Chapter II

CHAPTER II

RATES LIMITS DERIVATIVES

4. Rate of Increase.Slope. In the study of any quantity,its rate of increase (or decrease), when some related quantitychanges, is very important for any complete understanding.Thus, the rate of increase of the speed of a boat when thepower applied is increased is a fundamental consideration.Graphically, the rate of increase of $y$ with respect to $x$ isshown by the rate of increase of the heightof a curve. If the curve is very flat, thereis a small rate of increase; if steep, a largerate.

The steepness, or slope, of a curve showsthe rate at which the dependent variable isincreasing with respect to the independentvariable.

When we speak of the slope of a curveat any point $P$ we mean the slope of its tangentat that point. To find this, we muststart, as in Analytic Geometry, with a secantthrough $P$.

Fig.3

Let the equation of the curve, Fig. 3, be$y=x^2$, and let the point $P$ at which the slope is to be found, bethe point (2, 4).

Let $Q$ be any other point on the curve, and let $\mathrm{\Delta }x$ representthe difference of the values of $x$ at the two points $P$ and Q.

Then, in the figure, $OA=2,AB=\mathrm{\Delta }x,$ and $OB=2+\mathrm{\Delta }x$.Moreover, since $y=x^2$ at every point, the value of $y$ at $Q$ is$BQ={(2+\mathrm{\Delta }x)}^2$.

The slope $S$ of the secant $PQ$ is the quotient of thedifferences $\mathrm{\Delta }y$ and $\mathrm{\Delta }x$:

$$S=\tan \mathrm{\angle }MPQ=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{MQ}{PM}=\frac{{(2+\mathrm{\Delta }x)}^2-4}{\mathrm{\Delta }x}=4+\mathrm{\Delta }x.$$

The slope $m$ of the tangent at $P$, that is $\tan \mathrm{\angle }MPT$, is thelimit of the slope of the secant as $Q$ approaches $P$.The slope of the secant is the average slope of the curve between the points$P$ and Q. The slope of the curve at the single point $P$ is the limit of thisaverage slope as $\mathrm{Q}$ approaches $P$.

But, since $S=4+\mathrm{\Delta }x$, it is clear that the limit of $S$ as $Q$approaches $P$ is 4, since $\mathrm{\Delta }x$ approaches zero when $Q$ approaches$P$; hence the slope $m$ of the curve is 4 at the point $P$.

At any other point the argument would be similar. If theco"ordinates of $P$ are $(a,a^2)$, those of $Q$ would be$[(a+\mathrm{\Delta }x),{(a+\mathrm{\Delta }x)}^2]$; and the slope of the secant would be the differencequotient $\mathrm{\Delta }y÷\mathrm{\Delta }x$:

$$S=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{{(a+\mathrm{\Delta }x)}^2-a^2}{\mathrm{\Delta }x}=\frac{2a\mathrm{\Delta }x+\overline{\mathrm{\Delta }}{x}^2}{\mathrm{\Delta }x}=2a+\mathrm{\Delta }x.$$

Hence the slope of the curve at the point $(a,a^s)$ is

$$m=\underset{\mathrm{\Delta }x\doteq 0}{lim}S=\underset{\mathrm{\Delta }x\doteq 0}{lim}\mathrm{\Delta }y/\mathrm{\Delta }x=\underset{\mathrm{\Delta }x\doteq 0}{lim}(2a+\mathrm{\Delta }x)=2a.$$

On the curve $y=x^2$, the slope at any point is numerically twicethe value of $x$.

When the slope can be found, as above, the equation of thetangent at $P$ can be written down at once, by Analytic Geometry,since the slope $m$ and a point $(a,b)$ on a line determineits equation:

$$(y-b)=m(x-a).$$

Hence, in the preceding example, at the point (2, 4), wherewe found $m=4$, the equation of the tangent is

$(y-4)=4(x-2)$, or $4x-y=4$.

At the point $(a,a^2)$ on the curve $y=x^2$, we found $m=2a$;hence the equation of the tangent there is

$(y-a^2)=2a(x-a)$, or 2 $ax-y=a^2$.

5. General Rules.

A part of the preceding work holds truefor any curve, and all of the work is at least similar. Thus,for any curve, the slope is

$$m=\underset{\mathrm{\Delta }x\doteq 0}{lim}S=\underset{\mathrm{\Delta }x\doteq 0}{lim}(\mathrm{\Delta }y/\mathrm{\Delta }x);$$

that is, the slope $m$ of the curve is the limit of thedifference quotient $\mathrm{\Delta }y/\mathrm{\Delta }x$.

The changes in various examples arise in the calculation ofthe difference quotient, $\mathrm{\Delta }y+\mathrm{\Delta }x$, or $S$.

