过早变化率

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You may recall the story about Jim and his track-star girlfriend Becca from the lesson on understanding limits. The two of them were discussing how they might calculate her speed at the instant that Jim snapped a photo of her. The end result of the discussion was Becca pointing out that it is not technically possible to calculate the exact speed of something in a specific instant.
::你也许记得吉姆和他的跟踪明星女友贝卡的故事 从理解极限的课上。他们两人正在讨论他们如何计算她的速度, 当吉姆拍下她的照片时。讨论的最终结果是贝卡指出,在技术上不可能在特定时刻计算某事的确切速度。

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By now, we have explored the related concepts of limits and lines tangent to a curve, so we know it is possible to effectively calculate instantaneous speed. What process would be involved with actually calculating Becca's speed at the instant the photo was taken? What are the technical difficulties with identifying instantaneous speed?
::现在,我们已经探索了相关的界限和线条与曲线相切的概念,因此我们知道可以有效计算瞬时速度。在拍摄照片时实际计算贝卡速度需要什么过程?在确定瞬时速度方面有什么技术困难?

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Instantaneous Rates of Change
::过早变化率

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The function $\begin{align*}f^{\prime} (x)\end{align*}$ that we defined in previous lessons is so important that it has its own name: the derivative.
::我们在先前的教训中定义的函数 f(x) 非常重要, 以至于它有自己的名称: 衍生物 。

The Derivative The function f ' is defined by the formula $\begin{align*}f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}\end{align*}$ where f ' is called the derivative of f with respect to x . The domain of f consists of all the values of x for which the limit exists.

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Based on the discussion that we have had in previous section, the derivative $\begin{align*}f^{\prime} \end{align*}$ represents the slope of the tangent line at point x . Another way of interpreting it would be that the function y = f ( x ) has a derivative $\begin{align*}f^{\prime} \end{align*}$ whose value at x is the instantaneous rate of change of y with respect to point x .
::根据我们在前一节中的讨论,衍生物f 表示x点正切线的斜度。另一种解释方法是,函数y = f(x)具有衍生物f ,其值在x点是y相对于x点的瞬时变化率。

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One of the two primary concepts of calculus involves calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of displacement with respect to time. If a person travels 120 miles in 4 hours, his speed is 120/4 = 30 mi/hr. This speed is called the average speed or the average rate of change of distance with respect to time. Of course the person who travels 120 miles at a rate of 30 mi/hr for 4 hr probably does not do so continuously. Though he probably slowed down or sped up during the 4-hour period, it does generally suffice to say that he traveled for 4 hours at an average rate of 30 miles per hour. However, if the driver strikes a tree, it would not be his average speed that determines his survival but his speed at the instant of the collision . Similarly, when a bullet strikes a target, it is not the average speed that is significant but its instantaneous speed at the moment it strikes. So here we have distinct kinds of speeds, average speed and instantaneous speed.
::微积分的两个基本概念之一涉及计算一个数量相对于另一个数量的变化速度。 例如, 速度被定义为时间迁移速度。 如果一个人在4小时内行走120英里, 速度为120/4 = 30 毫/ 小时。 这个速度被称为平均速度或距离平均变化速度。 当然, 以30 毫/ 小时的速度行走120英里, 4小时行驶的人可能不会持续。 尽管他可能在4小时内放慢速度或加速速度, 但一般而言, 可以说他以平均每小时30英里的平均速度行走4小时就足够了。 但是, 如果司机撞上一棵树, 其平均速度不是决定他的生存速度, 而是在碰撞时的速度。 同样, 当子弹击中目标时, 平均速度不是很大,而是在撞击时的瞬间速度。 因此我们这里有不同的速度、 平均速度和即时速度。

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The average speed of an object is defined as the object's displacement ∆ x divided by the time interval ∆ t during which the displacement occurs:
::对象的平均速度定义为对象的移位 x 除以移位发生的时间间隔 t :

			$\begin{align*}\text{Average speed }= v=\frac {\triangle x}{\triangle t}=\frac {x_1-x_0}{t_1-t_0}\end{align*}$

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Notice that the points ( t ₀ , x ₀ ) and ( t ₁ , x ₁ ) lie on the position versus time curve, as the figure below shows.
::请注意,如下图所示,点数(t0,x0)和点数(t1, x1)位于位置与时间曲线上。

