课程： 5.8 直接变化 | The Way To Learn

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直接变化

According to Newton's second law, the net force ( F ) of an object is equal to its mass ( m ) times its acceleration ( a ), where F is measured in Newtons, m is measured in kilograms, and a is measured in meters per second squared ( $\begin{aligned} \frac{m}{s^{2}} \end{aligned}$ ). A person is ice skating and their friend pushes them with a force of 24 Newtons. That causes them to accelerate at 2 meters per second squared . What force does the friend need to apply to get their friend moving with an acceleration of 6 meters per second squared ?
::根据牛顿的第二项法律,一个物体的净力(F)等于其质量(m)乘以加速度(a),在牛顿测量F,米以公斤计,米以每秒平方公尺(ms2)计,米以米计,人滑冰,朋友用24牛顿的力量推他们,导致他们加速速度达到每秒平方公尺2米。朋友需要用什么力量让朋友加快速度,每秒平方公尺6米?

lesson content

Direct Variation
::直接变化

Variation describes a relationship between independent and dependent variables where the variables are related by a constant .
::变式表示独立变量和依附变量之间的关系,变量因常数而相关。

In direct variation , the independent variable is equal to the dependent variable times a constant. We call this constant the constant of variation or the constant of proportionality . Verbally, we can describe this as " y varies directly as x " or " y is directly proportional to x ."
::在直接变化中, 独立的变量等于依附变量乘以常数。我们称此常数为随附变量乘以常数。我们称此常数为变数常数或相称性常数。我们可以用“ y 直接变化为 x ” 或“ y 直接与 x 成比例 ” 来形容此常数。

Direct Variation
::直接变化

\begin{aligned} y = k x, y = k x^{n}, k \neq 0, n > 0 \end{aligned}

::y=kx,y=kxn,k0,n>0

k is the constant of variation or constant of proportionality. We say, " y varies directly as $\begin{aligned} x^{n} \end{aligned}$ " or " y is directly proportional to x ).
::k 是变化的常数或相称性的常数。我们说, “ y 以 xn 直接变化 ” 或“ y 与 x 直接成比例 ” 。

Note that when $\begin{aligned} k > 0 \end{aligned}$ , when x goes up, so does y . However, k can also be negative, so as x goes up, y goes down.
::注意当 k>0 时, x 上升时, y 也会发生。但是, k 也可能是负的, 所以 x 上升时, y 会下降。

Also, if we consider the exponent to be 1, we have $\begin{aligned} y = k x \end{aligned}$ , a line with a of k that goes through the origin.
::还有,如果我们把前言看成是1, 我们有y=kx, 与K的直线, 贯穿源头。

by Mathispower4u explains direct variation and solves application problems.
::Mathispower4u 解释直接变异和解决应用程序问题。

Example 1
::例1

Determine if the set of data varies directly as $\begin{aligned} y = k x \end{aligned}$ . If so, find the direct variation equation .
::确定数据集是否与y=kx直接不同。如果是,请找到直接变异方程。

$\begin{aligned} x \end{aligned}$	4	8	16	20
$\begin{aligned} y \end{aligned}$	1	2	4	5

Solution: Looking at the set of data, the x -values increase. For the data to vary directly, the y- values would have to be the same constant multiple of each other. Let's find the constant for the first set of data.
::解答 : 查看数据集, x 值会增加。数据要直接变化, Y 值必须是相同的常数倍数。让我们为第一组数据找到常数。

\begin{aligned} y & = k x \\ 1 & = k (4) \\ \frac{1}{4} & = k \end{aligned}

::y=kx1=k(4)14=k

The equation for the first point is $\begin{aligned} y = \frac{1}{4} x \end{aligned}$ . Substitute the other values into this equation to make sure the constant of variation works for each pair of values.
::第一个点的方程是y=14x。将其他值替换到此方程, 以确保每对值的变量常数发挥作用。

\begin{aligned} 2 = \frac{1}{4} (8) & 4 = \frac{1}{4} (16) & 5 = \frac{1}{4} (20) \\ 2 = 2 & 4 = 4 & 5 = 5 \end{aligned}

Since the equation holds for all of the data, x varies directly as y.
::由于方程式持有所有数据, x 与 y 直接不同。

Example 2
::例2

The variables $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ vary directly, and $\begin{aligned} y = 10 \end{aligned}$ when $\begin{aligned} x = 2 \end{aligned}$ . Write an equation that relates $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ and find $\begin{aligned} y \end{aligned}$ when $\begin{aligned} x = 9 \end{aligned}$ .
::变量 x 和 y 直接变化, y= 10 当 x= 2 时, y= 10 。写入一个与 x 和 y 相关的方程式, 然后在 x= 9 时找到 y 。