This difference quotient. is always obtained, as above, by findingthe value of $y$ at $Q$, from the value of $x$ at $Q$,from the equation of the curve, then finding $\mathrm{\Delta }y$by subtracting from this the value of $y$ at $P$,and finally forming the difference quotient by dividing $\mathrm{\Delta }y$by $\mathrm{\Delta }x$.

$. Slope Negative or Zero. If the slope of the curve isnegative, the rate of increase in its height is negative, that is,the height is really decreasing with respect to the independentvariable.

If the slope is zero, the tangent to the curve is horizontal.this is what happens ordinarily at a highest point (maximum)or at a lowest point (minimum) on a curve.

Example 1. Thus the curve $y=x^2$, as we have just seen, has, at anypoint $x=a$, a slope $m=2a$. Since $m1s$ positive when $a$ is positive, thecurve is rising on the right of the origin; since $m$ is negative when $a$ isnegative, the curve is falling (that is, thee height $y$ decreases as $x$ increases)on the left of the origin. At the origin $m=0$; the origin is the lowestpoint (a minimum) on the curve, because the curve falls as we come towardthe origin and rises afterwards.

Example 2. Find the slope of the curve

$$y=x^2+3x-6$$(1)

at the point where $x=-2$; also in general at a point $x=a$. Use thesevalues to find the equationof the tangent at $x=2$; thetangent at any point; themaximum or minimumpoints if any exist.

When $x=-2$, we find $y=-7$, ( $P$ in Fig. 4); takingany second point $Q$, $(-2+\mathrm{\Delta }x,-7+\mathrm{\Delta }y)$, itsco"ordinates must satisfy the given equation, therefore

(2) $-7+\mathrm{\Delta }y={(-2+\mathrm{\Delta }x)}^2+3(-2+\mathrm{\Delta }x)-6$,

or

(3) $\mathrm{\Delta }y=-4+{\overline{\mathrm{\Delta }x}}^2+\mathrm{\Delta }x=-\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2$,

where ${\overline{\mathrm{\Delta }}}^{\eta }x$ means the square

of $\mathrm{\Delta }x$. Hence the slope ofthe secant $PQ$ i8

(4) $S=\mathrm{\Delta }y/\mathrm{\Delta }x=-1+\mathrm{\Delta }x$.

The slope $m$ of the curve is the limit of $S$ as $\mathrm{\Delta }x$ approaches zero; i.e.

(5) $m=\underset{\mathrm{\Delta }x\doteq 0}{lim}S=\underset{\mathrm{\Delta }x\doteq 0}{lim}{\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\doteq 0}{lim}(-1+\mathrm{\Delta }x)=-1$.

$$u\underset{¯}{\wedge }0\mathrm{A}ae\pm 0M\mathrm{A}a=0$$

It follows that the equation of the tangent at $(-2,-7)$ is

(6) $(y+7)=-1(x+2)$, or $x+y+9=0$.

Likewise, if we take the point $P(a,b)$ in any position on the curvewhatsoever, the equation (1) gives

(7) $b=a^2+3a-6$.

Any second point $Q$ has coordinates $(a+\mathrm{\Delta }x,b+\mathrm{\Delta }y)$ where $\mathrm{\Delta }x$ and $\mathrm{\Delta }y$ are the differences in $x$ and in $y$, respectively,between $P$ and $Q$. Since $Q$ also lies on the curve, thesecoordinates satisfy (1) :

(8) $b+\mathrm{\Delta }y={(a+\mathrm{\Delta }x)}^2+3(a+\mathrm{\Delta }x)-5$.

Subtracting the equation (7) from (8),$\mathrm{\Delta }y=2a\mathrm{\Delta }x+\overline{\mathrm{\Delta }}{x}^2+3\mathrm{\Delta }x$, whence $S=\mathrm{\Delta }y/\mathrm{\Delta }x=(2a+3)+\mathrm{\Delta }x$,and

(9) $m=\underset{x\doteq 0}{lim}S=\underset{x\doteq 0}{lim}{\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}=\underset{x\doteq 0}{lim}[(2a+3)+\mathrm{\Delta }x]=2a+3$.

Therefore the tangent at $(a,b)$ is

(10) $y-(a^2+3a-6)=(2a+3)(x-a)$, or $(2a+3)x-y=a^2+5$.

From (9) we observe that $m=0$, when $2a+3=0$, i.e.. when $a=-3/2$.For all values greater than $-3/2,m=(2a+3$) is positive; for allvalues less than $-3/2,m$ is negative. Hence the curve has a minimumat $(-3/2,-29/4)$ In Fig. 4, since the curve falls as we come towardthis point and rises afterwards.

Example 3. Consider the curve $y=x^2-12x+7.$If the value of $x$at any point $P$ is $a$, the value of $y$ is $a^2-12a+7$. If the value of $x$ at$Q$ is $a+\mathrm{\Delta }x$, the value of $y$ at $Q$ is ${(a+\mathrm{\Delta }x)}^2-12(a+\mathrm{\Delta }x)+7$.