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This expression is also the expression for the slope of a secant line connecting the two points. Thus we conclude that the average velocity of an object between time t ₀ and t ₁ is represented geometrically by the slope of the secant line connecting the two points ( t _o , x _o ) and ( t ₁ , x ₁ ). If we choose t ₁ close to t _o , then the average velocity will closely approximate the instantaneous velocity at time t _o .
::此表达式也是连接两个点的松动线斜坡的表达式。 因此, 我们得出结论, 时间 t0 和 t1 之间的对象的平均速度, 以连接两个点( 到, xo) 和 ( t1, x1) 的松动线的斜坡几何表示, 如果我们选择接近 t1, 那么平均速度将接近时速的瞬时速度 。

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Geometrically, the average rate of change is represented by the slope of a secant line (figure a , below) and the instantaneous rate of change is represented by the slope of the tangent line (figure b , below).
::从几何上看,平均变化率以偏移线的斜坡表示(下文图a),瞬时变化率以正切线的斜坡表示(下文图b)。

Average Rate of Change (such as the average velocity ) The average rate of change of y = f ( x ) over the time interval [ x 0 , x 1 ] is the slope m sec of the secant line to the points ( x o , f ( x 0 )) and ( x 1 , f ( x 0 )) on the graph (figure a): $\begin{align*}m_{sec}= \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*}$

Instantaneous Rate of Change The instantaneous rate of change of y = f ( x ) at the point x 0 is the slope m sec of the tangent line to the point x 0 on the graph (figure b): $\begin{align*}m_{tan} = f'(x_0) = \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{t_1-t_0}\end{align*}$

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked a question about the process of calculating Becca's speed in the race.
::之前有人问过你 计算贝卡赛跑速度的过程

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Speed is, by definition, a relation between a distance and the time required to cross that distance (recall $\begin{align*}d = rt\end{align*}$ from your science class). If the time required is zero , then you end up dividing by zero, which is an undefined function.
::根据定义,速度是指距离与穿越距离所需时间(从你的科学类中召回 d=rt)之间的关系。如果所需时间为零,则最终以零除以零,这是一个未定义的函数。

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However, using calculus, you can identify what the end behavior of the function would be if you were to get infinitely close, and thereby effectively calculate the speed at a given instant. In fact, that is exactly what you did in the examples above.
::然而,使用微积分, 您可以确定函数的结束行为是什么样的, 如果您能够完全接近, 从而有效地计算一个特定瞬间的速度。 事实上, 这正是您在上述示例中所做的 。

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Example 2
::例2

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Calculate: (a) The slope of the line tangent to $\begin{align*}y = x^2 +5\end{align*}$ at the point on the curve $\begin{align*}x = 4\end{align*}$  and (b) the equation of that line.
::计算 : (a) 曲线 x=4 和 (b) 曲线方程的正切值为 y=x2+5 的线型斜度。

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In the previous concept "Tangents to a Curve" we showed how to calculate the derivative (the slope of the tangent) of a function of the form $\begin{align*}y = x^2 -c\end{align*}$ .
::在先前的“曲线图案”概念中,我们演示了如何计算y=x2-c形式的函数的衍生物(正切值的斜度)。

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1. The slope of the tangent to the curve $\begin{align*}y = x^2 +5\end{align*}$ at $\begin{align*}x = 4\end{align*}$  is:
   ::x=4时的 y=x2+5 曲线的正切坡为:

$\begin{align*}2(4) = 8\end{align*}$

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Recall from algebra the "slope - intercept form" of a straight line: $\begin{align*}y = mx + b\end{align*}$ , where m is the slope of the line.
::从代数中记起y=mx+b这一直线的“斜体-截取形式”,m是线的斜坡。

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2. Given $\begin{align*}m = 8\end{align*}$ from above, we have $\begin{align*}y = 8x + b\end{align*}$ . All we need is to solve for b to have the full equation.
   ::根据上面的 m=8,我们有y=8x+b。我们需要的就是解决b 的全方程。

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Substitute $\begin{align*}4\end{align*}$ (the point we are finding the slope of) in for $\begin{align*}x\end{align*}$ in the equation of the curve $\begin{align*}y = x^2 +5\end{align*}$ to identify a corresponding $\begin{align*}y\end{align*}$
::y=x2+5 的方程中 x x 的替代值 4( 我们所找到的点) , 以识别相应的 y

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$\begin{align*}y = (4)^2 +5\end{align*}$
::y=(4)2+5

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$\begin{align*}y = 21\end{align*}$
::y=21岁

