课程： 9.1 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/sec/sec. If an object with an acceleration of 8 meters/sec/sec has a force of 24 Newtons, what is the object's force when its acceleration is 12 meters/sec/sec?
::根据牛顿的第二项法律,一个物体的净功率(F)等于其质量(m)乘以加速度(a),在牛顿测量F,米以公斤计量,米/秒/秒/秒测量。如果加速度为8米/秒/秒/秒的物体有24牛顿的功率,当加速度为12米/秒/秒时,物体的功力是什么?

Direct Variation
::直接变化

We say that a set of data is related directly if the independent and dependent variables both grow large or small together. For example, the equation of the line $\begin{align*}y=2x\end{align*}$ would represent a direct variation relationship. As $\begin{align*}x\end{align*}$ gets bigger, so would $\begin{align*}y\end{align*}$ . In fact, direct variation equation is $\begin{align*}y=kx, k \neq 0,\end{align*}$ which looks just like the equation of a line without a $\begin{align*}y\end{align*}$ - intercept . We call $\begin{align*}k\end{align*}$ the constant of variation and $\begin{align*}y\end{align*}$ is said to vary directly with $\begin{align*}x\end{align*}$ . $\begin{align*}k\end{align*}$ can also be written $\begin{align*}k=\frac{y}{x}\end{align*}$ .
::我们说,如果独立变量和依附变量同时变大或变小,一组数据是直接相关的。例如,y=2x的等式将代表一种直接的变异关系。当x变大时,y=2x的等式也会是直接的。事实上,直接变异等式是y=kx,k0,它看起来就像没有y-intercept的线的等式。我们称之为 k 变量的常数, y 据说与 x. k 可以写为 k=yx 。

The variables $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ vary directly, and $\begin{align*}y = 10\end{align*}$ when $\begin{align*}x = 2\end{align*}$ . Let's write an equation that relates $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ and find $\begin{align*}y\end{align*}$ when $\begin{align*}x = 9\end{align*}$ .
::变量 x 和 y 直接变化, y= 10 当 x= 2 时, y= 10 。让我们写一个与 x 和 y 相关的方程式, 然后在 x= 9 时找到 y 。

Using the direct variation equation, we can substitute in $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ and solve for $\begin{align*}k\end{align*}$ .
::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}

$\begin{align*}y &= kx \\ 10 &= k(2) \\ 5 &= k\end{align*}$
::y=kx10=k(2)5=k

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

Now, let's determine if the following set of data varies directly. If so, let's find the direct variation equation.
::现在,让我们来决定下一组数据是否直接变化。如果是的话,让我们找到直接变化方程式。

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

Looking at the set of data, the $\begin{align*}x\end{align*}$ values increase. For the data to vary directly, the $\begin{align*}y\end{align*}$ values would also have to increase, and they do. To find the equation, use the first point and find $\begin{align*}k\end{align*}$ .
::查看数据集时, x 值会增加。要直接改变数据,y 值也会增加,而y 值也会增加。要找到方程,请使用第一点并找到 k。

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

$\begin{align*}y &= kx\\ 1 &= k(4)\\ \frac{1}{4} &= k\end{align*}$
::y=kx1=k(4)14=k

So, the equation for the first point is $\begin{align*}y=\frac{1}{4}x\end{align*}$ . Plug each point into the equation to make sure it works.
::所以,第一个点的方程是y=14x。将每个点插到方程中, 以确保它有效。

$\begin{align*}2= \frac{1}{4}(8) \qquad 4= \frac{1}{4}(16) \qquad 5= \frac{1}{4}(20)\end{align*}$

The number of calories, $\begin{align*}C\end{align*}$ , a person burns working out varies directly with length of time it was done, $\begin{align*}t\end{align*}$ (in minutes). A 150 pound person can burn 207 calories swimming laps for 30 minutes. Let's write a variation model for $\begin{align*}C\end{align*}$ as a function of $\begin{align*}t\end{align*}$ . Then, let's determine how long it will take that person to burn 520 calories.
::热量、 C、一个人的烧伤数量随时间长短而异, t (以分钟计) 150磅的人可以烧207卡路里游泳圈30分钟。让我们写C的变异模型作为 t的函数。然后,让我们来决定这个人烧掉520卡路里需要多长时间。

Plug in what you know to the direct variation model and solve for $\begin{align*}k\end{align*}$ .
::插入您所知道的直接变异模型和 K 的解答。

\begin{aligned} C & = k t \\ 207 & = k (30) & The model for a 150 -pound person is C = 6.9 t . \\ 6.9 & = k \end{aligned}

$\begin{align*}C&=kt \\ 207&=k(30) && \text{The model for a} \ 150\text{-pound person is} \ C=6.9 t. \\ 6.9&=k\end{align*}$
::C=kt207=k(30) 150磅人的模型为C=6.9t.6.9=k

To find how long it will take to this person to burn 520 calories, solve for $\begin{align*}t\end{align*}$ .
::找到这个人烧掉520卡路里要花多长时间解决 t

\begin{aligned} 520 & = 6.9 t & It will take 75.4 minutes to burn 520 calories . \\ 75.4 & = t \end{aligned}

$\begin{align*}520&=6.9 t && \text{It will take} \ 75.4 \ \text{minutes to burn} \ 520 \ \text{calories}.\\ 75.4&=t\end{align*}$
::520=6.9t 燃烧520卡路里需要75.4分钟。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the object's force when its acceleration is 12 meters/sec/sec.
::早些时候,当物体加速度为12米/秒/秒时,你被要求寻找物体的力量。

