题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A curve $C$ has parametric equations

\begin{align*} x=&\,\sin^2t,\\ y=&\,2\tan t,\\[4mm] &0\le t<\frac{\pi}{2}. \end{align*}

The point $P$ with parameter $t=\dfrac{\pi}{4}$ lies on $C$ .

The line $l$ is the normal to $C$ at $P$ , as shown in Figure 3.

(a) Show, using calculus, that an equation for $l$ is

8y+2x=17

(5)

The region $S$ , shown shaded in Figure 3, is bounded by $C$ , $l$ and the $x$ -axis.

(b) Find, using calculus, the exact area of $S$ .

(6)

题目中文翻译

本题中你必须写出解题过程的所有步骤。

完全依赖计算器技术的解法不被接受。

曲线 $C$ 的参数方程为

\begin{align*} x=&\,\sin^2t,\\ y=&\,2\tan t,\\[4mm] &0\le t<\frac{\pi}{2}. \end{align*}

参数为 $t=\dfrac{\pi}{4}$ 的点 $P$ 在曲线 $C$ 上。

如图 3 所示，直线 $l$ 是曲线 $C$ 在点 $P$ 处的法线。

(a) 用微积分证明，直线 $l$ 的一个方程为以下等式：

8y+2x=17

(5)

图 3 中阴影所示区域 $S$ 由 $C$ 、 $l$ 以及 $x$ 轴围成。

(b) 用微积分求 $S$ 的精确面积。

(6)

解答

(a)

解法一

思路

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先把 $t=\frac{\pi}{4}$ 代入参数方程求出点 $P$ 。然后分别求 $\frac{\mathrm{d}x}{\mathrm{d}t}$ 和 $\frac{\mathrm{d}y}{\mathrm{d}t}$ ，由参数方程的求导公式得到切线斜率；法线斜率是其负倒数，最后用点斜式整理成题目要求的方程。

答题过程

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At $t=\frac{\pi}{4}$ ,

x=\sin^2\frac{\pi}{4}=\frac{1}{2}

and

y=2\tan\frac{\pi}{4}=2.

Therefore,

P=\bigg(\frac{1}{2},2\bigg).

Differentiating the parametric equations,

\frac{\mathrm{d}x}{\mathrm{d}t} =2\sin t\cos t =\sin2t

and

\frac{\mathrm{d}y}{\mathrm{d}t} =2\sec^2t.

Hence

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}t} \div\frac{\mathrm{d}x}{\mathrm{d}t}\\ =&\,\frac{2\sec^2t}{2\sin t\cos t}. \end{align*}

At $t=\frac{\pi}{4}$ ,

\bigg.\frac{\mathrm{d}x}{\mathrm{d}t}\bigg|_{t=\frac{\pi}{4}}=1

and

\bigg.\frac{\mathrm{d}y}{\mathrm{d}t}\bigg|_{t=\frac{\pi}{4}}=4.

Therefore,

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac41=4.

The gradient of the normal is therefore $-\frac14$ . Using the point $P$ ,

\begin{align*} y-2=&\,-\frac14\bigg(x-\frac12\bigg),\\ 8y-16=&\,-2x+1,\\ 8y+2x=&\,17. \end{align*}

Thus the required equation of $l$ is

\boxed{8y+2x=17}.

解法二

思路

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先利用 $\tan^2t=\frac{\sin^2t}{1-\sin^2t}$ 消去参数，得到 $x,y$ 的笛卡尔关系；再进行隐式求导，同样求出 $P$ 点的切线与法线斜率。

答题过程

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Since

x=\sin^2t

and

y^2=4\tan^2t,

we obtain

\begin{align*} y^2 =&\,4\frac{\sin^2t}{\cos^2t}\\ =&\,\frac{4x}{1-x}. \end{align*}

Equivalently,

y^2(1-x)=4x.

Differentiating implicitly with respect to $x$ ,

\begin{align*} 2y(1-x)\frac{\mathrm{d}y}{\mathrm{d}x} =&\,4+y^2. \end{align*}

At $P=(\frac12,2)$ ,

2\frac{\mathrm{d}y}{\mathrm{d}x}=4+4,

\frac{\mathrm{d}y}{\mathrm{d}x}=4.

