Course: 9.3 联合变动 | The Way To Learn

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 联合变动

Collapse Expand
联合变动
The volume of a cylinder varies jointly with the square of the radius and the height. If the volume of the cylinder is $\begin{aligned} 64 {units}^{3} \end{aligned}$ and radius is $\begin{aligned} 4 units \end{aligned}$ , what is the height of the cylinder?
::圆柱体的体积与半径的正方形和高度不同。如果圆柱体的体积为64个单位3,半径为4个单位,圆柱体的高度是多少?

Joint Variation
::联合变动

The last type of variation is called joint variation . This type of variation involves three variables, usually $\begin{aligned} x, y \end{aligned}$ and $\begin{aligned} z \end{aligned}$ . For example, in geometry, the volume of a cylinder varies jointly with the square of the radius and the height. In this equation the constant of variation is $\begin{aligned} π \end{aligned}$ , so we have $\begin{aligned} V = π r^{2} h \end{aligned}$ . In general, the joint variation equation is $\begin{aligned} z = k x y \end{aligned}$ . Solving for $\begin{aligned} k \end{aligned}$ , we also have $\begin{aligned} k = \frac{z}{x y} \end{aligned}$ .
::最后一种变异类型称为联合变异。这种变异类型包含三个变量, 通常是 x,y 和 z。例如, 在几何学中, 圆柱体的体积与半径的正方形和高度是相异的。在这个方程中, 变量的常数是 , 所以我们有 Vr2h 。一般来说, 联合变异方程是 z=kxy 。 k 溶解时, 我们也有 k=zxy 。

Let's write an equation for the following relationships.
::让我们为以下关系写一个方程式

$\begin{aligned} y \end{aligned}$ varies inversely with the square of $\begin{aligned} x \end{aligned}$ .
::y 与 x 平方反向变化。

$\begin{aligned} y = \frac{k}{x^{2}} \end{aligned}$
::y=kx2 y=kx2

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and the square root of $\begin{aligned} y \end{aligned}$ .
::z 与 x 和 y 的平方根并存。

$\begin{aligned} z = k x \sqrt{y} \end{aligned}$
::z=kxy z=kxy

$\begin{aligned} z \end{aligned}$ varies directly with $\begin{aligned} x \end{aligned}$ and inversely with $\begin{aligned} y \end{aligned}$ .
::z 与 x 直接不同,y 反差。

$\begin{aligned} z = \frac{k x}{y} \end{aligned}$
::z=kxy z=kxy

Now, let's solve the following problems.
::现在,让我们解决以下的问题。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 3, y = 8, \end{aligned}$ and $\begin{aligned} z = 6 \end{aligned}$ , find the variation equation. Then, find $\begin{aligned} z \end{aligned}$ when $\begin{aligned} x = - 2 \end{aligned}$ and $\begin{aligned} y = 10 \end{aligned}$ .
::如果 x=3,y=8,z=6,则查找变异方程。然后,在 x=2 和 y=10 时查找z。

Using the equation when it is solved for $\begin{aligned} k \end{aligned}$ , we have:
::当 K 解决时使用方程式, 我们有:

$\begin{aligned} k = \frac{z}{x y} = \frac{6}{3 \cdot 8} = \frac{1}{4} \end{aligned}$ , so the equation is $\begin{aligned} z = \frac{1}{4} x y \end{aligned}$ .
::k=zxy=63_8=14, 方程式为z=14xy。

When $\begin{aligned} x = - 2 \end{aligned}$ and $\begin{aligned} y = 10 \end{aligned}$ , then $\begin{aligned} z = \frac{1}{4} \cdot - 2 \cdot 10 = - 5 \end{aligned}$ .
::当 x2 和 y= 10, 然后z= 14 210 5 时。

Geometry Connection The volume of a pyramid varies jointly with the area of the base and the height with a constant of variation of $\begin{aligned} \frac{1}{3} \end{aligned}$ . If the volume is $\begin{aligned} 162 u n i t s^{3} \end{aligned}$ and the area of the base is $\begin{aligned} 81 u n i t s^{2} \end{aligned}$ , find the height.
::金字塔的体积与基点面积和高度不尽相同,高度变化不变,为13。如果体积为162个单位3,基点面积为81个单位2,请找到高度。

Find the joint variation equation first.
::先找到联合变异方程

$\begin{aligned} V = \frac{1}{3} B h \end{aligned}$
::V=13 小时

Now, substitute in what you know to solve for the height.
::现在代替你所知道的解决高度问题的方法

$\begin{aligned} 162 & = \frac{1}{3} \cdot 81 \cdot h \\ 162 & = 27 h \\ 6 & = h \end{aligned}$

::162=1381h162=27 h6=h

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the height of the cylinder.
::早些时候,你被要求找到圆柱的高度。

The formula for the volume of a cylinder, $\begin{aligned} V = π r^{2} h \end{aligned}$ , is a joint variation equation in which the constant $\begin{aligned} k = π \end{aligned}$ .
::圆柱体体积的公式Vr2h是常数 k的组合变方程。

