题目

Problem

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

\begin{align*} f'(x)=30+\frac{6-5x^2}{\sqrt{x}} \end{align*}

Given that the point $P(4,-8)$ lies on $C$ ,

(a) find the equation of the normal to $C$ at $P$ , giving your answer in the form $y=mx+c$ , where $m$ and $c$ are constants.

(4)

(b) Find $f(x)$ , giving each term in its simplest form.

(5)

解答

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法线斜率是切线斜率的负倒数。切线斜率是 $f'(4)$ ，所以先代入 $x=4$ 。

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At $x=4$ ,

\begin{align*} f'(4) =&\,30+\frac{6-5(4)^2}{\sqrt4}\\ =&\,30+\frac{6-80}{2}\\ =&\,30-37\\ =&\,-7. \end{align*}

So the gradient of the normal is

\begin{align*} \frac17. \end{align*}

Using $P(4,-8)$ ,

\begin{align*} y+8=&\,\frac17(x-4)\\ y=&\,\frac17x-\frac47-8\\ y=&\,\frac17x-\frac{60}{7}. \end{align*}

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先把 $f'(x)$ 写成幂函数之和，再积分。最后用 $P(4,-8)$ 求积分常数。

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Rewrite $f'(x)$ :

\begin{align*} f'(x) =&\,30+\frac{6-5x^2}{\sqrt{x}}\\ =&\,30+6x^{-1/2}-5x^{3/2}. \end{align*}

Integrating,

\begin{align*} f(x) =&\,30x+6\left(\frac{x^{1/2}}{1/2}\right) -5\left(\frac{x^{5/2}}{5/2}\right)+c\\ =&\,30x+12x^{1/2}-2x^{5/2}+c. \end{align*}

Use $P(4,-8)$ :

\begin{align*} -8 =&\,30(4)+12(4)^{1/2}-2(4)^{5/2}+c\\ =&\,120+24-64+c\\ =&\,80+c. \end{align*}

\begin{align*} c=-88. \end{align*}

Therefore

\begin{align*} f(x)=30x+12\sqrt{x}-2x^{5/2}-88. \end{align*}