Hence

$$S=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{[{(a+\mathrm{\Delta }x)}^2-12(a+\mathrm{\Delta }x)+7]-[a^2-12a+7]}{\mathrm{\Delta }x}$$

$$=(3a^2+3a\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2)-12,$$

and

$$m=\underset{\mathrm{\Delta }\doteq 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=3a^2-12.$$

For example, if $x=1,y=-4$; at this point $(1,-4)$the slope is S. $1^2-12=-9$:and the equation of the tangent is$(y+4)=-9(x-1)$, or

$9x+y-5=0$.

Since 3 $a^2-12$ is negativewhen $a^2\mathrm{\&}lt;4$, the curve is falling when $a$ lies between $-2$and $+2$. Since $3a^2-12$ ispositive when $a^2\mathrm{\&}gt;4$, the curve is rising whien $x\mathrm{\&}lt;-2$ and when $x\mathrm{\&}gt;+2$.At $x=\pm 2$, the slope is zero. At $x=+2$ there is a minimum (see Fig.5 $)$, since the curve i8 falling before this point and rising afterwards. At$x=-2$ there is a maximum. At $x=+2,y={(2)}^2-12\cdot 2+7=-9$,which is the lowest value of $y$ near that point. At $x=-2,y=23$, thehighest value near it.

This information is quite useful in drawing an accurate figure. Weknow also that the curve rises faster and faster to the right of $x=2$.

Draw an accurate figure of your own on a large scale.

EXERCISES III.–SLOPES OF CURVES

1. Find the slope of the curve $y=x^2+2$ at the point where $x=1$.Find the equation of the tangent at that point. Verify the fact that theequation obtained i8 a straight line, that it has the correct slope, and thatit passes through the point $(1,3)$.

2. Draw the curve $y=x^2+2$ on a large scale. Through the point$(1,3)$ draw secants which make $\mathrm{\Delta }x=1,3,0.1,0.01$, respectively.Calculate the slope of each of these secants and show that the values areapproaching the value of the slope of the curve at (1, 3).

3. Find the slope of the curve and the equation of the tangent toeach of the following curves at the point mentioned. Verify each answeras in Ex. 1.

(a) $y=3x^2;(1,3)$. (d) $y=x^2+4x-5;(1,0)$.

(b)$y=2x^2-5;(2,3)$(̇e) $y=x^3+x^2;(1,2)$.

(c) $y=x^3;(1,1)$. (f) $y=x^3-3x+4;(2,6)$.

4. Find the slope of the curve $y=x^2-3x+1$ at any point $x=a$;from this find the highest (maximum) or lowest (minimum) point (ifany), and show in what portions the curve is rising or falling.

5. Draw the following curves, using for greater accuracy the precisevalues of $x$ and $y$ at the highest (maximum) and the lowest (minimum)points, and the knowledge of the values of $x$ for which the curve rises orfalls. The slope of the curve at the point where $x=0$ is also useful in$(b),(c),(e),(g)$.

(a) $y=x^2+5x+2$. (d) $y=x^4$. (g) $y=2x^3-8x$.

(b) $y=x^3$. (e)$y=-x^2+3x$. (h) $y=x^3-6x+5$.

(c) $y=x^3-3x+4$. (f) $y=3+12x-x^3$. (i) $y=x^3+x^2$.

6. Show that the slope of the graph of $y=ax+b$ is always $m=a$,(1) geometrically,(2) by the methods of §6.

7. Show that the lowest point on $y=x^2+px+q$ is the point where$x=-p/2,$ (1) by Analytic Geometry, (2) by the methods of §6.

8. The normal to a curve at a point ls defined in Analytic Geometryto be the perpendicular to the tangent at that point. Its slope $n$ is shownto be the negative reciprocal of the slope $m$ of the tangent: $n=-1/m$.Find the slope of the normal, and the equation of the normal in Ex. 1;in each of the equations under Ex. 8.

9. The slope $m$ of the curve $y=x^2$ at any point where $x=a$ is$m=2a$. Show that the slope is $+1$ at the point where $a=1/2$. Find thepoints where the slope has the value $-1,2,10$. Note that if the curveis drawn by taking different scales on the two axes, the slope no longermeans the tangent of the angle made with the horizontal axis.

10. Find the points on the following curves where the slope has thevalues assigned to it;

(a) $y=x^2-3x+6;(m=1,-1,2)$.

(b) $y=x^3;(m=0.+1,+6)$.

(c) $y=x^3-3x+4;(m=9,1)$.

11. Show that the curve $y=x^3-0.03x+2$ has a minimum at$(0.1,1.998)$ and a maximum at $(-0.1,2.002)$. Draw the curve near the point$(0,2)$ on a very large scale.

