평균변화율과 미분계수

$\lim_{x\to {1}^{-}}\frac{1-1}{x-1}=0\ne 1=\lim_{x\to {1}^{+}}\frac{x-1}{x-1}$

(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) Observe that $v(t) = 16-32t$ is a linear function with negative slope $-32$, and thus decreasing. Since the function is of interest from $t = 0$ to $t=2$, the velocity is greatest at the left endpoint, when $t=0$. This greatest velocity is $v(0) = 16$ feet per second. (b) Since $v$ is a linear function, its slope is constant, and hence the slope of every tangent line to the function is the same as the slope of the line the function generates. Thus, $v'(1) = -32$. (c) Because $v$ is measured in feet per second and $t$ in seconds, the units on $v'(1)$ are “feet per second per second.” Knowing that $v'(1) = -32$, this tells us that the ball's velocity is decreasing at a rate of 32 feet per second per second; that is, for whatever the ball's velocity is at $t = 1$, we would expect the velocity to decrease by (exactly) $32$ feet per second over the next one second. (Normally this total decrease would be approximate, not exact, but since the velocity function is linear, we can compute the exact total change.) (d) The instantaneous rate of change of velocity is precisely the acceleration of the ball. Indeed, $v'(t)$ tells us how the ball's velocity changes, and that it does so at a constant rate of $-32$ feet per second per second. This is also consistent with the fact that acceleration due to gravity is constant.