课程： 14.8 过早变化率

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过早变化率

When you first learned about slope you learned the mnemonic device “rise over run” to help you remember that to calculate the slope between two points you use the following formula:
::当您第一次了解斜坡时, 学会了“ 向上上升” 的内温装置, 以便帮助您记住, 要计算两个点之间的斜坡, 您使用以下公式 :

\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}

::m=y2 - y1x2 - x1

In Calculus, you will learn that for curved functions, it makes more sense to discuss the slope at one precise point rather than between two points. The slope at one point is called the slope of the tangent line and the slope between two separate points is called a secant line.
::在微积分中,你会知道,对于曲线函数来说,在一个精确点而不是两个点之间讨论斜坡比较合理。一个点的斜坡称为正切线的斜坡,两个分开点之间的斜坡称为偏移线。

Consider a car driving down the highway and think about its speed. You are probably thinking about speed in terms of going a given distance in a given amount of time. The units could be miles per hour or feet per second, but the units always have time in the denominator. What happens when you consider the instantaneous speed of the car at one instant of time? Wouldn’t the denominator be zero?
::想想开着一辆汽车从高速公路上行驶,想想它的速度。你也许在考虑在一定的时间内行驶一定距离的速度。单位可以是每小时英里或每秒英尺,但单位总是有时间的分母。当你一瞬间考虑汽车的瞬时速度时会发生什么? 分母不是零吗?

Instantaneous Rate of Change
::过早变化率

The slope at a point $\begin{aligned} P \end{aligned}$ represents the instantaneous rate of change at that point. The slope at a point $\begin{aligned} P \end{aligned}$ (otherwise known as the slop of the tangent line) can be approximated by the slope of secant lines as the “run” of each secant line approaches zero. A secant line is a line that passes through two distinct points on a function. A tangent line to a function at a given point is the straight line that just touches the curve at that point.
::P点的斜坡代表该点的瞬时变化速率。P点的斜坡(又称正切线的斜坡)可以被伸缩线的斜坡所近似,即每个伸缩线的“运行”接近零。Asecant 线条是一个函数通过两个不同点的线条。在给定点的函数的正切线是刚刚触动曲线的直线。

Because you are interested in the slope as the “run” approaches zero, this is a limit question. One of the main reasons that you study limits in calculus is so that you can determine the slope of a curve at a point or the slope of the tangent line.
::因为您对斜坡的“ 运行” 接近零表示感兴趣,所以这是一个限制问题。您研究微积分限制的主要原因之一是,您可以在正切线的某个点或斜坡上确定曲线的斜坡。

For a graph of lines, it is easy to estimate the slopes of the tangent lines since the slope of the tangent is the same as the slope of the line. Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a table.
::对于线条图,很容易估计正切线的斜坡,因为正切线的斜坡与线的斜坡相同。以下函数的斜坡估计值为 -3, -2,2-1,0,1,1,2,3。

By mentally drawing a tangent line at the following $\begin{aligned} x \end{aligned}$ values you can estimate the following slopes.
::通过在精神上绘制以下x值的正切线,您可以估计以下的斜坡。

$\begin{aligned} x \end{aligned}$	*slope*
-3	0
-2	0
-1	-1
0	-1
1	2
2	0
3	0

If you graph these points you will produce a graph of what’s known as the derivative of the original function. A derivative is a function of the slopes of the original function.
::如果您用图形绘制这些点, 您将会生成一个图形, 显示什么是原始函数的衍生物。衍生物是原始函数的斜坡函数。

Examples
::实例

Example 1
::例1

Earlier, you were asked about the instantaneous speed of a car. If you write the ratio of distance to time and use limit notation to allow time to go to zero you do seem to get a zero in the denominator.
::早些时候, 有人询问您汽车的瞬时速度。如果您写入距离与时间的比例, 并使用限值标记, 允许时间到零, 您的分母中似乎为零。

