미분법 전체 문제

$\frac{25}{19\pi }$ ft/min

$\frac{2}{45\pi }$ ft/min

The distance is increasing at a rate of $62.50$ ft/sec.

The distance is decreasing at a rate of $11.99$ ft/sec.

(a) $5/(2\pi) \approx 0.796$cm/s (b) $1/(4\pi)\approx 0.0796$ cm/s (c) $1/(40\pi)\approx 0.00796$ cm/s

(a) $50.92$ ft/min (b) $0.509$ ft/min (c) $0.141$ ft/min

As the tank holds about 523.6ft$^3$, it will take about 52.36 minutes.

(a) The balloon is 105ft in the air. (b) The balloon is rising at a rate of 17.45ft/min. (Hint: convert all angles to radians.)

The cone is rising at a rate of $0.003$ft/s.

Velocity is the derivative of position with respect to time. So, the velocity of the particle is given by $s'(t)$. We need to find $s'(t)$, and determine when it is zero.

To differentiate, we us the chain rule.

$$\begin{aligned} s'(t)&=e^{ t^3-7t^2+8t}\cdot\frac{\mathrm{d}}{\mathrm{d}t}\{ t^3-7t^2+8t\}\\ &=e^{{t^3-7t^2+8t}}\cdot(3t^2-14t+8) \\ \text{To determine where this function is zero, we factor:} \\ &=e^{{t^3-7t^2+8t}}\cdot(3t-2)(t-4) \end{aligned}$$

So, the velocity is zero when $e^{{t^3-7t^2+8t}}=0$, when $3t-2=0$, and when $t-4=0$. Since $e^{{t^3-7t^2+8t}}$ is never zero, this tells us that the velocity is zero precisely when $t=\frac{2}{3}$ or $t=4$.

The question asks for $s''(1)$. We start our differentiation using the quotient rule:

$$\begin{aligned} s'(t)&=\frac{e^t(t^2+1)-e^t(2t)}{(t^2+1)^2}\\ &=\frac{e^t(t^2-2t+1)}{(t^2+1)^2} \\ \text{Using the quotient rule again,} \\ s''(t)&=\frac{(t^2+1)^2 \frac{\mathrm{d}}{\mathrm{d}t}\{e^t(t^2-2t+1)\}-e^t(t^2-2t+1) \frac{\mathrm{d}}{\mathrm{d}t}\{(t^2+1)^2\}}{(t^2+1)^4}\\ &=\frac{(t^2+1)^2\cdot\left[ e^t(2t-2)+e^t(t^2-2t+1)\right]- e^t(t^2-2t+1) \cdot 2(t^2+1)(2t)}{(t^2+1)^4}\\ &=\frac{e^t(t^2+1)^2(t^2-1)-4te^t(t-1)^2(t^2+1)}{(t^2+1)^4}\\ s''(1)&=0 \end{aligned}$$

The velocity of the particle is given by $h'(t)=\sin t$. Note $0<1<\pi$, so $h'(1)>0$–the particle is rising (moving in the positive direction, in this case “up"). The acceleration of the particle is $h''(t)=\cos t$. Since $0<1<\frac{\pi}{2}$, $h''(t)>0$, so $h'(t)$ is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.

For this problem, remember that velocity has a sign indicating direction, while speed does not.

The velocity of the particle is given by $h'(t)=3t^2-2t-5$. At $t=1$, the velocity of the particle is $-4$, so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is $h''(t)=6t-2$, so when $t=1$, the acceleration is (positive) $4$ units per second per second. That means the velocity (currently $-4$ units per second) is becoming a bigger number–since the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of $-3$ represents a slower motion than a velocity of $-4$.) So, the particle is slowing down at $t=1$.

Since $A=0$, the equation relating the variables tells us:

$$\begin{aligned} 0&=\log\left(C^2+D^2+1\right)\\ 1&=C^2+D^2+1\\ 0&=C^2+D^2\\ 0&=C=D \\ \text{This will probably be useful information. Since we're also given the value of a derivative, let's differentiate the equation relating the variables implicitly with respect to $t$. For ease of notation, we will write $\displaystyle \frac{\mathrm{d}A}{\mathrm{d}t}=A'$, etc.} \\ A'B+AB'&=\frac{2CC'+2DD'}{C^2+D^2+1} \\ \text{At $t=10$, $A=C=D=0$:} \\ A'B+0&=\frac{0+0}{0+0+1}\\ A'B&=0 \\ \text{at $t=10$, $A'=2$ units per second:} \\ 2B&=0\\ B&=0. \end{aligned}$$

(a) Observe that the units on $g(v)$ are “kilometers per liter”. Since the units on $f(v)$ are “liters per kilometer”, it follows that $g$ is the reciprocal of $f$: $g(v) = \frac{1}{f(v)}$. By the quotient rule, we then know that $$g'(v) = \frac{f(v) \cdot 0 - 1 \cdot f'(v)}{f(v)^2} = -\frac{f'(v)}{f(v)^2}\,.$$ Further, using the given values we can say that $g(80) = \frac{1}{f(80)} = \frac{1}{0.05} = 20$ kilometers per liter, and $g'(80) = -\frac{f'(80)}{f(80)^2} = -\frac{0.0004}{0.05^2} = -0.16$. Note that the units on $g'(80)$ are “kilometers per liter per kilometer per hour”. (b) Since the units on $h(v)$ are “liters per hour”, and the units on $f(v)$ are “liters per kilometer”, we note that if we multiply $f(v)$ by $v$ (whose units are “kilometers per hour”), we get that the units on $v \cdot f(v)$ match the units of $h(v)$. Thus, $$h(v) = v \cdot f(v)\,.$$ By the product rule, it follows that $$h'(v) = v \cdot \frac{d}{dv}[f(v)] + \frac{d}{dv}[v] \cdot f(v) = v \cdot f'(v) + 1 \cdot f(v)\,.$$ Using the known function and derivative values of $f$, we determine that $h(80) = 80 \cdot 0.05 = 4$ liters per hour and $h'(80) = 80 \cdot 0.0004 + 0.05 = 0.082$ liters per hour per kilometer per hour. (c) To explain the meaning of these function and derivative values to someone who doesn't know calculus, we could say something like the following:

