题目

Problem

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

Given that

(a) find an equation of the normal to $C$ at $P$ .

(2)

(b) find $f(x)$ , writing your answer in simplest form.

(6)

解答

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切线方程是 $y=3x+4$ ，所以切线斜率是 $3$ 。法线斜率是负倒数，即 $-\frac13$ 。点 $P$ 的 $x$ 坐标是 $4$ ，把它代入切线方程可求出 $P$ 的坐标。

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The tangent has equation

\begin{align*} y=3x+4, \end{align*}

so the gradient of the tangent is $3$ . Therefore the gradient of the normal is

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

At $x=4$ ,

\begin{align*} y=&\,3(4)+4\\ =&\,16. \end{align*}

So $P=(4,16)$ . Hence the normal is

\begin{align*} y-16=-\frac13(x-4). \end{align*}

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已知的是二阶导数，所以要积分两次。第一次积分后会出现常数，用切线斜率 $f'(4)=3$ 求出。第二次积分后再出现常数，用点 $P=(4,16)$ 求出。

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Rewrite

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

Integrate once:

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

Since the tangent gradient at $P$ is $3$ ,

\begin{align*} f'(4)=&\,3. \end{align*}

\begin{align*} 3=&\,2(4)^2+2(4)^{\frac12}+c\\ 3=&\,32+4+c\\ c=&\,-33. \end{align*}

Therefore

\begin{align*} f'(x)=2x^2+2x^{\frac12}-33. \end{align*}

Integrate again:

\begin{align*} f(x) =&\,\int \left(2x^2+2x^{\frac12}-33\right)\,dx\\ =&\,\frac23x^3+\frac43x^{\frac32}-33x+d. \end{align*}

Use $P=(4,16)$ :

\begin{align*} 16 =&\,\frac23(4)^3+\frac43(4)^{\frac32}-33(4)+d\\ =&\,\frac{128}{3}+\frac{32}{3}-132+d\\ =&\,\frac{160}{3}-132+d. \end{align*}

Hence

\begin{align*} d =&\,16-\frac{160}{3}+132\\ =&\,\frac{48}{3}-\frac{160}{3}+\frac{396}{3}\\ =&\,\frac{284}{3}. \end{align*}

Therefore

\begin{align*} f(x)=\frac23x^3+\frac43x^{\frac32}-33x+\frac{284}{3}. \end{align*}