We can therefore plug in the given values and solve for h , the height.
::因此,我们可以插入给定值并解决 h, 高度。

%5C%5C%0A64%5Cpi%20%3D%2016%5Cpi%5C%5C%0A4%20%3D%20h">
$\begin{aligned} V = π r^{2} h \\ 64 π = π 4^{2} (h) \\ 64 π = 16 π (h) \\ 4 = h \end{aligned}$

::Vr2h644264164=h

Therefore, the cylinder has a height of 4 units.
::因此,气瓶的高度为4个单位。

Example 2
::例2

Write the equation for $\begin{aligned} z \end{aligned}$ , that varies jointly with $\begin{aligned} x \end{aligned}$ and the cube of $\begin{aligned} y \end{aligned}$ and inversely with the square root of $\begin{aligned} w \end{aligned}$ .
::写入 z 的方程,该方程与 x 和 y 的立方体相异,与 w 的平方根反差。

$\begin{aligned} z = \frac{k x y^{3}}{\sqrt{w}} \end{aligned}$
::z z=kxy3w

Example 3
::例3

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} y \end{aligned}$ and $\begin{aligned} x \end{aligned}$ . If $\begin{aligned} x = 25, z = 10, \end{aligned}$ and $\begin{aligned} k = \frac{1}{5} \end{aligned}$ , find $\begin{aligned} y \end{aligned}$ .
::x=25,z=10,k=15,找到y。

The equation would be $\begin{aligned} z = \frac{1}{5} x y \end{aligned}$ . Solving for $\begin{aligned} y \end{aligned}$ , we have:
::等式为z=15xy。解决y,我们有:

$\begin{aligned} 10 & = \frac{1}{5} \cdot 25 \cdot y \\ 10 & = 5 y \\ 2 & = y \end{aligned}$

::10=1525y10=5y2=y

Example 4
::例4

Kinetic energy $\begin{aligned} P \end{aligned}$ (the energy something possesses due to being in motion) varies jointly with the mass $\begin{aligned} m \end{aligned}$ (in kilograms) of that object and the square of the velocity $\begin{aligned} v \end{aligned}$ (in meters per seconds). The constant of variation is $\begin{aligned} \frac{1}{2} \end{aligned}$ .
::动能P(由于运动而拥有的能量)与该物体的质量m(公斤)和速度的正方形(每秒以米计)不同。变化的常数为12。

Write the equation for kinetic energy.
::写动能方程式

$\begin{aligned} P = \frac{1}{2} m v^{2} \end{aligned}$
::P=12 mv2

If a car is travelling 104 km/hr and weighs 8800 kg, what is its kinetic energy?
::如果一辆汽车每小时行驶104公里,重量8800公斤,其动能是什么?

The second portion of this problem isn’t so easy because we have to convert the km/hr into meters per second.
::这个问题的第二部分并不容易, 因为我们必须将公里/小时转换成每秒的米数。

$\begin{aligned} \frac{104 k m}{h r} \cdot \frac{h r}{3600 s} \cdot \frac{1000 m}{k m} = 0.44 \frac{m}{s} \end{aligned}$
::104公里3600秒1000公里=0.44米

Now, plug this into the equation from part a.
::现在,把这个插进A部分的方程中

$\begin{aligned} P & = \frac{1}{2} \cdot 8800 k g \cdot {(0.44 \frac{m}{s})}^{2} \\ = 1955.56 \frac{k g \cdot m^{2}}{s^{2}} \end{aligned}$

::P=128800 kg__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Typically, the unit of measurement of kinetic energy is called a joule. A joule is $\begin{aligned} \frac{k g \cdot m^{2}}{s^{2}} \end{aligned}$ .
::通常,动能测量单位称为焦耳,焦耳为kgm2s2。

Review
::回顾

For questions 1-5, write an equation that represents relationship between the variables.
::对于问题1-5,写一个代表变量之间关系的方程式。

$\begin{aligned} w \end{aligned}$ varies inversely with respect to $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ .
::x 和 y 的逆差。

$\begin{aligned} r \end{aligned}$ varies inversely with the square of $\begin{aligned} q \end{aligned}$ .
::r 与q平方反差。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ and inversely with $\begin{aligned} w \end{aligned}$ .
::z 与 x 和 y 合并变化,与 w 反向变化。

$\begin{aligned} a \end{aligned}$ varies directly with $\begin{aligned} b \end{aligned}$ and inversely with $\begin{aligned} c \end{aligned}$ and the square root of $\begin{aligned} d \end{aligned}$ .
::a 与 b 直接不同,与 c 和 d 的平方根相反。

$\begin{aligned} l \end{aligned}$ varies directly with $\begin{aligned} m \end{aligned}$ , and inversely with $\begin{aligned} p \end{aligned}$ .
::我与M直接不同,与p相反。

