미분법 전체 문제

First, let's consider the behaviour of exponential functions $a^x$ based on whether $a$ is greater or less than 1. As we know, $\displaystyle \lim_{x\to\infty}a^x=\left\{\begin{array}{ll} \infty & a>1\\0&a<1 \end{array}\right.$ and $\displaystyle \lim_{x\to-\infty}a^x=\left\{\begin{array}{ll} 0 & a>1\\\infty&a<1 \end{array}\right.$. Our function has $\displaystyle \lim_{x \to \infty} f(x)=\infty$ and $\displaystyle \lim_{x \to -\infty} f(x)=0$, so we conclude $a>1$: thus $(d)$ and also $(b)$ hold. (We could have also seen that $(b)$ holds because $a^x$ is defined for all real numbers.)

It remains to decide whether $a$ is greater or less than $e$. (If $a$ were equal to $e$, then $f'(x)$ would be the same as $f(x)$.) We saw in the text that $\frac{\mathrm{d}}{\mathrm{d}x}\{a^x\}=C(a)a^x$ for the function $C(a)=\displaystyle \lim_{h \to 0} \dfrac{a^h-1}{h}$. We know that $C(e)=1$. (Actually, we chose $e$ to be the number that has this property.) From our graph, we see that $f'(x)<f(x)$, so $C(a)<1=C(e)$. In other words, $\displaystyle \lim_{h \to 0} \dfrac{a^h-1}{h}<\displaystyle \lim_{h \to 0} \dfrac{e^h-1}{h}$; so, $a<e$. Thus $(e)$ holds.

The graph $f(x)=\sin x$ has horizontal tangent lines precisely at those points where $\cos x=0$. This must be true, since $\frac{\mathrm{d}}{\mathrm{d}x}\{\sin x\}=\cos x$: where the derivative of sine is zero, cosine itself is zero.

The graph $f(x)=\sin x$ has maximum slope at those points where $\cos x$ has a maximum. This makes sense, because $f'(x)=\cos x$: the maximum values of the slope of sine correspond to the maximum values of cosine.

We compute the derivative of $\cos(x)+2x$ as being $-\sin(x)+2$, which evaluated at $x=\frac{\pi}{2}$ yields $-1+2=1$. Since we also compute $\cos(\pi/2)+2(\pi/2)=0+\pi$, then the equation of the tangent line is

$$\begin{aligned} y - \pi = 1\cdot (x-\pi/2). \end{aligned}$$

We compute the derivative of $\tan(x)$ as being $\sec^2(x)$, which evaluated at $x=\frac{\pi}{4}$ yields $2$. Since we also compute $\tan(\pi/4)=1$, then the equation of the tangent line is

$$\begin{aligned} y - 1 = 2\cdot (x-\pi/4). \end{aligned}$$

We compute the derivative $y' = \cos(x)-\sin(x)+e^x$, which evaluated at $x=0$ yields $1-0+1 = 2$. Since we also compute $y(0)=0+1+1=2$, the equation of the tangent line is

$$\begin{aligned} y - 2 = 2(x-0) \end{aligned}$$

ie $y=2x+2$.

At $(0,4)$ and $(0,-4)$, the curve looks to be horizontal, if you zoom in: a tangent line here would have derivative zero. At the origin, the curve looks like its tangent line is vertical, so $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ does not exist.

(a) No. A function must pass the vertical line test: one input cannot result in two (or more) outputs. Since one value of $x$ sometimes corresponds to two values of $y$ (for example, when $x=\pi/4$, $y$ is $\pm 1/\sqrt{2}$), there is no function $f(x)$ so that $y=f(x)$ captures every point on the circle.

Remark: $y=\pm\sqrt{1-x^2}$ does capture every point on the unit circle. However, since one input $x$ sometimes results in two outputs $y$, this expression is not a function.

(b) No, for the same reasons as (a). If $f'(x)$ is a function, then it can give at most one slope corresponding to one value of $x$. Since one value of $x$ can correspond to two points on the circle with different slopes, $f'(x)$ cannot give the slope of every point on the circle. For example, fix any $0<a<1$. There are two points on the circle with $x$-coordinate equal to $a$. At the upper one, the slope is strictly negative. At the lower one, the slope is strictly positive.