12. Draw each of the following curves on an appropriate scale; ineach case show that the peculiar twist of the curve through its maximumand minimum would have been overlooked in ordinary plotting bypoints:

(a) $y=48x^3-x+1$.

[HINT. Use a very small vertical scale and a rather large horizontalscale. The slope at $x=0$ is also useful.]

(b) $y=x^3-30x^2+297x$.

[HINT. Use an exceedingly small vertical scale and a moderatehorizontal scale. The slope at $x=10$ is also useful.]

7. Speed.

An important case of rate of change of a quantity is therate at which a body moves,–its speed.

Consider the motion of a body falling from rest under theinfluence of gravity. During the first second it passes over16 ft., during the next it passes over 48 ft., during the thirdover 80 ft. In general, if $t$ is the number of seconds, and $s$the entire distance it has fallen, $s=16t^2$ if the gravitationalconstant $g$ be taken as 32. The graph of this equation (seeFig. 6) is a parabola with its vertex at the origin.

The speed, that is the rate of increase of the space passedover, is the slope of this curve, i.e.

$$\underset{\mathrm{\Delta }t\doteq 0}{lim}\mathrm{\Delta }s/\mathrm{\Delta }t.$$

This may be seen directly in another way. The averagespeed for an interval of time $\mathrm{\Delta }t$ is found by dividing thedifference between the space passed over at the beginning and atthe end of that interval of time by the difference in time: i.e.the average speed is the difference quotient $\mathrm{\Delta }s÷\mathrm{\Delta }t$. By thespeed at a given instant we mean the limit of the average speedover an interval $\mathrm{\Delta }t$ beginning or ending at that instant as thatinterval approaches zero, i.e.

$$\mathrm{speed}=\underset{\mathrm{\Delta }t\doteq 0}{lim}\mathrm{\Delta }s/\mathrm{\Delta }t.$$

Taking the equation $s=16t^2$, if $t=1/2$, $s=4$.(See point P in Fig. 6). After a lapse of time $\mathrm{\Delta }t$, thenew values are $t=1/2+\mathrm{\Delta }t$, and $s=16{(t=1/2+\mathrm{\Delta }t)}^2$(Q in Fig. 6)

Then

$$\mathrm{\Delta }s=16{(t=1/2+\mathrm{\Delta }t)}^2-4=16\mathrm{\Delta }t+16{\overline{\mathrm{\Delta }t}}^2,$$

$$\frac{\mathrm{\Delta }t}{\mathrm{\Delta }s}=16+16\mathrm{\Delta }t.$$

Whence

$$\mathrm{speed}=\underset{\mathrm{\Delta }t\doteq 0}{lim}\frac{\mathrm{\Delta }t}{\mathrm{\Delta }s}=\underset{\mathrm{\Delta }t\doteq 0}{lim}(16+16\mathrm{\Delta }t)=16;$$

that is, the speed at the end of the first half second is 16 ft. per second.

Likewise, for any value of $t$, say $t=T$, $s=16T^2$, while for$t=T+\mathrm{\Delta }t$, $s=16{(T+\mathrm{\Delta }t)}^2$; hence

$$\mathrm{average}\mathrm{speed}=\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}=\frac{16{(T+\mathrm{\Delta }t)}^2-16T^l}{\mathrm{\Delta }t}=32T+16\mathrm{\Delta }t$$

and

$$\mathrm{speed}=\underset{\mathrm{\Delta }t\doteq 0}{lim}\frac{\mathrm{\Delta }t}{\mathrm{\Delta }s}=32T.$$

Thus, at the end of two seconds, $T=2$, and the speed is $32\cdot 2=64$,in feet per second.

8. Component Speeds. Any curve may be regarded as thepath of a moving point. If a point $P$ does move along a curve,both $x$ and $y$ are fixed when the time $t$ is fixed. To specifythe motion completely, we need equations which give the valuesof $x$ and $y$ in terms of $t$.

The horizontal speed is the rate of increase of $x$ withrespect to the time. This may be thought of as the speed ofthe projection $M$ of $P$ on the $x$-axis. As shown in §7,this speed is the limit of the difference quotient$\mathrm{\Delta }x÷\mathrm{\Delta }t$ as $\mathrm{\Delta }t\doteq 0$.

Likewise, the vertical speed is the limit of the difference quotient$\mathrm{\Delta }y÷\mathrm{\Delta }t$ as $\mathrm{\Delta }t\doteq 0$. Since the slope$m$ of the curve is the limit of $\mathrm{\Delta }y÷\mathrm{\Delta }x$ as$\mathrm{\Delta }x\doteq 0$; and since

$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }t}÷\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t},$$

it follows that

$$m=(\mathrm{𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙}\mathrm{𝑠𝑝𝑒𝑒𝑑})÷(\mathrm{ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙}\mathrm{𝑠𝑝𝑒𝑒𝑑});$$

that is, the slope of the curve is the ratio of the rate of increaseof $y$ to the rate of increase of $x$.