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Substitute the values for x , y , and m that we now have into our $\begin{align*}y = mx + b\end{align*}$ form of the equation of the tangent line:
::将 X、 y 和 m 的值替换为 Y = mx+bform 的正切线等式 :

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$\begin{align*}21 = 8(4) +b\end{align*}$
::21=8(4)+b

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$\begin{align*}-11 = b\end{align*}$
::- 11=b

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Use the calculated values for m and b to complete the equation:
::使用 m 和 b 的计算值来完成方程:

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$\begin{align*}y = 8x -11\end{align*}$
::y=8x- 11 y=8x- 11

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Example 3
::例3

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Suppose that $\begin{align*}y = x^2 -3\end{align*}$ .
::假设y=x2-3。

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(a) Find the average rate of change of y with respect x over the interval [0, 2] and (b) find the instantaneous rate of change of y with respect x at the point x = −1.
:a) 在X点=-1时,在[0,2]和(b)间隔期间查找y相对于x的平均变化率;(b)在x点=-1时查找y相对于x的瞬时变化率。

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1. Applying the formula above for secant with $\begin{align*}f(x) = x^2 -3\end{align*}$ and $\begin{align*}x_0 = 0\end{align*}$ and $\begin{align*}x_1 = 2\end{align*}$ , yields
   ::对 f( x) =x2-3 和 x0=0 和 x1=2 的分离值应用上面的公式, 产量

		$\begin{align*}m_{sec}\end{align*}$	$\begin{align*}= \frac {f(x_1)-f(x_0)}{t_1-t_0}\end{align*}$
			$\begin{align*}= \frac {f(2)-f(0)}{2-0}\end{align*}$
			$\begin{align*}= \frac {1-(-3)}{2}\end{align*}$
			$\begin{align*}= 2\end{align*}$

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This means that the average rate of change of y is 2 units per unit increase in x over the interval [0,2].
::这意味着每单位2个单位平均变化率在间隔[0]之间增加[2]。

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2. Recall, for functions of the form $\begin{align*}y = x^2 +c\end{align*}$ that $\begin{align*}f^{\prime} (x) = 2x\end{align*}$ ( $\begin{align*}f^{\prime} (x)\end{align*}$ is "f prime of x ," meaning "the slope (m) of the line tangent to x ")
   ::重命名 y=x2+chatf_(x)=2x(f*(x)is "fprimeofx" 的窗体函数, 意指“ 正切向量线的斜度(m) ”

	m tan	= $\begin{align*}f^{\prime} (x_{0})\end{align*}$
		= $\begin{align*}f^{\prime} (-1)\end{align*}$
		= $\begin{align*}2(-1)\end{align*}$
		= $\begin{align*}-2\end{align*}$

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This means that the instantaneous rate of change is negative. That is, y is decreasing at the point x = -1. It is decreasing at a rate of 2 units per unit increase in x .
::这意味着瞬时变化率为负。 也就是说, 在点值= - 1 时, yis 正在下降, 每单位增加2 个单位 。

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Example 4
::例4

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Find the derivative of $\begin{align*}f(x) = \sqrt{x}\end{align*}$ and the equation of the tangent line at x ₀ = 1.
::在 x0 = 1 时查找 f(x) x 的衍生物和正切线的方程式。

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Using the definition of the derivative,
::使用衍生物的定义,

	$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}} {h}\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}} {h} \left( \frac{\sqrt{x + h} + \sqrt{x}} {\sqrt{x + h} + \sqrt{x}} \right )\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {h} \frac{x + h - x} {\sqrt{x + h} + \sqrt{x}}\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {\sqrt{x + h} + \sqrt{x}}\end{align*}$
		$\begin{align*}= \frac{1} {2\sqrt{x}}\end{align*}$

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Thus, the slope of the tangent line at x ₀ = 1 is
::因此,x0=1的正切线斜坡为

		$\begin{align*}f'(1)\end{align*}$	$\begin{align*}= \frac {1}{2\sqrt{1}}=\frac {1}{2}\end{align*}$

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For x ₀ = 1, we can find y ₀ by simply substituting into f ( x ):
::对于 x0 = 1, 我们只需将 Y0 替换为 f( x) 即可找到 y0 :

		$\begin{align*}f(x_0)\end{align*}$	$\begin{align*}\equiv y_0\end{align*}$
		$\begin{align*}f(1)\end{align*}$	$\begin{align*}= \sqrt{1}=1\end{align*}$
		$\begin{align*}y_0\end{align*}$	$\begin{align*}= 1\end{align*}$

