题目

Problem

Figure 1 shows a sketch of part of the curve $C$ with equation

\begin{align*} y = \frac{9x - x^2}{2\sqrt{x}}, \quad x > 0 \end{align*}

The point $P$ is the stationary point on $C$ .

(a) Find, using calculus, the $x$ coordinate of $P$ .

(4)

The finite region $R$ , shown shaded in Figure 1, is bounded by the curve $C$ , the $x$ -axis and the lines $x = 1$ and $x = 9$ .

(b) Using calculus, calculate the exact area of $R$ .

(5)

题目中文翻译

图 1 是曲线 $C$ 的草图，曲线 $C$ 的方程为 $y = \dfrac{9x - x^2}{2\sqrt{x}}$ ， $x > 0$ 。

点 $P$ 是 $C$ 上的驻点。

(a) 用微积分求 $P$ 的 $x$ 坐标。

图 1 中阴影所示的有限区域 $R$ 由曲线 $C$ 、 $x$ 轴以及直线 $x = 1$ 和 $x = 9$ 围成。

(b) 用微积分计算 $R$ 的精确面积。

解答

(a)

解法一

思路

展开

先把 $y$ 拆成两个幂函数之和，再逐项求导。令 $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ 即可解出驻点的 $x$ 坐标。

答题过程

展开

Rewrite the equation of $C$ as

\begin{align*} y = \frac{9}{2}x^{1/2} - \frac{1}{2}x^{3/2}. \end{align*}

Differentiating with respect to $x$ ,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{9}{2} \cdot \frac{1}{2}x^{-1/2} - \frac{1}{2} \cdot \frac{3}{2}x^{1/2}\\[2mm] =&\,\frac{9}{4}x^{-1/2} - \frac{3}{4}x^{1/2}. \end{align*}

At the stationary point $P$ , $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ , so

\begin{align*} \frac{9}{4}x^{-1/2} - \frac{3}{4}x^{1/2} = 0. \end{align*}

Multiplying both sides by $4x^{1/2}$ (valid since $x > 0$ ),

\begin{align*} 9 - 3x = 0 \end{align*}

\begin{align*} x = 3. \end{align*}

Therefore the $x$ coordinate of $P$ is $3$ .

(b)

解法一

思路

展开

区域 $R$ 由曲线 $C$ 与 $x$ 轴在 $x = 1$ 到 $x = 9$ 之间围成。对 $y = \dfrac{9x - x^2}{2\sqrt{x}}$ 积分即可。先把被积函数拆成幂函数形式，再逐项积分。

答题过程

展开

The region $R$ is bounded above by the curve $C$ and below by the $x$ -axis, from $x = 1$ to $x = 9$ .

\begin{align*} A = \int_1^9 \frac{9x - x^2}{2\sqrt{x}} \,\mathrm{d}x \end{align*}

Rewriting the integrand,

\begin{align*} A = \int_1^9 \left(\frac{9}{2}x^{1/2} - \frac{1}{2}x^{3/2}\right) \mathrm{d}x \end{align*}

Integrating term by term,

\begin{align*} A =&\,\left[\frac{9}{2} \cdot \frac{2}{3}x^{3/2} - \frac{1}{2} \cdot \frac{2}{5}x^{5/2}\right]_1^9\\[2mm] =&\,\left[3x^{3/2} - \frac{1}{5}x^{5/2}\right]_1^9. \end{align*}

Evaluating at the upper limit $x = 9$ ,

\begin{align*} 3(9)^{3/2} - \frac{1}{5}(9)^{5/2} =&\, 3 \times 27 - \frac{1}{5} \times 243\\[2mm] =&\, 81 - \frac{243}{5}\\[2mm] =&\, \frac{405 - 243}{5}\\[2mm] =&\, \frac{162}{5}. \end{align*}

Evaluating at the lower limit $x = 1$ ,

\begin{align*} 3(1)^{3/2} - \frac{1}{5}(1)^{5/2} =&\, 3 - \frac{1}{5}\\[2mm] =&\, \frac{14}{5}. \end{align*}

Therefore

\begin{align*} A = \frac{162}{5} - \frac{14}{5} = \frac{148}{5}. \end{align*}

The exact area of $R$ is $\dfrac{148}{5}$ .

IAL 2024 May R Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程