Solution: Using the direct variation equation, we can substitute in $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ and solve for $\begin{aligned} k \end{aligned}$ .
::解答: 使用直接变异方程式, 我们可以替换 x 和 y , 并解决 k 。

\begin{aligned} y & = k x \\ 10 & = k (2) \\ 5 & = k \end{aligned}

::y=kx10=k(2)5=k

Therefore , the equation is $\begin{aligned} y = 5 x \end{aligned}$ . To find $\begin{aligned} y \end{aligned}$ when $\begin{aligned} x \end{aligned}$ is 9, we have $\begin{aligned} y = 5 \cdot 9 = 45 \end{aligned}$ .
::因此,方程式是y=5x。在 x 9 时查找y,我们有y=5=9=45。

by CK-12 demonstrates how to write direct variation equation s given one point.
::通过 CK-12 演示如何写出给定一个点的直接变异方程。

Example 4
::例4

$\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ vary directly. When $\begin{aligned} x = - 8, y = 6 \end{aligned}$ . Find the equation and determine $\begin{aligned} x \end{aligned}$ when $\begin{aligned} y = 12 \end{aligned}$ .
::x 和 y 直接变化。当 x8,y=6 时, 查找方程, 确定 x y=12 时。

Solution: First, solve for $\begin{aligned} k \end{aligned}$ .
::解答: 第一, K的解答。

\begin{aligned} k = \frac{y}{x} = \frac{6}{- 8} = - \frac{3}{4} ⟶ y = - \frac{3}{4} x \end{aligned}

::kyx= 6 - 8\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\34x

Now, substitute in 12 for $\begin{aligned} y \end{aligned}$ and solve for $\begin{aligned} x \end{aligned}$ .
::现在,以 y 12 取代 y , 用 x 解析。

\begin{aligned} 12 & = - \frac{3}{4} x \\ - \frac{4}{3} \cdot 12 & = x \\ - 16 & = x \end{aligned}

::12-16=xxxxxxxxxxxxxxxxxxxxxxxx

Example 5
::例5

A person is ice skating and their friend pushes them with a force of 102 Newtons. That causes them to accelerate at 2 $\begin{aligned} \frac{m}{s^{2}} \end{aligned}$ . What force does the friend need to apply to get their friend moving with an acceleration of 3 $\begin{aligned} \frac{m}{s^{2}} \end{aligned}$ ?
::一个人在滑冰,朋友用102牛顿的力推他们,导致他们加速2毫秒2秒。朋友需要用什么力量才能让朋友加速3毫秒2秒的移动?

Solution: Writing Newton's second law as an equation, we have $\begin{aligned} F = m a \end{aligned}$ . This is an example of a direct variation equation where $\begin{aligned} y = F \end{aligned}$ , $\begin{aligned} m = k \end{aligned}$ , and $\begin{aligned} a = x \end{aligned}$ . Using the direct variation equation, we can substitute in $\begin{aligned} F \end{aligned}$ and $\begin{aligned} a \end{aligned}$ and solve for $\begin{aligned} m \end{aligned}$ .
::解答: 将牛顿的第二定律写成一个方程式, 我们有 F=ma 。这是一个直接变异方程式的例子, 即 y= F, m=k 和 a=x 。使用直接变异方程式, 我们可以用 F 和 a 来替代 m , 并解决 m 。

\begin{aligned} F & = m a \\ 102 & = m (2) \\ 51 & = m \end{aligned}

::F=ma102=m(2)51=m

The mass of the object, or constant of proportionality, is 51 kg . Now, we can use this in the direct variation equation to find the force necessary to increase the skater's acceleration.
::对象质量或相称性常数为51千克。现在, 我们可以在直接变异方程中使用它来找到增加滑冰者加速度所需的力量。

\begin{aligned} F & = m a \\ F & = (51) (3) \\ F & = 153 \end{aligned}

::F=MAF=(51)(3)F=153

The force needed is 153 N (newtons), or a strong push.
::需要的兵力是153 N(牛顿),或大力推动。

Example 6
::例6

Taylor’s income varies directly with the number of hours he works. If he worked 60 hours last week and made $900, how much does he make per hour? Set up a direct variation equation.
::泰勒的收入与其工作时数直接不同。如果他上周工作了60小时,赚了900美元,他每小时能挣多少? 设置一个直接的变数方程式。

Solution: We want to find Taylor’s hourly wage, which is the constant of variation. $\begin{aligned} k = \frac{900}{60} = 15 \end{aligned}$ , he makes $15/hour. The equation would be $\begin{aligned} y = 15 x \end{aligned}$ .
::解决方案:我们想要找到泰勒的小时工资, 即变化不变的每小时工资。 k=90060=15, 他每小时挣15美元。等式是y=15x 。

Summary
::摘要

The direct variation equation is $\begin{aligned} y = k x^{n} \end{aligned}$ where $\begin{aligned} k \neq 0 \end{aligned}$ and $\begin{aligned} n > 0 \end{aligned}$ .
::直接变异方程式是 y=kxn, k0 和 n>0 。

Review
::回顾

For problems 1-4, use the given $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ values to write a direct variation equation and find $\begin{aligned} y \end{aligned}$ given that $\begin{aligned} x = 12 \end{aligned}$ .
::对于问题1-4, 使用给定的 x 和 Y 值来写入直接变异方程式, 并查找 y, 给定的 x 和 Y 值 x= 12 。