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Thus the equation of the tangent line is
::因此,正切线的方程式是

		$\begin{align*}y-y_0\end{align*}$	$\begin{align*}= m(x-x_0)\end{align*}$
		$\begin{align*}y-1\end{align*}$	$\begin{align*}= \frac {1}{2}(x-1)\end{align*}$
		$\begin{align*}y\end{align*}$	$\begin{align*}= \frac {1}{2}x+\frac {1}{2}\end{align*}$

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Example 5
::例5

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Find the derivative of $\begin{align*}f(x) = \frac{x} {x + 1}\end{align*}$ .
::查找 f(x) =xx+1 的衍生物。

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Using the derivative formula:
::使用衍生公式:

	$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \lim_{h \rightarrow 0} \frac{\left [\frac{x + h} {x + h + 1} - \frac{x} {x + 1}\right ]}{h}\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{x+h}{x+h+1}-\frac{x}{x+1}\right]\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{(x + h) (x + 1) - x (x + h + 1)} {(x + h + 1) (x + 1)}\right ]\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{x^2 + x + hx + h - x^2 - xh - x} {(x + h + 1) (x + 1)}\right ]\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{h} {(x + h + 1) (x + 1)}\right ]\end{align*}$
		$\begin{align*}= \lim_{h \rightarrow 0} \frac{1} {(x + h + 1) (x + 1)}\end{align*}$
		$\begin{align*}= \frac{1} {(x + 1)^2}\end{align*}$

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Example 6
::例6

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A rocket is propelled upwards and reaches a height of h( t ) = 4.9 t ² in t seconds.
::火箭向上推进,在秒内达到h(t)=4.9t2的高度。

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1. How high does it reach in 35 seconds?
   ::35秒内能达到多高?

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The height of the rocket at 35 secs is $\begin{align*}4.9(35)^2 = 6002.5m\end{align*}$
::火箭高度35秒为4.9(35)2=6002.5米

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2. What is the average velocity of the rocket during the first 35 seconds?
   ::火箭在最初35秒的平均速度是多少?

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$\begin{align*}V_{avg} = \frac{6002.5m}{35s} = 171.5 m/s\end{align*}$
::Vavg=6002.5m35s=171.5m/s

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3. What is the instantaneous velocity of the rocket at the end of the 35 seconds?
   ::35秒后火箭的瞬时速度是多少?

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To find the instantaneous speed, we need to find the derivative of $\begin{align*}h(t) = 4.9t^2\end{align*}$
::要找到瞬时速度, 我们需要找到 h( t) =4. 9t2 的衍生物

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$\begin{align*}h'(t) = 9.8t\end{align*}$ Using the instantaneous rate of change formula from above
::h(t)=9.8t 使用上方瞬时变化速率公式

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$\begin{align*}9.8 \cdot 35s = 343 m/sec\end{align*}$
::9.8-35s=343米/秒

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Example 7
::例7

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A particle moves in the positive direction along a straight line so that after t nanoseconds, its traversed distance is given by $\begin{align*}\chi (t) = 9.9 t^3\end{align*}$ nanometers.
::粒子沿着直线向正向移动,这样在t 纳米秒后,其跨度距离由 (t)=9.93 纳米计给定。

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1. What is the average velocity of the particle during the first 2 nanoseconds?
   ::粒子在前2毫秒的平均速度是多少?

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The particle moves $\begin{align*}9.9t^3nm\end{align*}$ in $\begin{align*}t\end{align*}$ secs
::粒子在t秒内移动9.9吨3nm

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$\begin{align*}\therefore 9.9 (2^3) = \frac{79.2nm}{2s} \to \frac{39.6 nm}{1s}\end{align*}$
::9.9(23)=79.2nm2s39.6nm1s

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2. What is the instantaneous velocity of the particle at t = 2 nanoseconds?
   ::T = 2 毫秒的粒子瞬时速度是多少?

