Using Linear Models to Solve Real-World Application Problems | 用直线模型解决实际应用问题

📚 Using Linear Models to Solve Real-World Application Problems | 用直线模型解决实际应用问题

Linear models are one of the most powerful tools in mathematics. They allow us to describe real-world situations with a simple straight-line equation, making it possible to predict, compare, and make decisions effectively.

线性模型是数学中最强大的工具之一。它使我们能够用简单的直线方程描述现实世界中的情境，从而有效地进行预测、比较和决策。

1. What is a Linear Model? | 什么是线性模型？

A linear model is a mathematical equation of the form y = mx + c that describes a relationship where a change in one variable produces a proportional change in another variable.

线性模型是形如 y = mx + c 的数学方程，它描述了一个变量的变化引起另一个变量成比例变化的关系。

For example, if a car travels at a constant speed of 60 km/h, the distance d it covers after t hours is given by d = 60t. Here, the relationship between distance and time is perfectly linear.

例如，如果一辆汽车以 60 公里/小时的恒定速度行驶，它在 t 小时后行驶的距离 d 由 d = 60t 给出。在这里，距离与时间之间的关系是完全线性的。

2. The Equation of a Line: y = mx + c | 直线方程：y = mx + c

The general equation of a straight line is y = mx + c, where m is the gradient (slope) and c is the y-intercept.

直线的一般方程为 y = mx + c，其中 m 是斜率（梯度），c 是 y 截距。

y = mx + c

m = the gradient: the rate of change of y with respect to x

m = 斜率：y 相对于 x 的变化率
c = the y-intercept: the value of y when x = 0

c = y 截距：当 x = 0 时 y 的值

In real-world applications, understanding what m and c mean in context is far more important than simply plotting a graph.

在实际应用中，理解 m 和 c 在具体情境中的含义远比简单地绘制图形重要得多。

3. Identifying Linear Relationships from Data | 从数据中识别线性关系

When we are given a set of data, we can check whether a linear model is appropriate by examining the differences between consecutive y-values.

当我们得到一组数据时，可以通过检查相邻 y 值之间的差值来判断线性模型是否适用。

If the differences are constant (the first differences are the same), then the data follows a linear pattern.

如果这些差值恒定（一阶差分相同），则该数据遵循线性规律。

x	0	1	2	3
y	5	8	11	14

The y-values increase by 3 each time, so the gradient m = 3. When x = 0, y = 5, so c = 5. The model is y = 3x + 5.

y 值每次增加 3，因此斜率 m = 3。当 x = 0 时，y = 5，所以 c = 5。该模型为 y = 3x + 5。

4. Finding the Line: From Two Points to the Equation | 求直线方程：由两点确定直线

Given two points (x₁, y₁) and (x₂, y₂), we can find the gradient first using the formula:

已知两点 (x₁, y₁) 和 (x₂, y₂)，我们首先使用公式求斜率：

m = (y₂ − y₁) ÷ (x₂ − x₁)

Then substitute one point into y = mx + c to find c.

然后将其中一个点代入 y = mx + c 求出 c。

Example: A phone plan charges a fixed monthly fee plus a charge per minute. If 50 minutes cost $25 and 120 minutes cost $46, find the linear model.

示例：某电话套餐收取固定的月费加上每分钟的通话费。如果 50 分钟费用为 25 美元，120 分钟费用为 46 美元，求线性模型。

m = (46 − 25) ÷ (120 − 50) = 21 ÷ 70 = 0.3

Using (50, 25): 25 = 0.3 × 50 + c → c = 10. The model is y = 0.3x + 10, where y is the cost and x is the number of minutes.

代入 (50, 25)：25 = 0.3 × 50 + c → c = 10。模型为 y = 0.3x + 10，其中 y 是费用，x 是通话分钟数。

5. Interpreting the Gradient in Context | 在具体情境中理解斜率

The gradient m represents the rate of change: how much y changes for each one-unit increase in x.

斜率 m 表示变化率：当 x 每增加一个单位时，y 变化多少。

In a distance-time graph, the gradient is the speed.

在距离-时间图像中，斜率就是速度。
In a cost-quantity graph, the gradient is the unit price.

在费用-数量图像中，斜率就是单价。
In a temperature-time graph, the gradient is the rate of heating or cooling.

在温度-时间图像中，斜率就是升温或降温的速率。

A positive gradient indicates an increasing relationship; a negative gradient indicates a decreasing relationship.

斜率为正表示递增关系；斜率为负表示递减关系。

6. Interpreting the y-Intercept | 理解 y 截距的含义

The y-intercept c is the value of y when x = 0. In many situations, it represents a fixed or starting value.

y 截距 c 是当 x = 0 时 y 的值。在许多情境中，它代表固定值或初始值。

For example, in the equation y = 0.3x + 10 for the phone plan, the y-intercept 10 represents the fixed monthly fee — the cost even if you use 0 minutes.

