평균변화율과 미분계수

The rate of change in this case will be the relationship between the heat added and the temperature change. $\displaystyle\lim_{h \rightarrow 0} \dfrac{T(j+h)-T(j)}{h} \approx \dfrac{T(j+1)-T(j)}{1}=T(j+1)-T(j)$, or the change in temperature after the application of one joule. (This is closely related to heat capacity and to specific heat — there's a nice explanation of this on Wikipedia.)

As usual, it is instructive to think about the definition of the derivative: $$P'(T) = \displaystyle\lim_{h \rightarrow 0}\dfrac{P(T+h)=P(T)}{h} \approx \dfrac{P(T+1)-P(t)}{1} = P(T+1)-P(T).$$ This is the difference in population between two hypothetical populations, raised one degree in temperature apart. So, it is the number of extra individuals that exist in the hotter experiment (with the understanding that this number could be negative, as one would expect in conditions that are hotter than the bacteria prefer). So $P'(T)$ is the number of bacteria added to the colony per degree.

$R'(t)$ is the rate at which the wheel is rotating measured in rotations per second. To convert to degrees, we multiply by 360: $\boxed{360R'(t)}$.

(a) The change in population is $P(2000) - P(1980) = 197 800 - 181 843 = 15 957$ people. (b) Since there were $15 957$ people added to Grand Rapids over the 20 years from 1980 to 2000, in an average year the population grew by $\frac{15,957}{20} \approx 798$ people/year. (c) $\frac{f(b)-f(a)}{b-a}=\frac{\Delta f}{\Delta x}$ measures the slope of a secant line through the points $(a,f(a))$ and $(b,f(b))$. (d) $$AV_{[0,20]}=\frac{P(20)-P(0)}{20-0}=\frac{197 800 - 181 843}{20}=\frac{15 957}{20} \approx 798$$ people per year. (e) $$\begin{aligned}AV_{[5,10]} &=\frac{P(10) - P(5)}{10-5} = \frac{189 116.72 - 185 444.20}{10-5} \approx 734.50 \\ AV_{[5,9]} &= \frac{188 376.44 - 185 444.20}{9-5} \approx 733.06 \\ AV_{[5,8]} &= \frac{187 639.06 - 185,444.20}{8-5} \approx 731.62 \\ AV_{[5,7]} &= \frac{186 904.57 - 185,444.20}{7-5} \approx 730.19 \\ AV_{[5,6]} &= \frac{186 172.95 - 185,444.20}{6-5} \approx 728.7535\end{aligned}$$ The units on each of these quantities are “people per year”.

(a) First, $P(7)-P(0) = 115(1.014)^7-115(1.014)^0 \approx 0.1175$ billion people is the total population change. It follows that the average rate of change over this time period is $AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679$ billion people per year. The instantaneous rate of change of the population at $t = 7$ is $$P'(7) = \lim_{h \to 0} AV_{[7,7+h]} = \lim_{h \to 0} \frac{115(1.014)^{7+h}-115(1.014)^7}{h}\,,$$ and approximating this quantity with small values of $h$, it follows that $P'(7) \approx 0.1762$. Hence, the instantaneous rate of change at $t = 7$ is greater than the average rate of change of $P$on the interval $[0,7]$. (b) Using the formula for average rate of change, $$AV_{[19,29]}=\frac{1.15(1.014)^{29}-1.15(1.014)^{19}}{29-19} \approx 0.02234$$ billion people/year. (c) We will say that today's date is July 1, 2015, which means that $t = 22.5$, so we want to compute $P'(22.5)$. The limit definition tells us that $$P'(22.5) = \lim_{h \to 0} \frac{115(1.014)^{22.5+h}-115(1.014)^{22.5}}{h}\,.$$ Estimating this limit by using small values of $h$, we determine that $P'(22.5) \approx 0.02186$, which is measured in billions of people per year. (d) We want the equation of the tangent line that has slope $P'(22.5) \approx 0.02186$ and that passes through the point $(22.5, P(22.5))$. Using point-slope form, it follows that the line is given by $$y - 1.57236 = 0.02186(t-22.5)\,.$$

(a) The table below gives values of the difference quotient $\frac{f(a+h)-f(a)}{h}$ for $a=2$ and several small values of $h$. This table indicates that $f'(2)$ is approximately $1$.

