미분 전체 문제

a. $2({h}^{2}+6h+12);$ b. $\text{(i)}\,25.22$ m/s, $\text{(ii)}\,24.12$ m/s, $\text{(iii)}\,24.01$ m/s, $\text{(iv)}\,24$ m/s; c. $24$ m/s

a. $P(t)=0.03983+0.4280$ b. ${P}^{'}(t)=0.03983.$ The population is increasing. c. ${P}^{''}(t)=0.$ The rate at which the population is increasing is constant.

$0$

$(2,1)$

$p'(t) = \displaystyle\lim_{h \rightarrow 0}\frac{p(t+h)-p(t)}{h} \approx \frac{p(t+1)-p(t)}{1} = p(t+1)-p(t)$, or the difference in profit caused by the sale of the $(t+1)^{\mathrm{st}}$ widget. So, $p'(t)$ is the profit from the $(t+1)^{\mathrm{st}}$ widget. That is, $p'(t)$ is the profit per additional widget sold, when $t$ widgets are being sold. This is called the marginal profit per widget, when $t$ widgets are being sold.

How quickly the temperature is changing per unit change of depth, measured in degrees per metre. In an ordinary body of water, the temperature near the surface ($d=0$) is pretty variable, depending on the sun, but deep down it is more stable (unless there are heat sources). So, one might reasonably expect that $|T'(d)|$ is larger when $d$ is near 0.

The rate of change of velocity is acceleration. (If your velocity is increasing, you're accelerating; if your velocity is decreasing, you have negative acceleration.)

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.

Let $s(t)$ be the distance your keys have travelled since they left your hand. The rate at which they are travelling, $s'(t)$, is decreasing by 0.25 metres per second. That is, $s''(t)=-0.25$. Therefore, $s'(t)=-0.25t+c$ for some constant $c$. Since $c=s'(0)=2$, we see $$s'(t)=2-0.25t$$ Then $s(t)$ has $2-0.25t$ as its derivative, so $s(t)=2t-\frac{1}{8}t^2+d$ for some constant $d$. At time $t=0$, the keys have not yet gone anywhere, so $0=s(0)=d$. Therefore, $$s(t)=2t-\frac{1}{8}t^2$$

The keys reach your friend when $s(t)=2$ and $t>0$. That is:

$$\begin{aligned} 2&=2t-\frac{1}{8}t^2\\ 0&=\frac{1}{8}t^2-2t+2\\ t&=8\pm4\sqrt{3} \end{aligned}$$

We need to figure out which of these values of $t$ is really the time when the keys reach your friend. The keys travel this way from $t=0$ to the time they reach your friend. (Then $s(t)$ no longer describes their motion.) So, we need to find the first value of $t$ that is positive with $s(t)=2$. Since $8-4\sqrt{3}>0$, this is the first time $s(t)=2$ and $t>0$. So, the keys take $$8-4\sqrt{3}\approx 1~\text{second}$$ to reach your friend.

Suppose the deceleration provided by the brakes is $d~\frac{\mathrm{km}}{\mathrm{hr}^2}$. Then if $v(t)$ is the velocity of the car, $v(t)=120-dt$ (at $t=0$, the velocity is 120, and it decreases by $d$ kph per hour). The car stops when $0=v(t)$, so $t=\frac{120}{d}$ hours.

Let $s(t)$ be the distance the car has travelled $t$ hours after applying the brakes. Then $s'(t)=v(t)$, so $s(t)=120t-\frac{d}{2}t^2+c$ for some constant $c$. Since $0=s(0)=c$, $$s(t)=120t-\frac{d}{2}t^2$$

The car needs to stop in 100 metres, which is $\frac{1}{10}$ kilometres. We already found that the stopping time is $t=\frac{120}{d}$. So:

$$\begin{aligned} \frac{1}{10}&=s\left(\frac{120}{d}\right)\\ \frac{1}{10}&=120\left(\frac{120}{d}\right)-\frac{d}{2}\left(\frac{120}{d}\right)^2\\ \\ \text{Multiplying both sides by $d$:} \\ \frac{d}{10}&=120^2-\frac{120^2}{2}\\ d&=5\cdot 120^2 \end{aligned}$$

So, the brakes need to apply 72 000 kph per hour of deceleration.

