미분 전체 문제

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: $\boxed{360R'(t)}$.

Population growth is rate of change of population. Population in year $2000+t$ is given by $P(t)=P_0+b(t)-d(t)$, where $P_0$ is the initial population of the town. Then $P'(t)$ is the expression we're looking for, and $P'(t)=b'(t)-d'(t)$.

It is interesting to note that the initial population does not obviously show up in this calculation. It would probably affect $b(t)$ and $d(t)$, but if we know these we do not need to know $P_0$ to answer our question.

Let $w(t)$ and $l(t)$ be the width and length of the rectangle. Given in the problem is that $w'(t)=2$ and $l'(t)=5$. Since both functions have constant slopes, both must be lines. Their slopes are given, and their intercepts are $w(0)=l(0)=1$. So, $w(t)=2t+1$ and $l(t)=5t+1$.

The area of the rectangle is $A(t)=w(t)\cdot l(t)$, so using the product rule, the rate at which the area is increasing is $A'(t)=w'(t)l(t)+w(t)l'(t)=2(5t+1)+5(2t+1)=20t+7$ square metres per second.

We know the top speed of the caribou, so we can use this to give the minimum possible number of hours the caribou spent travelling during its migration. If the caribou travels at 70kph, it will take $(5000 \mathrm{km})\left(\dfrac{1 \mathrm{hr}}{70 \mathrm{km}}\right) \approx 71.4 \mathrm{hrs}$ to travel 5000 kilometres. Probably the caribou wasn't sprinting the whole time, so probably it took it longer than that, but we can only say for sure that the caribou spent at least about 71.4 hours migrating.

If $f(x)$ is the position of the crane at time $x$, measured in hours, then (if we let $x=0$ be the beginning of the day) we know that $f(24)-f(0)=240$. Since $f(x)$ is the position of the bird, $f(x)$ is continuous and differentiable. So, the MVT says there is a $c$ in $(0,24)$ such that $f'(x)=\dfrac{f(24)-f(0)}{24-0}=\dfrac{240}{24}=10$. That is, at some point $c$ during the day, the speed of the crane was exactly 10 kph.

The acceleration is constant, which means the rate of change of the velocity is constant. So, since it took 10 seconds for the velocity to increase by 1 metre per second (from $1~\frac{\mathrm{m}}{\mathrm{s}}$ to $2~\frac{\mathrm{m}}{\mathrm{s}}$), then it always takes 10 seconds for the velocity to increase by 1 metre per second.

So, it takes 10 seconds to accelerate from $2~\frac{\mathrm{m}}{\mathrm{s}}$ to $3~\frac{\mathrm{m}}{\mathrm{s}}$. To accelerate from $3~\frac{\mathrm{m}}{\mathrm{s}}$ to $13~\frac{\mathrm{m}}{\mathrm{s}}$ (that is, to change its velocity by 10 metres per second), it takes $10\times 10=100$ seconds.

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.

Let $v(t)$ be the velocity of the object. From the given information:

• $v(0)$ is some value, call it $v_0$,
• $v(1)=2v_0$ (since the speed doubled in the first second),
• $v(2)=2(2)v_0$ (since the speed doubled in the second second),
• $v(3)=2(2)(2)v_0$, and so on. So, for general $t$: $$v(t)=2^tv(0)$$ To find its acceleration, we simply differentiate. Recall $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\{2^x\}=2^x\log 2$, where $\log$ denotes logarithm base $e$. $$a(t)=2^tv_0\log 2$$ Remark: we can also write $a(t)=v(t)\log 2$. The acceleration doubles every second as well.

(a) To say that $f(50) = 0.75$ means that if a patient takes a dose of $50$ ml of a drug, the patient will experience a body temperature change of $0.75$ degrees F. (b) Since the units of $f$ are degrees F and the units of $q$ are ml, it follows that the units of $s(q) = f'(q)$ are “degrees Fahrenheit per milliliter.” (c) If $f'(50) = -0.02$ degrees Fahrenheit per milliliter, this means that the instantaneous rate of change of the temperature change function is $-0.02$ degrees F per ml at the dose-level of $50$ ml. This means that for a patient taking a $50$ ml dose, adding one more ml to the dose leads us to expect a temperature change that is about $0.02$ degrees less than the temperature change induced by a $50$ ml dose.

(a) To find the average rate of change of $h$ on the interval $[40000,55000]$, we divide the change in $h$ values (in dollars) on this interval by the change in $m$ values (in miles) to obtain $$AV_{[40000,55000]} = \frac{h(55000)- h(40000)}{55000-40000} = \frac{13200-15500}{15000} = -\frac{2300}{15000} \approx -0.153$$ dollars per mile. This tells us that from the mileage of $40 000$ to $50 000$ miles, on average the car's value decreases by $0.153$ dollars per mile driven. (b) Since $40 000$ and $70 0000$ are equidistant from $55 000$, we use the central difference quotient to approximate the derivative of $h$ at $55 0000$ to obtain $$h'(55000) \approx \frac{h(70000) - h(40000)}{70000-40000} = \frac{11100-15500}{30000} = - \frac{4400}{30000} \approx -0.147$$ dollars per mile. This approximation tells us that at the instant the car has been driven $55 000$ miles, the car's value is decreasing at an instantaneous rate of $0.147$ dollars per mile. That is, during $55 0001$st mile, we'd expect the car's value to drop by $0.147$ dollars. (c) As we pile up miles on our car, we expect the value to go down. However, we should expect the depreciation to be more extreme at first, but less in magnitude as we drive more miles and our car gets older. Note, though, the the value of the car is always decreasing and $h'(m)$ is always negative. Thus, we should expect that the rate of change of $h$ is increasing (becoming less negative) as $m$ increases and that $h'(30000) < h'(80000)$. (d) The value of a car is never negative. So we know that $h(m) \geq 0$ for all $m$. A new car decreases in value very quickly at first, and as the car gets older with more mileage, its value will depreciate to near zero. So $h$ is a decreasing function. Since the value of our car always decreases, it follows that $h'(m) < 0$ for all $m > 0$. The fact that the depreciation slows as $m$ increases implies that $h'$ is an increasing (getting less negative) function. Since $h'$ is increasing, it follows that $h$ is bends upward; we will study this issue more in Section 1.6. So the graph of $h$ might have the general shape of the graph of $y = e^{-x}$ for positive values of $x$.