题目

Problem

The curve $C$ has equation

y=x^2\cos\left(\frac{1}{2}x\right)\qquad 0<x\le \pi

The curve has a stationary point at the point $P$ .

(a) Show, using calculus, that the $x$ coordinate of $P$ is a solution of the equation

x=2\arctan\left(\frac{4}{x}\right)

(4)

Using the iteration formula

x_{n+1}=2\arctan\left(\frac{4}{x_n}\right)\qquad x_1=2

(b) find the value of $x_2$ and the value of $x_6$ , giving your answers to 3 decimal places.

(3)

题目中文翻译

曲线 $C$ 的方程为

y=x^2\cos\left(\frac{1}{2}x\right)\qquad 0<x\le \pi

该曲线在点 $P$ 处有一个驻点。

(a) 用求导证明，点 $P$ 的 $x$ 坐标满足方程

x=2\arctan\left(\frac{4}{x}\right)

使用迭代公式

x_{n+1}=2\arctan\left(\frac{4}{x_n}\right)\qquad x_1=2

(b) 求 $x_2$ 和 $x_6$ 的值，答案精确到小数点后 3 位。

解答

The curve is

y=x^2\cos\left(\frac12x\right)

Differentiate using the product rule:

\begin{aligned} \frac{dy}{dx} &=2x\cos\left(\frac12x\right) +x^2\left[-\sin\left(\frac12x\right)\right]\cdot \frac12\\ &=2x\cos\left(\frac12x\right)-\frac12x^2\sin\left(\frac12x\right) \end{aligned}

At a stationary point,

\frac{dy}{dx}=0

2x\cos\left(\frac12x\right)-\frac12x^2\sin\left(\frac12x\right)=0

Since

0<x\leq \pi

we may divide by $x$ :

2\cos\left(\frac12x\right)-\frac12x\sin\left(\frac12x\right)=0

Rearrange:

2\cos\left(\frac12x\right)=\frac12x\sin\left(\frac12x\right)

Multiply by $2$ :

4\cos\left(\frac12x\right)=x\sin\left(\frac12x\right)

Therefore

\frac{\sin\left(\frac12x\right)}{\cos\left(\frac12x\right)}=\frac{4}{x}

\tan\left(\frac12x\right)=\frac{4}{x}

Taking arctan on both sides gives

\frac12x=\arctan\left(\frac{4}{x}\right)

Hence

\boxed{x=2\arctan\left(\frac{4}{x}\right)}

as required.

The iteration formula is

x_{n+1}=2\arctan\left(\frac{4}{x_n}\right)

with

x_1=2

For $x_2$ :

x_2=2\arctan\left(\frac42\right)

x_2=2.214297\ldots

Therefore

\boxed{x_2=2.214}

to 3 decimal places.

Continuing the iteration:

\begin{aligned} x_2&=2.214297\ldots\\ x_3&=2.130426\ldots\\ x_4&=2.162810\ldots\\ x_5&=2.150239\ldots\\ x_6&=2.155109\ldots \end{aligned}

Hence

\boxed{x_6=2.155}

to 3 decimal places.