题目

Problem

The equation of a curve is $y = 4x^2 + \frac{9}{x^2} - 8$ .

(a) A point $P$ is moving along the curve in such a way that its $y$ -coordinate is decreasing at 5 units per second.

Find the rate at which the $x$ -coordinate of point $P$ is changing when $x = 2$ .

[4]

(b) Find the coordinates of the stationary points of the curve and determine their nature.

[5]

题目中文翻译

一条曲线的方程为 $y = 4x^2 + \frac{9}{x^2} - 8$ 。

(a) 点 $P$ 沿曲线运动，并且其 $y$ 坐标以每秒 5 个单位的速率减小。

当 $x = 2$ 时，求点 $P$ 的 $x$ 坐标的变化率。

[4]

(b) 求曲线驻点的坐标，并判断其性质。

[5]

解答

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下降的纵坐标意味着 $\frac{\mathrm{d}y}{\mathrm{d}t}=-5$ 。用链式法则把它除以曲线在 $x=2$ 处的斜率。

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Differentiating gives $\frac{\mathrm{d}y}{\mathrm{d}x}=8x-18/x^3$ . At $x=2$ , this is $16-18/8=55/4$ . Hence

-5=\frac{55}4\frac{\mathrm{d}x}{\mathrm{d}t} \quad\Longrightarrow\quad \frac{\mathrm{d}x}{\mathrm{d}t}=-\frac4{11}

units per second.

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令一阶导数为零，得到关于 $x^4$ 的方程。两个实根的平方相同，因此纵坐标相同。

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For $x\ne0$ , the stationary-point condition $8x-18/x^3=0$ gives $x^4=9/4$ . Thus $x=\pm\sqrt{3/2}$ . At either value,

y=4\Big(\frac32\Big)+\frac9{3/2}-8=4.

The second derivative is $8+54/x^4$ , which is positive at both points. Thus both $(-\sqrt{3/2},4)$ and $(\sqrt{3/2},4)$ are minima.