9. Continuous Functions.In §§4-8, we have supposed thatthe curves used were smooth. The functions which we havebad have all been representable by smooth curves; exceptperhaps at isolated points, to a small change in the value ofone co"ordinate, there has been a correspondingly small changein the value of the other co"ordinate. Throughout this text,unless the contrary is expressly stated, the functions dealt withwill be of the same sort. Such functions are called continuous.(See §10, p. 17.)

The curve $y=1/x$ is continuous except at the point $x=0$; $y=\tan x$is continuous except at the points $x=\pm \pi /2,\pm 3\pi /2,$ etc.Such exceptional points occur frequently; we do not discard a curve because of them, but it is understood that any of our results may fail at such points.

EXERCISES IV.–SPEED

1. From the formula $s=16t^2$, calculate the values of $s$ when $t=1,2,1.1,1.01,1.001$. From these values calculate the average speed between$t=1$ and $t=2$; between $t=1$ and $t=1.1$ ; between $t=1$ and $t=1.01$;between $t=1$ and $t=1.001$. Show that these average speeds aresuccessively nearer to the speed at the instant $t=1$.

2. Calculate as in Ex. 1 the average speed for smaller and smallerintervals of time after $t=2$; and show that these approach the speed at theinstant $t=2$.

3. A body thrown vertically downwards from any height with anoriginal velocity of 100 ft. per second, passes over in time $t$(in seconds) a distance $s$ (in feet) given by the equation$s=100t+16t^2$ (if $g=32$, as in §7). Flnd the speed $v$ at thetime $t=1$; at the time $t=2$; at the time $t=4$; at the time $t=T$.

4. In Ex. $3$ calculate the average speeds for smaller and smallerintervals of time after $t=0$; and show that they approach the originalspeed $v_0=100$. Repeat the calculations for intervals beginning with $t=2$.

5. Calculate the speed of a body at the times indicated in thefollowing possible relations between $s$ and $t$:

(a)$s=t^2;t=1,2,10,T$. (c) $s=-16t^2+160t;t=0,2,5$.

(b) $s=16t^2-100t;t=0,2,T$. (d) $s=t^3-3t+4;t=0,1/2,1$.

6. The relation (c) in Ex. 5 holds (approximately, since $g=32$approximately) for a body thrown upward with an initial speed of 160 ft.per second, where $s$ means the distance from the starting point countedpositive upwards. Draw a graph which represents this relation betweenthe values of $s$ and $t$.

In this graph mark the greatest value of $s$. What is the value of $v$ atthat point? Find exact values of $s$ and $t$ for this point.

7. A body thrown horizontally with an original speed of 4 ft. persecond falls in a vertical plane curved path so that the values of itshorizontal and its vertical distances from its original position are respectively,$x=4t,y=16t^2$, where $y$ is measured downwards. Show that the verticalspeed is $32T$, and that the horizontal speed is 4, at the instant $t=T$.Eliminate $t$ to show that the path is the curve $y=x^2$.

8. Show by Ex. 7 and §8 that the slope of the curve $y=x^2$ at thepoint where $t=1$, i.e. $(4,16)$, is $32÷4$, or 8. Write theequation of the tangent at that point.

9. Show that the slope of the curve $y=x^2$ (Ex. 7) at the point$(a,a^2)$, i.e. $t=a/4$, is $2a$, from Ex. 7 and §8; and also directly by means of §6.

10. If a body moves so that its horizontal and its vertical distancesfrom the starting point are, respectively, $x=16t^2,y=4t$, show that itspath is the curve $y^2=x$; that its horizontal speed and its vertical speedare, respectively, $32T$ and 4, at the instant $t=T$.

11. From Ex. 10 and §8 show that the slope of the curve $y^2=x$ at thepoint $(16,4)$, i.e. when $t=1$, is $4÷32=1/8$. Write the equation of the tangent at that point.

12. From Ex. 10 and §8 show that the slope of the curve $y^2=x$ at thepoint where $t=T$ is $4÷(32T)=1/(8T)=1/(2k)$, where $k$ is the valueof $y$ at the point. Compare this result with that of Ex. 8.

10. Limits. Infinitesimals.

We have been led in what precedes to make use of limits.Thus the tangent to a curve atthe point $P$ is defined by saying that its slope is the limit ofthe slope of a variable secant through $P$; the speed at a giveninstant is the limit of the average speed; the difference of thetwo values of $x,\mathrm{\Delta }x$, was thought of as approaching zero; andso on. To make these concepts clear, the following precisestatements are necessary and desirable.

When the difference between the variable $x$ and a constant $a$ becomes and remains less, in absolute value, than any preassigned positive quantity, however small, then $a$ is the limit of the variable $x$.