例如，在电话套餐方程 y = 0.3x + 10 中，y 截距 10 代表固定的月费——即使你使用 0 分钟也需要支付的费用。

In a distance-time graph starting from a point already 5 km from home, the intercept would be 5.

在距离-时间图像中，如果起始位置离家已有 5 公里，截距就是 5。

7. Real-World Example 1: Taxi Fare Problem | 实际应用一：出租车计费问题

A taxi company charges a flag-fall fee of $3.50 plus $0.80 per kilometre. Write a linear model and find the cost of a 12 km journey.

某出租车公司收取起步费 3.50 美元，外加每公里 0.80 美元。写出线性模型，并计算 12 公里行程的费用。

The model is y = 0.8x + 3.5, where x is the distance in kilometres and y is the total cost in dollars.

模型为 y = 0.8x + 3.5，其中 x 是以公里为单位的距离，y 是以美元为单位的总费用。

For x = 12: y = 0.8 × 12 + 3.5 = 9.6 + 3.5 = 13.1. The journey costs $13.10.

当 x = 12 时：y = 0.8 × 12 + 3.5 = 9.6 + 3.5 = 13.1。该行程费用为 13.10 美元。

We can also reverse the process: if a passenger pays $18.30, how far did they travel?

我们也可以反向计算：如果乘客支付了 18.30 美元，他行驶了多远？

18.3 = 0.8x + 3.5 → 0.8x = 14.8 → x = 18.5 km.

18.3 = 0.8x + 3.5 → 0.8x = 14.8 → x = 18.5 公里。

8. Real-World Example 2: Water Tank Draining | 实际应用二：水箱排水问题

A water tank initially contains 240 litres. Water drains out at a constant rate of 15 litres per minute.

一个水箱初始含有 240 升水。水以每分钟 15 升的恒定速率排出。

The linear model is V = 240 − 15t, where V is the volume in litres and t is the time in minutes.

线性模型为 V = 240 − 15t，其中 V 是以升为单位的体积，t 是以分钟为单位的时间。

The gradient is −15, meaning the volume decreases by 15 litres each minute. The intercept is 240, the initial volume.

斜率为 −15，表示体积每分钟减少 15 升。截距为 240，即初始体积。

To find when the tank is empty, set V = 0: 0 = 240 − 15t → t = 16 minutes.

求水箱何时排空，令 V = 0：0 = 240 − 15t → t = 16 分钟。

9. Predictions and Extrapolation | 预测与外推

Once we have a linear model, we can use it to make predictions by substituting a value into the equation.

一旦我们得到了线性模型，就可以通过将数值代入方程来进行预测。

However, it is important to distinguish between interpolation (predicting within the range of observed data) and extrapolation (predicting beyond the observed range).

然而，区分内插（在观察数据范围内进行预测）和外推（在观察范围之外进行预测）非常重要。

Interpolation is usually safe because we are predicting within a tested range. Extrapolation is risky because real-world relationships may change outside the observed data.

内插通常比较安全，因为我们是在经过验证的范围内进行预测。外推则存在风险，因为现实世界中的关系在观察数据之外可能会发生变化。

For example, a linear model of population growth may work well for small towns but break down as resources become limited.

例如，一个线性人口增长模型在小城镇中可能效果很好，但随着资源变得有限，模型就会失效。

10. Limitations of Linear Models | 线性模型的局限性

Linear models assume a constant rate of change. In reality, many situations are not perfectly linear: growth may slow down, costs may decrease with bulk discounts, and external factors may intervene.

线性模型假设变化率恒定。在现实中，许多情况并非完全线性：增长可能会减慢，成本可能因批量折扣而降低，外部因素也可能进行干预。

Additionally, linear models are only valid within a certain domain. For example, a model y = 100 − 2x becomes meaningless when x > 50, since y would become negative.

此外，线性模型只在一定的定义域内有效。例如，当 x > 50 时，模型 y = 100 − 2x 就变得没有意义，因为 y 会变成负数。

Always consider whether a linear model is reasonable for the situation before using it to make predictions.

在使用线性模型进行预测之前，务必考虑该模型在这种情况下是否合理。

11. Exam Strategies and Common Pitfalls | 考试策略与常见误区

Always identify which variable is x and which is y before writing the equation.

在写方程之前，务必先确定哪个变量是 x，哪个变量是 y。
Check the units: if x is in hours and y is in kilometres, the gradient has units km/h.

检查单位：如果 x 以小时为单位，y 以公里为单位，则斜率的单位是公里/小时。
Do not confuse the gradient with the intercept. Read the question carefully to see which value is needed.

不要混淆斜率和截距。仔细阅读题目，看清需要的是哪个值。
When finding the equation from two points, always write the answer in the required form.

当由两点求方程时，始终以要求的格式写出答案。

Linear models connect mathematics to everyday life. Mastering them not only earns marks in exams but also builds critical thinking skills for solving practical problems.

线性模型将数学与日常生活联系起来。掌握它们不仅能在考试中得分，还能培养解决实际问题的批判性思维能力。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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Using Linear Models to Solve Real-World Application Problems | 用直线模型解决实际应用问题