(c) We differentiate:

$$\begin{aligned} 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \end{aligned}$$

But there is a $y$ in the right-hand side of this equation, and it's not clear how to get it out. Our answer in (b) tells us that, actually, we can't get it out, if we want the right-hand side to be a function of $x$. The derivative cannot be expressed as a function of $x$, because one value of $x$ corresponds to multiple points on the circle.

Remark: since $y=\pm\sqrt{1-x^2}$, we could try writing $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}=\pm\frac{x}{\sqrt{1-x^2}}$$ but this is not a function of $x$. Again, in a function, one input leads to at most one output, but here one value of $x$ will usually lead to two values of $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Since $f'(x)<0$, we need a decreasing function. This only applies to (ii), (iii), and (v). Since $f''(x)>0$, that means $f'(x)$ is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so $f''(x)=0$–therefore, it's not (v). In (iii), the slope is decreasing, because near $a$ the curve is quite flat ($f'(x)$ near zero) but near $b$ the curve is very steeply decreasing ($f'(x)$ is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing ($f'(x)$ is a strongly negative number) and becomes flatter and flatter ($f'(x)$ nears 0), so $f'(x)$ is increasing–in other words, $f''(x)>0$. So, (ii) is the only curve that has $f'(x)<0$ and $f''(x)>0$.

As is so often the case, we use a right triangle in this problem to relate the quantities.

$$\begin{aligned} \sin\theta&=\frac{D}{2}\\ D&=2\sin\theta\\ \text{Using the chain rule, we differentiate both sides with respect to time, $t$.} \\ \frac{\mathrm{d}D}{\mathrm{d}t}&=2\cos\theta\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t} \\ \text{So, if $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=0.25$ radians per hour and $\theta = \dfrac{\pi}{4}$ radians, then} \\ (a)\qquad\frac{\mathrm{d}D}{\mathrm{d}t}&=2\cos\left(\frac{\pi}{4}\right)\cdot 0.25=2\left(\frac{1}{\sqrt{2}}\right)\frac{1}{4}=\frac{1}{2\sqrt{2}}~\text{metres per hour}. \end{aligned}$$

Setting aside part (b) for a moment, let's think about (c). If $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}$ have different signs, then because $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}=2\cos\theta\cdot\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$, that means $\cos\theta<0$. We have to have a nonnegative depth, so $D>0$ and $D=2\sin\theta$ implies $\sin \theta >0$. If $\sin\theta\ge0$ and $\cos \theta<0$, then $\theta\in(\pi/2,\pi]$. On the diagram, that looks like this:

That is: the water has reversed direction. This happens, for instance, when a river empties into the ocean and the tide is high. Skookumchuck Narrows provincial park, in the Sunshine Coast, has reversing rapids.

Now, let's return to (b). If the rope is only 2 metres long, and the river rises higher than 2 metres, then our equation $D=2\sin\theta$ doesn't work any more: the buoy might be stationary underwater while the water rises or falls (but stays at or above $2$ metres deep).

(a) When the point is at $(0,-2)$, its $y$-coordinate is not changing, because it is moving along a horizontal line. So, the rate at which the particle moves is simply $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$. Let $\theta$ be the angle an observer would be looking at, in order to watch the point. Since we know $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$, a reasonable plan is to find an equation relating $\theta$ and $x$, and then differentiate implicitly with respect to $t$. To do this, let's return to our diagram.

When the point is a little to the right of $(0,-2)$, then we can make a triangle with the origin, as shown. If we let $\theta$ be the indicated angle, then $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=1$ radian per second. (It is given that the observer is turning one radian per second, so this is how fast $\theta$ is increasing.) From the right triangle in the diagram, we see $$\tan\theta=\frac{x}{2}$$ Now, we have to take care of a subtle point. The diagram we drew only makes sense for the point when it is at a position a little to the right of $(0,-2)$. So, right now, we've only made a set-up that will find the derivative from the right. But, with a little more thought, we see that even when $x$ is negative (that is, when the point is a little to the left of $(0,-2)$), our equation holds if we are careful about how we define $\theta$. Let $\theta$ be the angle between the line connecting the point and the origin, and the $y$-axis, where $\theta$ is negative when the point is to the left of the $y$-axis.