If we write Newton's second law as an equation, we get $\begin{align*}F = ma\end{align*}$ . We can now see that this is an example of a direct variation equation, where $\begin{align*}y = F\end{align*}$ , $\begin{align*}m = k\end{align*}$ , and $\begin{align*}a = x\end{align*}$ . Using the direct variation equation, we can substitute in $\begin{align*}F\end{align*}$ and $\begin{align*}a\end{align*}$ and solve for $\begin{align*}m\end{align*}$ .
::如果我们把牛顿的第二定律写成一个方程, 我们就会得到F=ma。我们现在可以看到这是一个直接变异方程的例子, 即 y=F, m=k, 和 a=x。使用直接变异方程, 我们可以用 F 和 a 来替代 m 。

\begin{aligned} F & = m a \\ 24 & = m (8) \\ 3 & = m \end{aligned}

$\begin{align*}F &= ma \\ 24 &= m(8) \\ 3 &= m\end{align*}$
::F=ma24=m( 8)3=m

So the mass of the object is 3 kg but we're looking for its force when its acceleration is 12 meters/sec/sec. Hence, we use the formula again.
::所以天体的质量是3公斤但是当天体加速度是12米/秒/秒时我们在找它的力量

\begin{aligned} F & = m a \\ F & = (3) (12) \\ F & = 36 \end{aligned}

$\begin{align*}F &= ma \\ F &= (3)(12) \\ F &= 36\end{align*}$
::F=maF=(3)(12)F=36

Therefore, the object's force is 36 Newtons.
::因此,物体的力量是36牛顿。

Example 2
::例2

The variables $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ vary directly, and $\begin{align*}y=-6\end{align*}$ when $\begin{align*}x=-8\end{align*}$ . Find the equation and determine $\begin{align*}x\end{align*}$ when $\begin{align*}y=12\end{align*}$ .
::变量 x 和 y 直接变化, 当 x\\ 8 时为 y 6 。查找方程并确定 x y= 12 时为 y= 。

First, solve for $\begin{align*}k\end{align*}$ .
::首先,解决k。

$\begin{align*}k=\frac{y}{x}=\frac{-6}{-8}=\frac{3}{4} \rightarrow y=\frac{3}{4}x\end{align*}$
::kyx=6- 8=34_y=34x

Now, substitute in 12 for $\begin{align*}y\end{align*}$ and solve for $\begin{align*}x\end{align*}$ .
::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}

$\begin{align*}12&=\frac{3}{4}x \\ \frac{4}{3} \cdot 12&=x \\ 16&=x \\\end{align*}$
::12=34x43=12=x16=x

Example 3
::例3

Determine if the set below varies directly.
::确定下一组是否直接变化。

$\begin{align}x\end{align}$	1	2	3	4	5
$\begin{align}y\end{align}$	2	4	8	16	20

At first glance, it looks like both values increase together. Let’s check to see if $\begin{align*}k\end{align*}$ is the same for each set of points.
::乍一看,它看起来两种数值都同时增长。让我们来看看 k 是否对每组点都相同。

$\begin{align*}k=\frac{y}{x}=\frac{2}{1}=\frac{4}{2} \ne \frac{8}{3}\end{align*}$
::kyx=21=4283

At this point, we can stop because the point $\begin{align*}(3, 8)\end{align*}$ does not have the same ratio as the first two points. Therefore, this set of data does not vary directly.
::在这一点上,我们可以停止,因为点(3,8)与前两点的比率不同,因此,这组数据没有直接变化。

Example 4
::例4

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。

We want to find Taylor’s hourly wage, which is the constant of variation.
::我们想找到泰勒的小时工资,

$\begin{align*}k=\frac{900}{60}=15\end{align*}$ , he makes $15/hour. The equation would be $\begin{align*}y=15x\end{align*}$ .
::k=90060=15, 他每小时赚15美元。公式是y=15x。

Review
::回顾

For problems 1-4, use the given $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ values to write a direct variation equation and find $\begin{align*}y\end{align*}$ given that $\begin{align*}x =12\end{align*}$ .
::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 。

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

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

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

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

For problems 5-8, use the given $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ values to write a direct variation equation and find $\begin{align*}x\end{align*}$ given that $\begin{align*}y=2\end{align*}$ .
::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 。

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

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

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

$\begin{align*}x=\frac{2}{3},y=\frac{5}{6}\end{align*}$
::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。

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

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

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

Solve the following word problems using a direct variation equation.
::使用直接变异方程式解决以下字词问题。

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 (3500 calories) of body fat each week?
::根据体重和速度,凯特在跑5英里时烧了586卡路里。如果她只跑3英里,她会烧多少卡路里?如果她想每星期烧一磅(3500卡路里)体重,她每周需要跑多少英里(距离最近的里程)?

One 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?
::Mark和Bill在8小时中覆盖450英里的路程,包括中途站。如果他们保持同样的速度,在15小时的驾驶后,他们离起点(离最近的里程)还要多远?

About three hours into a fundraising car wash, the Mathletes Club earned $240 washing 48 cars. How much was charged for each carwash? How many more cars will they have to wash to reach their goal of earning $400?
::在大约三个小时的筹款洗车中,数学俱乐部赚了240美元洗48辆车。每洗1辆洗车要收多少钱?还要洗多少车才能达到挣400美元的目标?

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?
::Dorothy上周工作36小时挣了900美元,小时工资是多少?如果她一周内全时工作(40小时),她能挣多少钱?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

直接变化

Direct Variation ::直接变化

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)