$\begin{aligned} lim_{t i m e \to 0} (\frac{d i s t a n c e}{t i m e}) \end{aligned}$
::平时0( 远距离时间)

The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment in time. You are no longer restricted to finding slope from two separate points.
::限制的伟大之处在于您已经学会了在分母变为零时寻找极限的技术。汽车的即时速度基本上意味着在那个精确时刻速计读取的数字。您不再局限于从两个不同的点上找到斜坡。

Example 2
::例2

Estimate the slope of the function $\begin{aligned} f (x) = \sqrt{x} \end{aligned}$ at the point $\begin{aligned} (1, 1) \end{aligned}$ by calculating 4 successively close secant lines. If you had to guess what the slope was at the point $\begin{aligned} (1, 1) \end{aligned}$ what would you guess the slope to be?
::在点(1,1)上计算函数f(x)=x的斜度,方法是连续计算4条接连的伸缩线。如果你必须猜出点(1,1)的斜度是什么,你会猜出斜度是什么?

Calculate the slope between $\begin{aligned} (1, 1) \end{aligned}$ and 4 other points on the curve:
::计算曲线上(1,1)和4个其他点之间的斜度:

The slope of the line between $\begin{aligned} (5, \sqrt{5}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{1} = \frac{\sqrt{5} - 1}{5 - 1} \approx 0.309 \end{aligned}$
::在(5,5)和(1,1)之间线的斜坡为: m1=5-5-15-1-10.309

The slope of the line between $\begin{aligned} (4, 2) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{2} = \frac{2 - 1}{4 - 1} \approx 0.333 \end{aligned}$
::在(4,2)和(1,1)之间线的斜坡是: m2=2 - 14 - 1-10.333

The slope of the line between $\begin{aligned} (3, \sqrt{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{3} = \frac{\sqrt{3} - 1}{3 - 1} \approx 0.366 \end{aligned}$
::在(3,3)和(1,1)之间线的斜坡是:m3=3-13-1-1-0.366

The slope of the line between $\begin{aligned} (2, \sqrt{2}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{4} = \frac{\sqrt{2} - 1}{2 - 1} \approx 0.414 \end{aligned}$
::在(2,2)和(1,1)之间线的斜坡是:m4=2-12-10.414

Notice that the pattern in the previous example is leading up to $\begin{aligned} \frac{\sqrt{1} - 1}{1 - 1} \end{aligned}$ . Unfortunately, this cannot be computed directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.
::请注意, 上个示例中的模式正在导致 1 - 11 - 1 。不幸的是, 这无法直接计算, 因为分母为零。幸运的是, 您知道如何使用限制来评估。

To determine what the slope would be at the point $\begin{aligned} (1, 1) \end{aligned}$ , you would evaluate the limit $\begin{aligned} lim_{x \to 1} (\frac{\sqrt{x} - 1}{x - 1}) \end{aligned}$
::为确定点(1,1,1)的坡度,您将评估极限 limx%1(x-1x-1)

\begin{aligned} m & = lim_{x \to 1} (\frac{(\sqrt{x} - 1)}{(x - 1)} \cdot \frac{(\sqrt{x} + 1)}{(\sqrt{x} + 1)}) \\ = lim_{x \to 1} (\frac{(x - 1)}{(x - 1) (\sqrt{x} + 1)}) \\ = lim_{x \to 1} (\frac{1}{(\sqrt{x} + 1)}) \\ = \frac{1}{\sqrt{1} + 1} \\ = \frac{1}{2} \\ = 0.5 \end{aligned}

::m=limx%1 ((x-1)(x-1)(x-1)(x-1)(x+1)(x+1)))=limx%1 ((x-1)(x-1)(x+1)(x+1)))=limx%1(1(1(x+1)(1)(1)(11+1=12=0.5)

The slope of the function $\begin{aligned} f (x) = \sqrt{x} \end{aligned}$ at the point $\begin{aligned} (1, 1) \end{aligned}$ is exactly $\begin{aligned} m = \frac{1}{2} \end{aligned}$ .
::点(1,1)的函数 f(x)=x的斜度为 m=12。