We also use the expression “ $x$ approaches $a$ as a limit,” or,more simply, “ $x$ approaches $a.$” The symbol for limit is $lim$;the symbol for approaches is $\doteq$ thus we may write$limx=a$, or $x\doteq a$, or $lim(a-x)=0$, or $a-x\doteq 0$.

When the limit of a variable is zero, the variable is calledan infinitesimal. Thus $a-x$ above is an infinitesimal. Thedifference between any variable and its limit is always aninfinitesimal. When a variable $x$ approaches a limit $a$,any oontinuous function $f(x)$ approaches the limit $f(a)$: thus, if $y=f(x)$ and $b=f(a)$, we may write

$\underset{x\doteq a}{lim}y=b$, or$\underset{x\doteq a}{lim}f(x)=f(a)$.

This condition is the precise definition of continuity at thepoint $x=a$. (See §9, p. 14.)

11. Properties of Limits.The following properties of limitswill be assumed as self-evident; some of them have alreadybeen used in the articles noted below.

THEOREM A. The limit of the sum of two variables is the sumof the limits of the two variables. This is easily extended to thecase of more than two variables. (Used in §§4, 6, and 7.)

THEOREM B. The limit of the produot of two variables is theproduct of the limits of the variables. (Used in §§4, 6, and 7.)

THEOREM C. The limit of the quotient of one variable dividedby another is the quotient of the limits of the variables, providedthe limit of the divisor is not zero. (Used in §8.)

The exceptional case in Theorem $\mathrm{C}$ is really the mostinteresting and important case of all. The exception arisesbecause when zero occurs as a denominator, the divisioncannot be performed. In finding the slope of a curve, we consider$lim(\mathrm{\Delta }y/\mathrm{\Delta }x)$ as $\mathrm{\Delta }x$ approaches zero; notice that this is precisely the case ruled out in Theorem C. Again, the speed is$lim(\mathrm{\Delta }s/\mathrm{\Delta }t)$ as $\mathrm{\Delta }t$ approaches zero. The limit of any suchdifference quotient is one of these exceptional cases.

Now it is clear that the slope of a curve (or the speed of anobject) may have a great variety of values in different cases:no one answer is sufficient for all examples, in the case of thelimit of a quotient when the denominator approaches zero.

THEOREM D. The limit of the ratio of two infinitesimalspends upon the law connecting them; otherwise it is quite inderminate.Of this the student will see many instances; forthee Differential Calculus consists of the consideration of justsuch limits. In fact, the very reason for the existence of theDifferential Calculus is that the exceptional case of Theorem$\mathrm{C}$ is important, and cannot be settled in an offhand manner.

The thing to be noted here is, that, no matter how small twoquantities may be, their ratio may be either small or large;and that, if the two quantities are variables whose limit iszero, the limit of their ratio may be either finite, zero, ornon-existent. In our work with such forms we shall try tosubstitute an equivalent form whose limit can be found.Obviously, to say that two variables are vanishing impliesnothing about the limit of their ratio.

12. Ratio of an Arc to its Chord.

Another important illustration of a ratio of infinitesimals isthe ratio of the chord ofa curve to its subtended arc:

$$R=\frac{\mathrm{chord}PQ}{\mathrm{arc}PQ}.$$

If $Q$ approaches $P$, both the arcand the chord approach zero. Atany stage of the process the arc isgreater than the chord; but as $Q$approaches $P$ this differencediminishes very rapidly, and theratio $R$ approaches 1:

$$\underset{Q\doteq P}{lim}R=\underset{PQ\doteq 0}{lim}\frac{\mathrm{chord}PQ}{\mathrm{arc}PQ}=1.$$

This property is self-evident because it amounts to the samething as the definition of the length of the curve; we ordinarilythink of the length of an are of a curve as the limit of thelength of an inscribed broken line, as the lengths of thesegments of the broken line approach zero. Thus, the lengthof circumference of a circle is defined to be the limit of theperimeter of an inscribed polygonas the lengths of all its sidesapproach zero. This would not betrue if the ratio of an arc to itschord did not approach $1$.

la. Ratio of the Sine of anAngle to the Angle. In a circle,the arc $PQ$ and the chord $PQ$ canbe expressed in terms of the angleat the center. Let $\alpha =\mathrm{\angle }QOP/2$;then $\mathrm{arc}PQ=2\alpha \times r$ if $\alpha$is measured in circular measure (seeTables, II, F, 3); and the chord $PQ=2r\sin a$,since $r\sin \alpha =AP$.

It follows that

$$\underset{\alpha \doteq 0}{lim}\frac{\mathrm{chord}PQ}{\mathrm{arc}PQ}=\underset{\alpha \doteq 0}{lim}\frac{2r\sin a}{2r\alpha }=\underset{\alpha \doteq 0}{lim}\frac{\sin \alpha }{\alpha }=1,$$

hence $\underset{\alpha \doteq 0}{lim}{\displaystyle \frac{\sin \alpha }{\alpha }}=1$,for we have just seen that the limit of the ratio of an infinitesimalchord to its arc is 1.

