题目

Problem

Figure 4 is a graph showing the path of a golf ball after the ball has been hit until it first hits the ground.

The vertical height, $h$ metres, of the ball above the ground has been plotted against the horizontal distance travelled, $x$ metres, measured from where the ball was hit.

The ball travels a horizontal distance of $d$ metres before it first hits the ground.

The ball is modelled as a particle travelling in a vertical plane above horizontal ground.

The path of the ball is modelled by the equation

h=1.5x-0.5xe^{0.02x}\qquad 0\leq x\leq d

Use the model to answer parts (a), (b) and (c).

(a) Find the value of $d$ , giving your answer to 2 decimal places.

(Solutions relying entirely on calculator technology are not acceptable.)

(3)

(b) Show that the maximum value of $h$ occurs when

x=50\ln\left(\frac{150}{x+50}\right)

(4)

Using the iteration formula

x_{n+1}=50\ln\left(\frac{150}{x_n+50}\right)\qquad \text{with}\qquad x_1=30

(ii) find, by repeated iteration, the horizontal distance travelled by the golf ball before it reaches its maximum height. Give your answer to 2 decimal places.

(3)

题目中文翻译

图 4 给出了高尔夫球从被击出到第一次落地之间路径的图像。

球离地面的竖直高度为 $h$ 米，已绘制成关于水平距离 $x$ 米的图像，其中 $x$ 从球被击出的地点开始测量。

球在第一次落地前一共飞行了 $d$ 米的水平距离。

将球视为在水平地面上方竖直平面内运动的质点。

球的路径模型方程为

h=1.5x-0.5xe^{0.02x}\qquad 0\leq x\leq d

使用该模型回答 (a)、(b) 和 (c)。

(a) 求 $d$ 的值，答案精确到小数点后 2 位。

（不接受依赖计算器技术的解法。）

(b) 证明 $h$ 的最大值出现在

x=50\ln\left(\frac{150}{x+50}\right)

时。

利用迭代公式

x_{n+1}=50\ln\left(\frac{150}{x_n+50}\right)\qquad \text{其中}\qquad x_1=30

(ii) 用重复迭代求高尔夫球到达最大高度前所飞行的水平距离，答案精确到小数点后 2 位。

解答

(a)

解法一

思路

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球第一次落地时 $h=0$ ，而对应的水平距离 $d>0$ 。代入后提出并约去因子 $d$ ，再使用自然对数解指数方程；这样完整展示代数过程，而不是只依赖计算器求根。

答题过程

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When the ball first reaches the ground, $x=d$ and $h=0$ . Hence

0=1.5d-0.5de^{0.02d}.

Since $d>0$ , division by $0.5d$ gives

e^{0.02d}=3.

Taking natural logarithms,

\begin{align*} 0.02d=&\,\ln 3, \\[2mm] d=&\,\frac{\ln 3}{0.02} \\[2mm] =&\,54.9306\ldots. \end{align*}

Therefore, to $2$ decimal places,

\boxed{d=54.93\text{ m}}.

(b)

解法一

思路

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对 $h$ 关于 $x$ 求导，其中 $xe^{0.02x}$ 要使用乘积法则。最大高度出现在区间内部的驻点，因此令导数为零，并逐步整理成题目要求的迭代形式。

答题过程

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Differentiating using the product rule,

\begin{align*} \frac{\mathrm{d}h}{\mathrm{d}x} =&\,1.5-0.5\big(e^{0.02x} +0.02xe^{0.02x}\big) \\[2mm] =&\,1.5-e^{0.02x}(0.5+0.01x). \end{align*}

At a stationary point,

1.5=e^{0.02x}(0.5+0.01x).

Therefore,

\begin{align*} e^{0.02x} =&\,\frac{1.5}{0.5+0.01x} \\[2mm] =&\,\frac{150}{x+50}. \end{align*}

Taking natural logarithms,

\begin{align*} 0.02x=&\,\ln\bigg(\frac{150}{x+50}\bigg), \\[2mm] x=&\,50\ln\bigg(\frac{150}{x+50}\bigg). \end{align*}

Also,

\frac{\mathrm{d}^2h}{\mathrm{d}x^2} =-e^{0.02x}(0.02+0.0002x)<0

for $0\leq x\leq d$ . Hence this stationary point is a maximum, as required.

(c)(i)

解法一

思路

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将初值 $x_1=30$ 代入题目给出的迭代公式，直接计算下一项。

答题过程

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Using $x_1=30$ ,

\begin{align*} x_2=&\,50\ln\bigg(\frac{150}{30+50}\bigg) \\[2mm] =&\,31.4304\ldots. \end{align*}

Therefore, to $2$ decimal places,

\boxed{x_2=31.43}.

(c)(ii)

解法一

思路

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继续重复使用迭代公式。数列在极限值两侧交替靠近，因此应继续计算，直到相邻项都稳定地舍入为相同的两位小数。

答题过程

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Repeated iteration gives

\begin{array}{c|c@{\qquad}c|c} n & x_n & n & x_n \\ \hline 1 & 30.000000 & 11 & 30.874910 \\ 2 & 31.430433 & 12 & 30.886583 \\ 3 & 30.544311 & 13 & 30.879367 \\ 4 & 31.091391 & 14 & 30.883828 \\ 5 & 30.752925 & 15 & 30.881070 \\ 6 & 30.962056 & 16 & 30.882775 \\ 7 & 30.832735 & 17 & 30.881721 \\ 8 & 30.912664 & 18 & 30.882372 \\ 9 & 30.863247 & 19 & 30.881969 \\ 10 & 30.893794 & 20 & 30.882218 \end{array}

The iteration converges to

x=30.882123\ldots.

Therefore, the horizontal distance travelled before the ball reaches its maximum height is

\boxed{30.88\text{ m}}.

IAL 2024 May Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)(i)

解法一

思路

答题过程

(c)(ii)

解法一

思路

答题过程