$h$	$-0.0001$	$-0.001$	$-0.01$	$0.01$	$0.001$	$0.0001$
$\frac{f(2+h)-f(2)}{h}$	$0.9999$	$0.9990$	$0.9900$	$1.0100$	$1.0010$	$1.0001$
The tangent line to $f$ at the point $(2,f(2))$ as shown in the following appears to have a slope that is approximately 1.
Finally, we verify these approximations using an algebraic approach. The definition of the derivative of $f$ at $x=a$ shows that
$$\begin{aligned}f'(2) &= \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} \\ &= \lim_{h \to 0} \frac{\left[(2+h)^2 - 3(2+h)\right] - (-2)}{h} \\ &= \lim_{h \to 0} \frac{\left[4+4h+h^2 - 6 - 3h\right] + 2}{h} \\ &= \lim_{h \to 0} \frac{h^2 + h}{h} \\ &= \lim_{h \to 0} \frac{h(h+1)}{h} \\ &= \lim_{h \to 0} h+1 \\ &= 1\end{aligned}$$
This confirms our conclusions using the numeric and geometric approaches.
(b) The following table gives values of the difference quotient $\frac{f(a+h)-f(a)}{h}$ for $a=1$ and several small values of $h$. This table indicates that $f'(1)$ is approximately $-1$.
$h$	$-0.0001$	$-0.001$	$-0.01$	$0.01$	$0.001$	$0.0001$
---	---	---	---	---	---	---
$\frac{f(1+h)-f(1)}{h}$	$-1.00010$	$-1.001001$	$-1.010101$	$-0.990099$	$-0.999001$	$-0.999900$
The tangent line to $f$ at the point $(1,f(1))$ as shown in the following figure appears to have a slope that is approximately $-1$.
Finally, we verify these approximations using an algebraic approach. The definition of the derivative of $f$ at $x=a$ shows that
$$\begin{aligned}f'(1) &= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - 1}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - 1}{h} \left(\frac{1+h}{1+h} \right) \\ &= \lim_{h \to 0} \frac{1-(1+h)}{h(1+h)} \\ &= \lim_{h \to 0} \frac{-h)}{h(1+h)} \\ &= \lim_{h \to 0} -\frac{1}{1+h} \\ &= -1\end{aligned}\,.$$ This confirms our conclusions using the numeric and geometric approaches.
(c) The table below gives values of the difference quotient $\frac{f(a+h)-f(a)}{h}$ for $a=1$ and several small values of $h$. This table indicates that $f'(1)$ is approximately $0.5$.
$h$	$-0.0001$	$-0.001$	$-0.01$	$0.01$	$0.001$	$0.0001$
---	---	---	---	---	---	---
$\frac{f(1+h)-f(1)}{h}$	$0.50001300$	$0.50012510$	$0.50125629$	$0.49875620$	$0.49987500$	$0.49999000$
The tangent line to $f$ at the point $(1,f(1))$ as shown in the figure below appears to have a slope that is approximately $0.5$.
Finally, we verify these approximations using an algebraic approach. The definition of the derivative of $f$ at $x=a$ shows that
$$\begin{aligned}f'(1) &= \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-\sqrt{1}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \left(\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} \right) \\ &= \lim_{h \to 0} \frac{(1+h)-1}{h(\sqrt{1+h}+1)} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{1+h}+1)} \\ &= \frac{1}{\sqrt{1+h}+1} \\ &= \frac{1}{2}\end{aligned}\,.$$ This confirms our conclusions using the numeric and geometric approaches.
(d) The data in the following table indicates that $\frac{f(1+h)-f(1)}{h}$ has values of $1$ for $h < 0$ and values of $-1$ for $h > 0$. This implies that $f'(1)$ does not exist.
$h$	$-0.0001$	$-0.001$	$-0.01$	$0.01$	$0.001$	$0.0001$
---	---	---	---	---	---	---
$\frac{f(1+h)-f(1)}{h}$	$1$	$1$	$1$	$-1$	$-1$	$-1$
The graph of $f$ in the figure below appears to show a sharp corner at the point $(1,f(1))$, also implying that there is no tangent line to $f$at $(1,f(1))$.
Now we verify these approximations using an algebraic approach. Notice first that
$$\begin{aligned}\frac{f(1+h)-f(1)}{h} &= \frac{[2-|(1+h)-1|] - 2}{h} \\ &= \frac{(2-|h|)-2}{h} \\ &= \frac{-|h|}{h}\end{aligned}$$
The definition of the absolute value function tells us that $|h| = h$ whenever $h \ge 0$, while $|h| = -h$ whenever $h < 0$. Thus, for $h > 0$, we have
$$\frac{f(1+h)-f(1)}{h} = \frac{-|h|}{h} = -\frac{h}{h} = -1$$
and for $h < 0$, we have
$$\frac{f(1+h)-f(1)}{h} = \frac{-|h|}{h} = -\frac{-h}{h} = 1\,.$$ This confirms that $\lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$ does not exist.
(e) The data in the following table indicate that $\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h}$ approaches $0$ as $h \to 0$.
$h$	$-0.0001$	$-0.001$	$-0.01$	$0.01$	$0.001$	$0.0001$
---	---	---	---	---	---	---
$\frac{f(1+h)-f(1)}{h}$	$0.00005$	$0.0005$	$0.005$	$-0.005$	$-0.0005$	$-0.00005$
The graph of $f$ in the figure below shows a horizontal tangent line at the point $(\frac{\pi}{2},f(\frac{\pi}{2}))$, which also suggests that $f'(\frac{\pi}{2}) = 0$.
Finally, we attempt to verify these approximations using an algebraic approach. Notice first that
$$\begin{aligned}\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} &= \frac{\sin(\frac{\pi}{2}+h) - \sin \frac{\pi}{2}}{h} \\ &= \frac{\sin(\frac{\pi}{2}+h) - 1}{h}\end{aligned}$$
Using the sum of two angles identity for the sine function, it follows that
$$\sin(\frac{\pi}{2}+h) = \sin \frac{\pi}{2} \cos(h) + \cos \frac{\pi}{2} \sin(h) = 1 \cdot \cos(h) + 0 \cdot \sin(h) = \cos(h)\,.$$ Hence,
$$\begin{aligned}f'(\frac{\pi}{2}) &= \lim_{h \to 0} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} \\ &= \lim_{h \to 0} \frac{\cos(h) - 1}{h}\end{aligned}$$
At this point, we encounter a limit that we are unable to evaluate further using algebraic means. Like the earlier limit above, we can estimate the limit using small values of $h$. Doing so confirms that this limit also approaches $0$, and we hence conclude that $f'(\frac{\pi}{2}) = 0$.