The normal therefore has gradient $-\frac14$ , and

\begin{align*} y-2=&\,-\frac14\bigg(x-\frac12\bigg),\\ 8y+2x=&\,17. \end{align*}

Therefore,

\boxed{8y+2x=17}.

同一消参路线也可保留显式式子后求导。由原参数范围， $\sin t$ 、 $\cos t$ 均非负，且 $\cos t>0$ ，所以

y=\frac{2\sqrt{x}}{\sqrt{1-x}}.

在 $P$ 附近，写成乘积求导：

y=2x^{\frac12}(1-x)^{-\frac12},

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,x^{-\frac12}(1-x)^{-\frac12}\\[4mm] &\,\hspace{2pt}+x^{\frac12}(1-x)^{-\frac32}. \end{align*}

Taking a common denominator,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{(1-x)+x}{\sqrt{x}(1-x)^{\frac32}}\\[4mm] =&\,\frac{1}{\sqrt{x}(1-x)^{\frac32}}. \end{align*}

At $x=\frac12$ , the denominator is $\frac14$ , so the tangent gradient is again $4$ . The same normal through $P$ therefore gives $8y+2x=17$ .

(b)

解法一

思路

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区域 $S$ 可分成两部分：从原点到 $P$ 的曲线下方面积，以及从 $P$ 到法线与 $x$ 轴交点的三角形面积。曲线部分用参数积分 $\int y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t$ ，上下限由 $t=0$ 和 $t=\frac{\pi}{4}$ 得到。

答题过程

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The area under $C$ from $O$ to $P$ is

\int_0^{\frac{\pi}{4}} y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t.

Let $A_C$ denote this area. Using $y=2\tan t$ and $\frac{\mathrm{d}x}{\mathrm{d}t}=2\sin t\cos t$ , the integrand simplifies as

\begin{align*} y\frac{\mathrm{d}x}{\mathrm{d}t} =&\,2\tan t(2\sin t\cos t)\\ =&\,4\sin^2t\\ =&\,2-2\cos2t. \end{align*}

Therefore,

A_C=\int_0^{\frac{\pi}{4}}(2-2\cos2t)\,\mathrm{d}t.

An antiderivative is $F(t)=2t-\sin2t$ . At the limits,

F\bigg(\frac{\pi}{4}\bigg)=\frac{\pi}{2}-1

and

F(0)=0.

Hence

A_C=\frac{\pi}{2}-1.

The line $l$ meets the $x$ -axis when $y=0$ . From

8y+2x=17,

its $x$ -intercept is $\frac{17}{2}$ . The area under $l$ from $P$ to this intercept is a triangle with base

\frac{17}{2}-\frac12=8

and height $2$ . Its area is therefore

\frac12(8)(2)=8.

Hence the exact area of $S$ is

\begin{align*} A_S=&\,\frac{\pi}{2}-1+8\\[4mm] =&\,\boxed{\frac{\pi}{2}+7}. \end{align*}

同一面积分割也可以用积分求法线下方面积，替代三角形公式：

y=-\frac{x}{4}+\frac{17}{8}.

Let

\begin{align*} &\,A_l\\ =&\,\int_{\frac12}^{\frac{17}{2}}\bigg(-\frac{x}{4}+\frac{17}{8}\bigg)\,\mathrm{d}x. \end{align*}

An antiderivative is

G(x)=-\frac{x^2}{8}+\frac{17x}{8}.

At the limits,

\begin{align*} G\bigg(\frac{17}{2}\bigg) =&\,-\frac{289}{32}+\frac{289}{16}\\[4mm] =&\,\frac{289}{32} \end{align*}

and

\begin{align*} G\bigg(\frac12\bigg) =&\,-\frac{1}{32}+\frac{17}{16}\\[4mm] =&\,\frac{33}{32}. \end{align*}

Thus

A_l=\frac{289-33}{32}=8,

and $A_S=A_C+A_l=\frac{\pi}{2}+7$ , as above.

P4 2023 Jan Q8

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程