This result is very important in later work; just here itserves as a new illustration of the ratio of infinitesimals:the ratio of the sine of an angle to the angle itself(measured in circular measure) approaches 1 as the angleapproaches zero.

14. Infinity.Theorem D accounts for the case when thenumerator as well as the denominator in Theorem C isinfinitesimal. There remains the case when the denominator onlyis infinitesimal. A variable whose reciprocal is infinitesimal issaid to become infinite as the reciprocal approaches zero.

Thus $y=1/x$ is a variable whose reciprocal is $x$. As $x$approaches zero, $y$ is said to become infinite. Notice howeverthat $y$ has no value whatever when $x=0$.Likewise $y=\sec x$ is a variable whose reciprocal, $\cos x$, isinfinitesimal as $x$ approaches $\pi /2$; hence we say that $\sec x$becomes infinite as $x$ approaches $\pi /2$.

In any case, it is clear that a variable which becomes infinitebecomes and remains larger in absolute value than anypreassigned positive number, however large.

The student should carefully notice that infinity is not anumber; when we say that ’‘ $\sec x$ becomes infinite as $x$approaches $\pi /2$,we do not mean that $\sec (\pi /2)$ has a value, wemerely tell what occurs when $x$ approaches $\pi /2$.

EXERCISES V.–LIMITS AND INPINITESSIMALS

1. Imagine a point traversing a line-segment in such fashion that ittraverses half the segment in the first second, half the remainder in thenext second, and so on; always half the remainder in the next followingsecond. Will it ever traverse the entire line ? Show that the remainderafter $t$ seconds is $1/2^t$, if the total length of the segment is 1.Is this infinitessimal? Why?

2. Show that the distance traversed by the point in Ex. 1 in $t$ secondsis $1/2+1/2^n+\mathrm{\cdots }+1/2^t$. Show that this sum is equal to $1-1/2^t$; hence show that its limit is 1. Show that in any case the limit of the distancetraversed ls the total distance, as $t$ increases indefinitely.

3. Show that the limit of $3-x^2$ as $x$ approaches zero is 3. State thisresult in the symbols used in §10. Draw the graph of $y=3-x^2$ andshow that $y$ approaches 3 as $x$ approaches zero.

4. Evaluate the following limits:

(a)$\underset{x\doteq 0}{lim}(2-6x+3x^2)$.(d)$\underset{x\doteq 1}{lim}{\displaystyle \frac{3-2x^2}{4+2x^2}}$.(g)$\underset{x\doteq 7}{lim}{\displaystyle \frac{x^2-3x+2}{x^2+2x+3}}$.

(b)$\underset{x\doteq 1}{lim}(2-6x+3x^2)$.(e)$\underset{x\doteq 0}{lim}{\displaystyle \frac{x_j}{1-x}}$.(h)$\underset{x\doteq 0}{lim}{\displaystyle \frac{a+bx}{c+dx}}$.

(c) $\underset{x\doteq k}{lim}(a+bx+c{x}^2)$.(f) $\underset{x\doteq 1}{lim}{\displaystyle \frac{1-x}{x}}$.(i) $\underset{x\doteq 0}{lim}{\displaystyle \frac{a+bx+c{x}^2}{m+nx+l{x}^2}}$.

5. If the numerator and denominator of a fraction contain a commonfactor, that factor may be canceled in finding a limit, since the value ofthe fraction which we use is not changed. Evaluate before and aftercanceling a common factor:

(a) $\underset{x\doteq 1}{lim}{\displaystyle \frac{(x+2)(x+1)}{(2x+3)(x+1)}}$.(b) $\underset{x\doteq 0}{lim}{\displaystyle \frac{x(x+2)}{(x+1)(x+2)}}$.

Evaluate after (not before) removing a common factor:

(c) $\underset{x\doteq 0}{lim}{\displaystyle \frac{x^2}{x}}$.(d) $\underset{x\doteq 1}{lim}{\displaystyle \frac{x^2-3x+2}{x^2-1}}$.(e) $\underset{x\doteq 1}{lim}{\displaystyle \frac{(x+2)(x-1)}{(2x+3)(x-1)}}$.

(f) $\underset{x\doteq 1}{lim}{\displaystyle \frac{\sqrt{x-1}}{x-1}}$(g) ${lim}_{x\doteq 0}\frac{x^2{(x+1)}^2}{x^3+2x^2}$(h) $\underset{x\doteq 0}{lim}{\displaystyle \frac{x^n}{x}}=\{\begin{array}{cc}0,n\hfill & \hfill gt;1\hfill \\ \mathrm{l},n=\mathrm{l}\hfill \end{array}$

6. Show that

$$\underset{x\doteq \mathrm{\infty }}{lim}\frac{2x^2+3}{x^2+4x+b}=2.$$

[HINT. Divide numerator and denominator by $x^2$; then suchterms as $3/x^2$ approach zero as $x$ becomes infinite.]