Write the variation equation and answer the given question in each problem.
::写下变异方程并回答每个问题中的问题。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 2, y = 3 \end{aligned}$ and $\begin{aligned} z = 4 \end{aligned}$ , write the variation equation and find $\begin{aligned} z \end{aligned}$ when $\begin{aligned} x = - 6 \end{aligned}$ and $\begin{aligned} y = 2 \end{aligned}$ .
::如果 x=2,y=3 和 z=4, 则在 x=6 和 y=2 时, 刻写变异方程并找到 z。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 5, y = - 1 \end{aligned}$ and $\begin{aligned} z = 10 \end{aligned}$ , write the variation equation and find $\begin{aligned} z \end{aligned}$ when $\begin{aligned} x = - \frac{1}{2} \end{aligned}$ and $\begin{aligned} y = 7 \end{aligned}$ .
::z 与 x 和 y 相异。如果 x= 5, y 1 和 z= 10, 请写变量方程, 并在 x 12 和 y = 7 时找到 z 。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 7, y = 3 \end{aligned}$ and $\begin{aligned} z = - 14 \end{aligned}$ , write the variation equation and find $\begin{aligned} y \end{aligned}$ when $\begin{aligned} z = - 8 \end{aligned}$ and $\begin{aligned} x = 3 \end{aligned}$ .
::如果 x=7,y=3 和 z14, 请写变量方程, 并在 z8 和 x=3 时查找 y 。

$\begin{aligned} z \end{aligned}$ varies jointly with $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 8, y = - 3 \end{aligned}$ and $\begin{aligned} z = - 6 \end{aligned}$ , write the variation equation and find $\begin{aligned} x \end{aligned}$ when $\begin{aligned} z = 12 \end{aligned}$ and $\begin{aligned} y = - 16 \end{aligned}$ .
::z 与 x 和 y 相异。如果 x= 8,y 3 和 z 6, 则写入变异方程, 并在 z = 12 和 y 16 时找到 x 。

$\begin{aligned} z \end{aligned}$ varies inversely with $\begin{aligned} x \end{aligned}$ and directly $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = 4, y = 48 \end{aligned}$ and $\begin{aligned} z = - 2 \end{aligned}$ , write the variation equation and find $\begin{aligned} x \end{aligned}$ when $\begin{aligned} z = 8 \end{aligned}$ and $\begin{aligned} y = 96 \end{aligned}$ .
::如果 x=4,y=48 和 z2, 写变量方程式, 然后在 z=8 和 y=96 时找到 x, z=8 和 y=96 。

$\begin{aligned} z \end{aligned}$ varies inversely with $\begin{aligned} x \end{aligned}$ and directly $\begin{aligned} y \end{aligned}$ . If $\begin{aligned} x = \frac{1}{2}, y = 5 \end{aligned}$ and $\begin{aligned} z = 20 \end{aligned}$ , write the variation equation and find $\begin{aligned} x \end{aligned}$ when $\begin{aligned} z = - 4 \end{aligned}$ and $\begin{aligned} y = 8 \end{aligned}$ .
::如果 x=12,y=5 和 z=20, 请写变量方程, 然后在 z4 和 y=8 时找到 x 。

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

If 20 volunteers can wash 100 cars in 2.5 hours, find the constant of variation and find out how many cars 30 volunteers can wash in 3 hours.
::如果20名志愿者能在2.5小时内洗100辆汽车,

If 10 students from the environmental club can clean up trash on a 2 mile stretch of road in 1 hour, find the constant of variation and determine how low it will take to clean the same stretch of road if only 8 students show up to help.
::如果环境俱乐部的10名学生能在1小时内在2英里长的公路上清理垃圾,

The work $\begin{aligned} W \end{aligned}$ (in joules) done when lifting an object varies jointly with the mass $\begin{aligned} m \end{aligned}$ (in kilograms) of the object and the height $\begin{aligned} h \end{aligned}$ (in meters) that the object is lifted. The work done when a 100 kilogram object is lifted 1.5 meters is 1470 joules. Write an equation that relates $\begin{aligned} W, m, \end{aligned}$ and $\begin{aligned} h \end{aligned}$ . How much work is done when lifting a 150 kilogram object 2 meters?
::升起物体时完成的工作 W( 焦耳) 与物体质量m( 公斤) 和物体升起时的高度h( 米) 不同。当100公斤物体升起1.5米时完成的工作为 1470 焦耳。写一个与W, m 和 h有关的方程式。当升起150公斤物体2米时做了多少工作?

The intensity $\begin{aligned} I \end{aligned}$ of a sound (in watts per square meter) varies inversely with the square of the distance $\begin{aligned} d \end{aligned}$ (in meters) from the sound’s source. At a distance of 1.5 meters from the stage, the intensity of the sound at a rock concert is about 9 watts per square meter. Write an equation relating $\begin{aligned} I \end{aligned}$ and $\begin{aligned} d \end{aligned}$ . If you are sitting 10 meters back from the stage, what is the intensity of the sound you hear?
::声音的强度I(以每平方米瓦特计)与声音源的距离(以米计)的正方形反差。在距离舞台1.5米的距离,摇滚音乐会的声音强度约为每平方米9瓦特。写一个与I和d有关的方程式。如果你坐在距离舞台10米的后面,听到的声音的强度是多少?

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。

Section outline

General

联合变动

Joint Variation ::联合变动

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)

Joint Variation
::联合变动

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)