• The given values of $f(80) = 0.05$ and $f'(80) = 0.0004$ tell us that when the car is going $80$ kilometers per hour, it is using $0.05$ liters of fuel for every kilometer it travels. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will use about an additional $0.0004$ liters of fuel per kilometer.
• The values of $g(80) = 20$ and $g'(80) = -0.16$ tell us that when the car is going $80$ kilometers per hour, it is able to travel $20$ kilometers for each liter of fuel used. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will be able to travel about $0.16$ fewer kilometers per liter of fuel.
• Finally, the values of $h(80) = 4$ and $h'(80) = 0.82$ tell us that when the car is going $80$ kilometers per hour, it is using $4$ liters of fuel for every hour it travels. In addition, for each additional kilometer per hour of velocity above $80$ kph, we expect that the car will ues about $0.082$ more liters of fuel per hour. Note that all three of these observations are saying fundamentally the same thing, just from different perspectives: the first regards fuel consumption measured in liters per kilometer; the second regards fuel consumption in kilometers per liter; and the last regards fuel consumption in liters per hour.

(a) The object's instantaneous velocity at $t=2$ is given by $h'(2)$. Since $h(t) = 3 + \frac{2\cos(t)}{1.2^t}$, by the quotient rule we can determine that $$\begin{aligned}h'(t) &= \frac{1.2^t \cdot (-2\sin(t)) - 2\cos(t) \cdot 1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{1.2^t (-2\sin(t) - 2\cos(t) \ln(1.2))}{(1.2^t)^2} \\ &= \frac{-2\sin(t) - 2\cos(t) \ln(1.2)}{1.2^t}\end{aligned}\,.$$ Thus, $h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575$ feet per second is the instanateous (vertical) velocity of the object at $t=2$. (b) To find the object's acceleration at $t=2$, we compute $h''(2)$. First, by the quotient rule, $$\begin{aligned}h''(t) &= \frac{1.2^t(-2\cos(t) + 2\sin(t) \ln(1.2)) - (-2\sin(t) - 2\cos(t) \ln(1.2))1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{-2\cos(t) + 2\sin(t) \ln(1.2)) + 2\sin(t)\ln(1.2) + 2\cos(t) \ln^2(1.2)}{1.2^t} \\ &= \frac{\cos(t)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(t))}{1.2^t}\end{aligned}\,.$$ Therefore, the acceleration at $t = 2$ is $h''(2) = \frac{\cos(2)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193$ feet per second per second. (c) At $t=2$, the object is falling and slowing down. We know the object is falling because its velocity is negative, and slowing down because its negative velocity is increasing (due to $h''(2)$ being positive) and thus getting less negative.

(a) We are asked to find $\frac{dV}{dh}$ (the derivative of $V$ with respect to the independent variable $h$) at the instant $h = 1$. If we first expand $V$ to write $$V = \frac{\pi}{3}\left(12h^2 - h^3\right)$$ we find that $$\frac{dV}{dh} = \frac{\pi}{3} \left(24h - 3h^2\right) = \pi \left(8h-h^2 \right)\,,$$ with units “cubic feet per foot”. Evaluating at $h=1$, we get $$\left. \frac{dV}{dh} \right|_{h=1} = \pi \left(8 \cdot 1 - 1^2 \right) = 7\pi$$ (b) Since $h(t) = \sin(\pi t) + 1$, the instantaneous rate of change of $h$ with respect to $t$ is $h'(t) = \pi \cos(\pi t)$, and therefore $h'(2) = \pi \cos(2\pi) = \pi$ feet per hour. (c) First we want to find $\frac{dV}{dt}$, the derivative of $V$ with respect to the independent variable $t$. Here we can write $V$ as a function of $t$ by writing $V(t) = \frac{\pi}{3}\left(12(h(t))^2 - (h(t))^3\right)$, where $h(t) = \sin(\pi t) + 1$. By the chain rule, it follows that $$\frac{dV}{dt} = \frac{\pi}{3}\left( 12 \cdot 2 h(t) \cdot h'(t) - 3 (h(t))^2 \cdot h'(t) \right)$$ and hence using $h(t) = \sin(\pi t) + 1$ and $h'(t) = \pi \cos(\pi t)$, $$\frac{dV}{dt} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(\pi t) + 1) \cdot \pi \cos(\pi t) - 3 (\sin(\pi t) + 1)^2 \cdot \pi \cos(\pi t) \right]$$ with units “ feet per hour”. Now we want to find $\frac{dV}{dt}$ when $t=2$. Using our most recent work, we have $$\left. \frac{dV}{dt} \right|_{t=2} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(2\pi) + 1) \cdot \pi \cos(2\pi) - 3 (\sin(2\pi) + 1)^2 \cdot \pi \cos(2\pi) \right] = 7 \pi^2$$ cubic feet per hour. (d) In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time. The first calculation is a direct one, while the second is more indirect, as it involves the composition of $V$ with $h$ as a function of time. In the latter case, both $h$ and $V$ are functions of time, which allows us to think not only of how $V$ is changing as a function of $h$, but also as a function of $t$.