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Using the formula for finding the derivative, we get $\begin{align*}\chi'(t) = 29.7t^2\end{align*}$
::使用寻找衍生物的公式, 我们得到\\( t)=29.7t2

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$\begin{align*}\therefore \chi'(2) = 118.8nm/s\end{align*}$
::*(2)=118.8nm/s

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$\begin{align*}\therefore\end{align*}$ The instantaneous velocity at $\begin{align*}t = 2\end{align*}$ is 118.8 nm/sec
::T=2的瞬时速度为118.8纳米/秒

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Review
::回顾

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Find the average rate of change:
::查找平均变化率:

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1. $\begin{align*} C = f(x) \end{align*}$ and $\begin{align*}f(x) = x^2 - 4x + 2\end{align*}$ . Find the average rate of change of (C) with respect to (x), when (x) is changed from x = 15, to x = 59.
   ::C=f(x)和f(x)=x2-4x+2。 查找(C)相对于(x)的平均变化率,当(x)从x=15改为x=59。
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2. $\begin{align*}H = f(x)\end{align*}$ and $\begin{align*} f(x) = x^2 - 5x + 201\end{align*}$ Find the average rate of change of (H), with respect to (x) when (x) is changed from x = 10 to x = 11.
   ::H=f(x)和f(x)=x2-5x+201 找出(H)相对于(x)的平均变化率,如果(x)从x=10改为x=11。
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3. $\begin{align*}N = f(x)\end{align*}$ and $\begin{align*}f(x) = 3x^2 - 4x - 1\end{align*}$ Find the average rate of change of (N) with respect to (x) when (x) is changed is changed from x = 20 to x = 64.
   ::N=f(x)和f(x)=3x2-4x-1 查找在(x)被更改时(N)相对于(x)的平均变化率从x=20改为x=64。
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4. $\begin{align*}H = f(x)\end{align*}$ and $\begin{align*} f(x) = x^2 + 10x + 201\end{align*}$ Find the average rate of change of (H) with respect to (x) when (x) is changed from x = 25 to x = 74.
   ::H=f(x)和f(x)=x2+10x+201 找寻(H)相对于(x)的平均变化率,当(x)从x=25改为x=74时。
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5. $\begin{align*}N = f(x)\end{align*}$ and $\begin{align*}f(x) = -5x^2 - 3x - 4\end{align*}$ Find the average rate of change of (N) with respect to (x) when (x) is changed from x = 30 to x = 54.
   ::N=f(x) 和 f(x) = 5x2-3x-4 查找在(x) 从x = 30 改为 x = 54. 时对(N) 的平均变化率。

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Find the instantaneous rate of change:
::查找瞬时变化率 :

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6. If $\begin{align*}C=f(x)\end{align*}$ and $\begin{align*}f(x) = -4x^2 + 2x + 5\end{align*}$ Find the instantaneous rate of change of (C) with respect to (x) when x = 25.
   ::如果C=f(x)和f(x)=4x2+2x+5,在x=25时查找(C)相对于(x)的瞬时变化率。
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7. If $\begin{align*} N = f(x)\end{align*}$ and $\begin{align*}f(x) = 3x^2 - x - 5\end{align*}$ Find the instantaneous rate of change of (N) with respect to x when x = 10.
   ::如果 N=f(x) 和 f(x) = 3x2-x-5 查找xx= 10时的(N)瞬时变化率。
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8. $\begin{align*} H = f(x)\end{align*}$ and $\begin{align*}f(x) = 4x^2 + 195\end{align*}$ Find the instantaneous rate of change of (H) with respect to x when x = 10.
   ::H=f(x)和f(x)=4x2+195 查找x=10时x(H)的瞬时变化率。
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9. $\begin{align*} N = f(x)\end{align*}$ and $\begin{align*}f(x) = -x^2 + x - 3\end{align*}$ Find the instantaneous rate of change of (N) with respect to x when x = 150.
   ::N=f(x) 和 f(x) x2+x-3 在 x = 150时查找 x (N) 的瞬时变化率。
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10. $\begin{align*} C = f(x)\end{align*}$ and $\begin{align*}f(x) = -3x^2 +4x - 4\end{align*}$ Find the instantaneous rate of change of (N) with respect to x when x = 20.
    ::C=f(x) 和 f(x) = 3x2+4x-4 相对于 x = 20时的(N)瞬时变化率。

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Use the definition of the derivative to find f ( x ) and then find the equation of the tangent line at x = x ₀ .
::使用衍生物的定义查找 f(x),然后在 x = x0 找到正切线的方程式。