Example 3
::例3

Sketch a complete cycle of a sine graph. Estimate the slopes at $\begin{aligned} 0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π \end{aligned}$ .
::绘制一个完整的正弦图周期。估计斜坡为 0, 2, 3, 2, 2。

$\begin{aligned} x \end{aligned}$	*Slope*
0	1
$\begin{aligned} \frac{π}{2} \end{aligned}$	0
$\begin{aligned} π \end{aligned}$	-1
$\begin{aligned} \frac{3 π}{2} \end{aligned}$	0
$\begin{aligned} 2 π \end{aligned}$	1

You should notice that these are the exact values of cosine evaluated at those points.
::您应该注意到,这些是在这些点上评估的 Cosine 的确切值。

Example 4
::例4

Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
::罗根乘自行车在20海里处旅行3小时,然后她乘车驾驶60英里,2小时。

Distance vs. Time:
::距离与时间 :

Rate vs. Time: (this is the graph of the derivative of the original function shown above)
::率与时间这是上面所示原始函数的衍生物图)

Example 5
::例5

Approximate the slope of $\begin{aligned} y = x^{3} \end{aligned}$ at $\begin{aligned} (1, 1) \end{aligned}$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
::使用左侧的分隔线,接近 y=x3at 的斜坡(1,1) 。实际的斜坡会大于还是小于估计值?

The slope of the line between $\begin{aligned} (0.7, {0.7}^{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{1} = \frac{{0.7}^{3} - 1}{0.7 - 1} \approx 2.19 \end{aligned}$
::线(0.7,0.73)和(1,1)之间的斜坡是:m1=0.73-10.7-12.19。

The slope of the line between $\begin{aligned} (0.8, {0.8}^{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{2} = \frac{{0.8}^{3} - 1}{0.8 - 1} \approx 2.44 \end{aligned}$
::在(0.8,0.83)和(1,1)之间线的斜坡是:m2=0.83-10.8-11-2.44。

The slope of the line between $\begin{aligned} (0.9, {0.9}^{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{3} = \frac{{0.9}^{3} - 1}{0.9 - 1} \approx 2.71 \end{aligned}$
::线(0.9.0.93)和(1.1)之间的斜坡为:m3=0.93-10.9-12.71。

The slope of the line between $\begin{aligned} (0.95, {0.95}^{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{1} = \frac{{0.95}^{3} - 1}{0.95 - 1} \approx 2.8525 \end{aligned}$
::在(0.95,0.9553)和(1,1)之间线的斜坡为:m1=0.953-10.95-11-2.8525

The slope of the line between $\begin{aligned} (0.975, {0.975}^{3}) \end{aligned}$ and $\begin{aligned} (1, 1) \end{aligned}$ is: $\begin{aligned} m_{1} = \frac{{0.975}^{3} - 1}{0.975 - 1} \approx 2.925625 \end{aligned}$
::线线(0.975,0.9753)和(1,1)之间的斜坡为:m1=0.9753-10.975-12.925625

The slope at $\begin{aligned} (1, 1) \end{aligned}$ will be slightly greater than the estimates because of the way the slope curves. The slope at $\begin{aligned} (1, 1) \end{aligned}$ appears to be about 3.
::由于坡度曲线的方式,(1,1)的坡度将略高于估计数。(1,1)的坡度似乎约为3。

Summary

The slope at a single point on a curve is called the slope of the tangent line, while the slope between two points is called a secant line.
::曲线上一个点的斜坡称为正切线的斜坡,两点之间的斜坡称为伸缩线。

Instantaneous rate of change is represented by the slope at a point, which can be approximated by the slope of secant lines as the "run" approaches zero.
::在一个点的斜坡代表着不时的变化速度,偏斜线的斜坡可以与“运行”接近零的斜坡相近。

Estimating the slopes of tangent lines on a graph can help determine the derivative of the original function.
::估计图中相切线的斜坡有助于确定原始函数的衍生物。

Review
::回顾

1. Approximate the slope of $\begin{aligned} y = x^{2} \end{aligned}$ at $\begin{aligned} (1, 1) \end{aligned}$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
::1. 使用左侧的分隔线,接近y=x2(1,1)的斜坡。实际斜坡是大于还是小于估计数?