$\begin{aligned} x = 3, y = 15 \end{aligned}$
::x=3,y=15x3,y=15

$\begin{aligned} x = 9, y = - 3 \end{aligned}$
::x=9,y3

$\begin{aligned} x = \frac{1}{2}, y = \frac{1}{3} \end{aligned}$
::x=12,y=13 x=13

$\begin{aligned} x = - 8, y = \frac{4}{3} \end{aligned}$
::x8,y=43

For problems 5-8, use the given $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ values to write a direct variation equation and find $\begin{aligned} x \end{aligned}$ given that $\begin{aligned} y = 2 \end{aligned}$ .
::对于问题 5-8, 使用给定的 x 和 y 值来写入直接变异方程, 并找到 x, 并给定的 y= 2 。

$\begin{aligned} x = 5, y = 4 \end{aligned}$
::x=5,y=4x=5,y=4

$\begin{aligned} x = 18, y = 3 \end{aligned}$
::x=18,y=3xx=18,y=3

$\begin{aligned} x = 7, y = - 28 \end{aligned}$
::x=7,y28

$\begin{aligned} x = \frac{2}{3}, y = \frac{5}{6} \end{aligned}$
::x=23,y=56

Determine if the following data sets vary directly.
::确定以下数据集是否直接变化。

$\begin{aligned} x \end{aligned}$	12	16	5	20
$\begin{aligned} y \end{aligned}$	3	4	1	5

$\begin{aligned} x \end{aligned}$	2	10	5	6
$\begin{aligned} y \end{aligned}$	14	70	35	42

$\begin{aligned} x \end{aligned}$	2	8	18	34
$\begin{aligned} y \end{aligned}$	3	12	27	51

Explore More
::探索更多

1. Based on her weight and pace, Kate burns 586 calories when she runs 5 miles. How many calories will she burn if she runs only 3 miles? How many miles (to the nearest mile) does she need to run each week if she wants to burn one pound (3,500 calories) of body fat each week?
::1. 根据体重和速度,Kate在跑5英里时烧掉586卡路里,如果她只跑3英里,她要烧掉多少卡路里?如果她想每星期烧掉一磅(3 500卡路里)身体脂肪,她每周需要跑多少英里(距离最近的里程)?

2. On a road trip, Mark and Bill cover 450 miles in 8 hours, including stops. If they maintain the same pace, how far (to the nearest mile) will they be from their starting point after 15 hours of driving?
::2. 在一次公路旅行中,Mark和Bill在8小时内覆盖450英里,包括停留点,如果保持同样的速度,在15小时的驾驶之后,他们离起点(离最近的里程)还要多远?

3. Dorothy earned $900 last week for working 36 hours. What is her hourly wage? If she works full time (40 hours) in a week how much will she make?
::3. Dorothy上周工作36小时挣了900美元,小时工资是多少?如果她每周全时工作(40小时),她能挣多少钱?

4. According to Hooke’s Law, the distance that a spring stretches varies directly as the weight placed at the end of the spring. ² If a weight of 10 g stretches a certain spring 6 cm, how far will the spring stretch with a weight of 25 g?
::4. 根据胡克法则,弹簧伸展的距离随弹簧末的重量而异。 2 如果10克的重量延展某一弹簧6厘米,弹簧的长度将有多长,重量为25克?

5. The price of gasoline purchased varies directly with the number of gallons purchased. If 10 gallons of gasoline are purchased for $31.40, what will the price of 15 gallons be?
::5. 购买汽油的价格与购买加仑价格直接不同,如果购买10加仑汽油为31.40美元,15加仑的价格是多少?

6. Ohm's Law states that the voltage in a circuit varies directly as the current that passes through the circuit, and the resistance is the constant of variation. ³ If the voltage in a circuit is 120 volts when the current is 4 amperes, what is the voltage in the circuit when the current is 7 amperes?
::6. 《奥姆法》规定,电路电压随着电流通过电路而直接变化,阻力是变化的常数。 3 如果电路电压是120伏,当电流是4安培尔时,电流是7安培尔时,电流的电压是多少?

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

PLIX
::PLIX

Try this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

References
::参考参考资料

1."Newton's Laws," last accessed May 22, 2017,
::1"纽顿法律" 2017年5月22日最后一次访问

2. ."Hooke's Law," last edited April 26, 2017,
::2017年4月26日上次编辑的《Hooke's Law》

3. "Ohm's Law," last edited May 17, 2017, https://en.wikipedia.org/wiki/Ohm%27s_law.
::2017年5月17日编辑, https://en.wikipedia. org/wiki/Ohm%27s_law。

章节大纲

常规

直接变化

Direct Variation ::直接变化

Direct Variation ::直接变化

Example 1 ::例1

Example 2 ::例2

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

PLIX ::PLIX

References ::参考参考资料

Direct Variation
::直接变化

Direct Variation
::直接变化

Example 1
::例1

Example 2
::例2

Example 4
::例4

Example 5
::例5

Example 6
::例6

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

PLIX
::PLIX

References
::参考参考资料