If the acceleration was constant, then it was $$\frac{17 500~\mathrm{mph}}{\frac{8.5}{60}~\mathrm{hr}} \approx 123 500 ~\frac{\mathrm{miles}}{\mathrm{hr}^2}$$

So, the velocity $t$ hours from liftoff is $$v(t)=123500t$$ Therefore, the position of the shuttle $t$ hours from liftoff (taking $s(0)=0$ to be its initial position) is $$s(t)=\frac{123500}{2}t^2=61750t^2$$ So, after $\frac{8.5}{60}$ hours, the shuttle has travelled $$s\left(\frac{8.5}{60}\right)=(61750)\left(\frac{8.5}{60}\right)^2\approx 1240 ~\mathrm{miles}$$ or a little less than 2000 kilometres.

The acceleration of a falling object due to gravity is $9.8$ metres per second squared. So, the object's velocity $t$ seconds after being dropped is $$v(t)=9.8t~\frac{\mathrm{m}}{\mathrm{s}}$$ We want $v(t)$ to be the speed of the peregrine's dive, so we should convert that to metres per second: $$325~\frac{\mathrm{km}}{\mathrm{hr}}\cdot \left(\frac{1000~\mathrm{m}}{1~\mathrm{km}}\right)\left(\frac{1~\mathrm{hr}}{3600~\mathrm{sec}}\right)=\frac{1625}{18}~\frac{\mathrm{m}}{\mathrm{s}}$$ The stone will reach this velocity when $$9.8t=\frac{1625}{18}\qquad\Rightarrow\qquad t=\frac{1625}{18(9.8)}$$

What is left to figure out is how far the stone will fall in this time. The position of the stone $s(t)$ has derivative $9.8t$, so $$s(t)=4.9t^2$$ if we take $s(0)=0$. So, if the stone falls for $\frac{1625}{18(9.8)}$ seconds, in that time it travels $$s\left(\frac{1625}{18(9.8)}\right)=4.9\left(\frac{1625}{18(9.8)}\right)^2\approx 416 ~\mathrm{m}$$ So, you would have to drop a stone from about 416 metres for it to fall as fast as the falcon.

Since gravity alone brings your cannon ball down, its acceleration is a constant $-9.8~\frac{\mathrm{m}}{\mathrm{s}^2}$. So, $v(t)=v_0-9.8t$ and thus its height is given by $s(t)=v_0t-4.9t^2$ (if we set $s(0)=0$).

We want to know what value of $v_0$ makes the maximum height 100 metres. The maximum height is reached when $v(t)=0$, which is at time $t=\frac{v_0}{9.8}$. So, we solve:

$$\begin{aligned} 100&=s\left(\frac{v_0}{9.8}\right)\\ 100&=v_0\left(\frac{v_0}{9.8}\right)-4.9\left(\frac{v_0}{9.8}\right)^2\\ 100&=\left(\frac{1}{9.8}-\frac{4.9}{9.8^2}\right)v_0^2\\ 100&=\frac{1}{2\cdot 9.8}v_0^2\\ v_0^2&=1960\\ v_0&=\sqrt{1960}\approx 44~\frac{\mathrm{m}}{\mathrm{s}} \end{aligned}$$

where we choose the positive square root because $v_0$ must be positive for the cannon ball to get off the ground.

The derivative of acceleration is constant, so the acceleration $a(t)$ has the form $mt+b$. We know $a(0)=-50~000$ and $a\left(\frac{3}{3600}\right)=-60~000$ (where we note that $3$ seconds is $\frac{3}{3600}$ hours). So, the slope of $a(t)$ is $\frac{-60~000+50~000}{\frac{3}{3600}}=-12~000~000$, which leads us to $$a(t)=-50~000-(12~000~000)t$$ where $t$ is measured in hours.

Since $v'(t)=a(t)=-50~000-(12~000~000)t$, we see $$v(t)=\frac{-12~ 000~000}{2}t^2-50~000t+c=-6~000~000t^2-50~000t+c$$ for some constant $c$. Since $120=v(0)=c$: $$v(t)=-6~000~000t^2-50~000t+120$$

Then after three seconds of braking,

$$\begin{aligned}v\left(\frac{3}{3600}\right) &=-6~000~000\left(\frac{3}{3600}\right)^2-50~000\left(\frac{3}{3600}\right)+120\\ &=-\frac{25}{6}-\frac{125}{3}+120\\ &\approx 74.2 \text{ kph} \end{aligned}$$

Remark: When acceleration is constant, the position function is a quadratic function, but we don't want you to get the idea that position functions are always quadratic functions–in the example you just did, it was the velocity function that was quadratic. Position, velocity, and acceleration functions don't have to be polynomial at all–it's only in this section, where we're dealing with the simplest cases, that they seem that way.