Since $x$ and $\theta$ are both negative when the point is to the left of the $y$-axis,

$$\begin{aligned} \tan|\theta|&=\frac{2}{|x|}\\ \tan(-\theta) &= \frac{-x}{2} \\ \text{So, since $\tan(-\theta)=-\tan(\theta)$:} \\ \tan\theta&=\frac{x}{2} \end{aligned}$$

So, we've shown that the relationship $\tan\theta=\dfrac{x}{2}$ holds when our point is at $(x,-2)$, regardless of the sign of $x$.

Moving on, since we are given $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and asked for $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$, we differentiate with respect to $t$:

$$\begin{aligned} \sec^2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t}&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t} \\ \text{When the point is at $(0,-2)$, since the observer is turning at one radian per second, also $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=1$. Also, looking at the diagram, $\theta=0$. Plugging in these values:} \\ \sec^2\left(0\right)\cdot(1)&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\\ 1&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\\ \frac{\mathrm{d}x}{\mathrm{d}t}&=2 \end{aligned}$$

So, the particle is moving at 2 units per second.

(b) When the point is at $(0,2)$, it is moving along a line with slope $-\frac{1}{2}$ and $y$-intercept $2$. So, it is on the line $$y=2-\frac{1}{2}x$$ That is, at time $t$, if the point is at $(x(t),y(t))$, then $x(t)$ and $y(t)$ satisfy $y(t)=2-\frac{1}{2}x(t)$. Implicitly differentiating with respect to $t$: $$\frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}$$ So, when $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=1$, $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=-\dfrac{1}{2}$. That is, its $y$-coordinate is decreasing at $\dfrac{1}{2}$ unit per second.

For the question “How fast is the point moving?", remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions (an $x$ coordinate and a $y$ coordinate), not one.

Remark: the solution below is actually pretty complicated. It is within your abilities to figure it out, but later on in your mathematical career you will learn an easier way, using vectors. For now, take this as a relatively tough exercise, and a motivation to keep learning: your intuition that there must be an easier way is well founded!

The point is moving along a straight line. So, to take care of complication (2), we can give its position as a point on the line. We can take the line as a sort of axis. We'll need to choose a point on the axis to be the “origin": $(2,1)$ is a convenient point. Let $D$ be the point's (signed) distance along the “axis" from $(2,1)$. When the point is a distance of one unit to the left of $(2,1)$, we'll have $D=-1$, and when the point is a distance of one unit to the right of $(2,1)$, we'll have $D=1$. Then $D$ changes with respect to time, and $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}$ is the velocity of the point. Since we know $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}$, a reasonable plan is to find an equation relating $x$, $y$, and $D$, and differentiate implicitly with respect to $t$. (This implicit differentiation takes care of complication (1).) Using the Pythagorean Theorem:

$$\begin{aligned} D^2&=(x-2)^2+(y-1)^2 \\ \text{Differentiating with respect to $t$:} \\ 2D\cdot\frac{\mathrm{d}D}{\mathrm{d}t}&=2(x-2)\cdot\frac{\mathrm{d}x}{\mathrm{d}t}+2(y-1)\cdot\frac{\mathrm{d}y}{\mathrm{d}t} \\ \text{We plug in $x=0$, $y=2$, $\frac{\mathrm{d}x}{\mathrm{d}t}=1$, $\frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{2}$, and $D=-\sqrt{(0-2)^2+(2-1)^2}=-\sqrt{5}$ (negative because the point is to the left of $(2,1)$):} \\ -2\sqrt{5}\cdot\frac{\mathrm{d}D}{\mathrm{d}t}&=2(-2)(1)+2(1)\left(-\frac{1}{2}\right)\\ \frac{\mathrm{d}D}{\mathrm{d}t}&=\frac{\sqrt{5}}{2}~\text{units per second} \end{aligned}$$

In the graph above, the concave-up sections are marked in red. These are where the graph has an increasing derivative; equivalently, where the graph lies above its tangent lines; more descriptively, where it curves like a smiley face.

Concave-down sections are marked in blue. These are where the graph has a decreasing derivative; equivalently, where the graph lies below its tangent lines; more descriptively, where it curves like a frowney face.

The most basic shape of the graph is given by the last two bullet points:

The curve is concave down over the interval $(-5,5)$, so let's give it a frowney-face curvature there.

Finally, when $x>5$ or $x<-5$, our curve should be concave up, so let's give it smiley-face curvature there, without changing its basic increasing/decreasing shape.

This finishes our sketch.