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11. f ( x ) = 6 x ² ; x ₀ = 3.
    ::f(x) = 6x2;x0 = 3。
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12. $\begin{align*}f(x) = \sqrt {x+2}; x_0 = 8\end{align*}$
    :xx) x+2;x0=8
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13. f ( x ) = 3x ³ - 2; x ₀ = -1
    ::f(x) = 3x3 - 2; x0 = - 1
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14. $\begin{align*}f(x) = \frac {1}{x+2}; x_0 = -1\end{align*}$
    ::f( x) = 1x+2; x0 @% 1
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15. f ( x ) = a x ² - b , (where a and b are constants); x ₀ = b
    ::f(x) = x2 - b, (a和b为常数);x0 = b
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16. f ( x ) = x ^1/3 ; x ₀ = 1.
    ::f(x) = x1/3; x0 = 1。
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17. Suppose that f has the property that f ( x + y ) = f ( x ) + f ( y ) + 3 xy and %7D%7Bh%7D%20%3D%204"> $\begin{align*}\lim_{h \to 0}\frac {f(h)}{h} = 4\end{align*}$ . Find f (0) and f '( x ).
    ::假设f 具有 f(x + y) = f(x) + f + 3xy 和 limh_0f h= 4 的属性。 查找 f( 0) 和 f'(x) 。
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18. Find dy / dx
    ::查找 dy/ dx

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Solve the rate of change problems.
::解决变化的速度问题。

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18. A packing company in the south makes "Mama's Spaghetti Sauce." The cost of producing x jars is J = f(x) dollars. What does f'(100) = 9999 mean in this context?
    ::南方的一家包装公司制造了“Mama's Spaghetti Sauce”。生产X罐子的成本是J=f(x)美元。在这种情况下f'(100)=999999是什么意思?
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19. A cherry pie is taken from an oven when its temperature is 202°F and placed on a table in a room where the temperature is 75°F. The temperature of the pie over x minutes is given by T = f(x). What does f'(100) = 102 mean in this context?
    ::当烤箱温度为202°F时,樱桃馅饼取自烤箱,放在一个温度为75°F的房间的桌子上。 饼在x分钟后的温度由T = f(x)给出。 在这方面f'(100)=102是什么意思?
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20. The number of virus, after (x) hours, in a controlled laboratory experiment is V = f(x). What are the units of f'(x)?
    ::在受控制的实验室实验中,在(x)小时之后,病毒数量为V=f(x)。 f'(x)的单位是什么?
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21. The number of people in the US affected by the common cold in the month of November is defined by N = f(x) where x is the day of the month. What is the meaning of f'(x) in this context?
    ::11月份美国受普通寒冷影响的人数由N=f(x)确定,x是月份的一天。f'(x)在这方面的含义是什么?
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22. The number of households in Florida affected by hurricane season in the month of July is defined by J = f(x) where x is the day of the month. $\begin{align*}f(x) = 2x^2 + x + 1\end{align*}$ Find the average rate of change of J with respect to x when days is changed from x = 5 to x = 34.
    ::佛罗里达州7月份受飓风季节影响的住户数目由J = f(x) = f(x) 确定,其中x是月份的一天。 f(x) = 2x2+x+1 找寻当日从x = 5 改为x = 34时,J 相对于x的平均变化率。
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23. A cake is taken from an oven when its temperature is 196°F and placed on a cooling rack in a room where the temperature is 75°F. The temperature of the cake over (x) minutes is given by H = f(x). $\begin{align*}f(x) = 4x^2 + 15x + 196\end{align*}$ Find the instantaneous rate of change of H with respect to x when x = 15.
    ::当烤箱温度为196°F时,蛋糕取自烤箱,放在一个温度为75°F的房间的冷却架上。 (x)分钟以上蛋糕的温度由H = f(x) f(x)=4x2+15x+196给定,当x =15时,找到H的瞬时变化速度相对于x的瞬时变化速度。
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24. A pan of meatloaf is taken from an oven when its temperature is 205°F and placed on a table in a room where the temperature is 75°F. The temperature of the meatloaf over x minutes is given by H = f(x). $\begin{align*}f(x) = 2x^2 + 5x + 205\end{align*}$ . Find the average rate of change of H with respect to x when minutes is changed from x = 5 to x = 54.
    ::当烤箱的温度为205°F时,从烤箱中取出一片肉饼,放在一个温度为75°F的房间的桌子上。x分钟的肉饼温度由H = f(x) f(x) = 2x2+5x+205给出。当分钟从 x = 5 改为 x = 54时,请找到相对于 x 的H 平均变化速度。

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Review (Answers)
::回顾(答复)

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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 。