2. Evaluate the following limit and explain how it confirms your answer to #1.
::2. 评估以下限度,并解释如何证实你对#1的答复。

$\begin{aligned} lim_{x \to 1} (\frac{x^{2} - 1}{x - 1}) \end{aligned}$
::limx%1 (x2- 1x- 1)

3. Approximate the slope of $\begin{aligned} y = 3 x^{2} + 1 \end{aligned}$ at $\begin{aligned} (1, 4) \end{aligned}$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
::3. 使用左侧偏移线(1,4)接近y=3x2+1的斜坡(1,4),实际斜坡是大于还是小于估计数?

4. Evaluate the following limit and explain how it confirms your answer to #3.
::4. 评价以下限制,并解释如何证实你对3号的回答。

$\begin{aligned} lim_{x \to 1} (\frac{3 x^{2} + 1 - 4}{x - 1}) \end{aligned}$
::limx%1( 3x2+1- 4x- 1)

5. Approximate the slope of $\begin{aligned} y = x^{3} - 2 \end{aligned}$ at $\begin{aligned} (1, - 1) \end{aligned}$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
::5. 使用左侧的分离线,接近y=x3-2(1,-1)的斜坡。实际斜坡是大于还是小于估计?

6. Evaluate the following limit and explain how it confirms your answer to #5.
::6. 评估以下限制,并解释如何证实你对5号的回答。

$\begin{aligned} lim_{x \to 1} (\frac{x^{3} - 2 - (- 1)}{x - 1}) \end{aligned}$
::立方公尺x1(x3--2)-(--1)x-1

7. Approximate the slope of $\begin{aligned} y = 2 x^{3} - 1 \end{aligned}$ at $\begin{aligned} (1, 1) \end{aligned}$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
::7. 在(1,1)时使用左侧的分隔线,接近y=2x3-1的斜坡。实际斜坡是大于还是小于估计?

8. What limit could you evaluate to confirm your answer to #7?
::8. 为了确认你对7号的回答,你能评估什么限度?

9. Sketch a complete cycle of a cosine graph. Estimate the slopes at $\begin{aligned} 0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π \end{aligned}$ .
::9. 绘制一个余弦图的完整周期,估计斜坡为0,2,3,2,2。

10. How do the slopes found in the previous question relate to the sine function? What function do you think is the derivative of the cosine function?
::10. 上一个问题中发现的斜坡与正弦函数有何关系?你认为余弦函数的衍生函数是什么?

11. Sketch the line $\begin{aligned} y = 2 x + 1 \end{aligned}$ . What is the slope at each point on this line? What is the derivative of this function?
::11. 将 y=2x+1 线平铺成一条线。这条线上每个点的斜坡是什么? 此函数的衍生物是什么?

12. Logan travels by bike at 30 mph for 2 hours. Then she gets in a car and drives 65 mph for 3 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
::12. 洛根乘自行车在30海里处旅行2小时,然后她乘车驾驶65英里处3小时。

13. Explain what a tangent line is and how it relates to derivatives.
::13. 解释何为不相干线及其与衍生物的关系。

14. Why is finding the slope of a tangent line for a point on a function the same as the instantaneous rate of change at that point?
::14. 为什么在某一函数上找到与该点瞬时变化率相同的点的正切线斜坡?

15. What do limits have to do with finding the slopes of tangent lines?
::15. 界限与寻找相左线的斜坡有什么关系?

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

过早变化率

Instantaneous Rate of Change ::过早变化率

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)