题目

Problem

A curve has equation $y=f(x)$ , where

\begin{align*} f''(x)=\frac{6}{x^{3/2}}+x,\qquad x>0. \end{align*}

The point $P(4,-50)$ lies on the curve.

Given that $f'(x)=-4$ at $P$ ,

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

(3)

(b) find $f(x)$ .

(8)

解答

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在点 $P$ 处，切线斜率是 $-4$ 。法线斜率是它的负倒数，所以是 $\frac14$ 。

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The gradient of the tangent at $P$ is $-4$ , so the gradient of the normal is

\begin{align*} \frac14. \end{align*}

Using $P(4,-50)$ ,

\begin{align*} y+50=&\,\frac14(x-4)\\ y=&\,\frac14x-51. \end{align*}

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已知的是二阶导数，所以要积分两次。第一次积分后用 $f'(4)=-4$ 求第一个常数；第二次积分后用 $f(4)=-50$ 求第二个常数。

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Write

\begin{align*} f''(x)=6x^{-3/2}+x. \end{align*}

Integrate once:

\begin{align*} f'(x) =&\,6\cdot\frac{x^{-1/2}}{-1/2}+\frac12x^2+k\\ =&\,-12x^{-1/2}+\frac12x^2+k. \end{align*}

Use $f'(4)=-4$ :

\begin{align*} -4 =&\,-12(4)^{-1/2}+\frac12(4)^2+k\\ =&\,-6+8+k\\ =&\,2+k. \end{align*}

\begin{align*} k=-6. \end{align*}

Thus

\begin{align*} f'(x)=-12x^{-1/2}+\frac12x^2-6. \end{align*}

Integrate again:

\begin{align*} f(x) =&\,-12\cdot\frac{x^{1/2}}{1/2} +\frac12\cdot\frac{x^3}{3} -6x+d\\ =&\,-24x^{1/2}+\frac16x^3-6x+d. \end{align*}

Use $f(4)=-50$ :

\begin{align*} -50 =&\,-24(4)^{1/2}+\frac16(4)^3-6(4)+d\\ =&\,-48+\frac{64}{6}-24+d\\ =&\,-72+\frac{32}{3}+d\\ =&\,-\frac{184}{3}+d. \end{align*}

Hence

\begin{align*} d=\frac{34}{3}. \end{align*}

Therefore

\begin{align*} f(x)=-24\sqrt{x}+\frac16x^3-6x+\frac{34}{3}. \end{align*}