题目

Problem

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

f(x)=\arcsin\left(\frac{x}{2}\right)\qquad -2\le x\le 2,\ -\frac{\pi}{2}\le y\le \frac{\pi}{2}

(a) Sketch $C$ .

(1)

(b) Given $x=2\sin y$ , show that

\frac{dy}{dx}=\frac{1}{\sqrt{A-x^2}}

where $A$ is a constant to be found.

(3)

The point $P$ lies on $C$ and has $y$ coordinate $\dfrac{\pi}{4}$ .

(c) Find the equation of the tangent to $C$ at $P$ . Write your answer in the form $y=mx+c$ , where $m$ and $c$ are constants to be found.

(3)

题目中文翻译

曲线 $C$ 的方程为 $y=f(x)$ ，其中

f(x)=\arcsin\left(\frac{x}{2}\right)\qquad -2\le x\le 2,\ -\frac{\pi}{2}\le y\le \frac{\pi}{2}

(a) 画出曲线 $C$ 的草图。

(b) 已知 $x=2\sin y$ ，证明

\frac{dy}{dx}=\frac{1}{\sqrt{A-x^2}}

其中 $A$ 是需要求出的常数。

点 $P$ 在曲线 $C$ 上，且其 $y$ 坐标为 $\dfrac{\pi}{4}$ 。

解答

The graph of

y=\arcsin\left(\frac{x}{2}\right)

passes through

(-2,-\frac{\pi}{2}),\qquad (0,0),\qquad (2,\frac{\pi}{2})

and is increasing throughout its domain.

We are given

x=2\sin y

Differentiate both sides with respect to $y$ :

\frac{\mathrm{d}x}{\mathrm{d}y}=2\cos y

Therefore

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{2\cos y}

Now

\sin y=\frac{x}{2}

\cos y=\sqrt{1-\sin^2 y} =\sqrt{1-\frac{x^2}{4}}

Since $-\dfrac{\pi}{2}\le y\le\dfrac{\pi}{2}$ , we take the non-negative square root.

Hence

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{2\sqrt{1-\frac{x^2}{4}}}

Simplifying,

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{\sqrt{4-x^2}}

Therefore

\boxed{A=4}

At $P$ ,

y=\frac{\pi}{4}

Using

x=2\sin y

we get

x=2\sin\frac{\pi}{4} =2\cdot\frac{\sqrt2}{2} =\sqrt2

P=\left(\sqrt2,\frac{\pi}{4}\right)

The gradient at $P$ is

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{\sqrt{4-(\sqrt2)^2}} =\frac{1}{\sqrt2} =\frac{\sqrt2}{2}

Using the point-gradient form of a line,

y-\frac{\pi}{4} =\frac{\sqrt2}{2}(x-\sqrt2)

Therefore

y=\frac{\sqrt2}{2}x-1+\frac{\pi}{4}

So the equation of the tangent is

\boxed{y=\frac{\sqrt2}{2}x+\frac{\pi}{4}-1}