frac43. \end{align*}

Integrating,

\begin{align*} f(x) =&\,\int\left(3x^{1/2}-9x^{-3/2}+\frac43\right)\,\mathrm{d}x\\ =&\,2x^{3/2}+18x^{-1/2}+\frac43x+c. \end{align*}

Since $P(9,20)$ lies on the curve,

\begin{align*} 20 =&\,2(9)^{3/2}+18(9)^{-1/2}+\frac43(9)+c\\ =&\,2(27)+18\left(\frac13\right)+12+c\\ =&\,54+6+12+c. \end{align*}

\begin{align*} c=-52. \end{align*}

Therefore

\begin{align*} f(x)=2x^{3/2}+18x^{-1/2}+\frac43x-52. \end{align*}

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法线斜率是切线斜率的负倒数。切线斜率由 $f'(9)$ 给出，然后用点 $P(9,20)$ 写直线方程。

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At $x=9$ ,

\begin{align*} f'(9) =&\,3\sqrt9-\frac{9}{9\sqrt9}+\frac43\\ =&\,9-\frac13+\frac43\\ =&\,10. \end{align*}

So the gradient of the normal is

\begin{align*} -\frac{1}{10}. \end{align*}

Using $P(9,20)$ ,

\begin{align*} y-20=&\,-\frac{1}{10}(x-9)\\ 10y-200=&\,-x+9\\ x+10y-209=&\,0. \end{align*}