题目

Problem

The diagram shows part of the curve with equation $y=\frac{4}{(2x-1)^2}$ and parts of the lines $x=1$ and $y=1$ .

The curve passes through the points $A(1,4)$ and $B\left(\frac{3}{2},1\right)$ .

(a) Find the exact volume generated when the shaded region is rotated through $360^\circ$ about the $x$ -axis.

[5]

(b) A triangle is formed from the tangent to the curve at $B$ , the normal to the curve at $B$ and the $x$ -axis.

Find the area of this triangle.

[6]

题目中文翻译

图中显示曲线 $y=\frac{4}{(2x-1)^2}$ 的一部分，以及直线 $x=1$ 和 $y=1$ 的一部分。

该曲线经过点 $A(1,4)$ 和 $B\left(\frac{3}{2},1\right)$ 。

(a) 求阴影区域绕 $x$ 轴旋转 $360^\circ$ 所生成立体的精确体积。

[5]

(b) 由曲线在 $B$ 点的切线、曲线在 $B$ 点的法线以及 $x$ 轴围成一个三角形。

求这个三角形的面积。

[6]

解答

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绕 $x$ 轴旋转后，截面是圆环，内半径为 $1$ ，外半径为曲线高度。

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The volume is

\begin{align*} V=&\,\pi\int_1^{3/2}\Big(\frac{16}{(2x-1)^4}-1\Big)\,\mathrm{d}x\\ =&\,\pi\Big[-\frac8{3(2x-1)^3}-x\Big]_1^{3/2}\\ =&\,\pi\Big(-\frac{11}6+\frac{11}3\Big)\\ =&\,\frac{11\pi}6. \end{align*}

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求 $B$ 处切线斜率，再取负倒数得到法线斜率。两条直线的横截距决定三角形的底。

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Differentiating gives $y'=-16(2x-1)^{-3}$ . At $B=(3/2,1)$ the tangent slope is $-2$ and the normal slope is $1/2$ . Their equations are

y-1=-2(x-3/2),\qquad y-1=\frac12(x-3/2).

Setting $y=0$ gives the intercepts $x=2$ and $x=-1/2$ , respectively. The base has length $5/2$ and the height is $1$ , so the area is $5/4$ .