7. Evaluate:

(a) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{2x+1}{3x+2}}$.(b) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{2x^2-4}{3x^2+2}}$.(c) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{ax+b}{mx+n}}$.

(d) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{x}{\sqrt{1+x^2}}}$.(e) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{1+x^2}}{\sqrt{x^2-1}}}$ .(f) $\underset{x\doteq \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{a{x}^2+bx+c}}{mx+n}}$.

8. Let $0$ be the center of a circle of radius $r=OB$, and let$a=\mathrm{\angle }\mathrm{𝐶𝑂𝐵}$ be an angle at the center. Let $BT$ beperpendicular to $OB$, and let $BF$ be perpendicular to$OC$. Show that $OF$ approaches $OC$ as $\alpha$approaches zero; likewise arc $OB\doteq 0$, arc $DB\doteq 0$,and $FC\doteq 0$, as $\alpha \doteq 0$.

9. In the figure of Ex. 8, show that theobvious geometric inequality $FB\mathrm{\&}lt;\mathrm{arc}CB\mathrm{\&}lt;BT$is equivalent to $r\sin \alpha \mathrm{\&}lt;r\cdot \alpha \mathrm{\&}lt;r\tan \alpha$, if $\alpha$ is measured in circularmeasure. Hence show that $\alpha /\sin \alpha$ lies between 1 and$1/\cos \alpha$, and therefore that $lim(\alpha /\sin \alpha )=1$as $\alpha \doteq 0$. (Verification of. §l3.)

10. In the flgure of Ex. 8, show that

$lim\alpha \doteq 0{\displaystyle \frac{FB}{r}}=0;\underset{\alpha \doteq 0}{lim}{\displaystyle \frac{OF}{r}}=1;\underset{\alpha \doteq 0}{lim}{\displaystyle \frac{BT}{r}}=0;\underset{\alpha \doteq 0}{lim}{\displaystyle \frac{FC}{r}}=0;\underset{\alpha \doteq 0}{lim}{\displaystyle \frac{\mathrm{arc}CB}{r}}=0$.

11. Show that the following quantities become infinite as theindependent variable approaches the value specified: in $(a)$ and $(b)$ draw the graph.

(a) $\underset{x\doteq 0}{lim}{\displaystyle \frac{1}{x^2}}$.(b) $\underset{x\doteq 1}{lim}{\displaystyle \frac{x+2}{x-1}}$

(c) $\underset{x\doteq 0}{lim}{\displaystyle \frac{r}{FC}}$, (Ex. 8).(d) $\underset{x\doteq 0}{lim}{\displaystyle \frac{f}{BT}}$, (Ex. 8).

(e) $\underset{x\doteq 0}{lim}{\displaystyle \frac{x^n}{x}},(n\mathrm{\&}lt;1)$.(f) $\underset{x\doteq 2}{lim}{\displaystyle \frac{2x+3}{\sqrt{x^2-3x+2}}}$.

12. As the chord of a circle approaches zero, which of the followingratios has a finite limit, which is infinitesimal, and which is becominginfinite: the chord to its arc; the radius to the chord; the sector of thearc to the triangle cut off by the chord; the area of the circle to thesector; the chord of twice the arc to the chord of thrice the arc; theradius of the circle to the chord of an arc a thousand times the given arc ?

13. Is the sum of two infinitesimals itself infinitesimal ? Is thedifference ? Is the product Is the quotient ? Is a constant times aninfinitesimal an infinitesimal?

14. If to each of two integers an infinitesimal is added, show that theratio of these sums differs from that of the integers by an infinitesimal.[See Ex. 4 $(h).$]

15. Show that the graph of $y=f(x)$ has a vertical asymptote if $f(x)$becomes infinite as $x=a$. Illustrate this by drawing the following graphs:

(a) $y={\displaystyle \frac{3x+2}{x-2}}$.(c) $y={\displaystyle \frac{1}{1-\cos x}}$.(e) $y={\displaystyle \frac{1}{\sqrt{1-x^2}}}$.

(b) $y={\displaystyle \frac{2x-1}{(x+1)(x-b)}}$.(d) $y={\displaystyle \frac{e^x+e^{-x}}{e^x-e^{-x}}}$.(f) $y={\displaystyle \frac{ax+b}{cx+d}}$.

15. Derivatives. While such illustrations as those in §12and Exercises V, above, are interesting and reasonably important,by far the most important cases of the ratio of two infinitesimalsare those of the type studied in §§4-8, in which eachof the infinitesimals is the difference of two values of a variable,such as $\mathrm{\Delta }y/\mathrm{\Delta }x$ or $\mathrm{\Delta }s/\mathrm{\Delta }t$. Such a difference qquotient $\mathrm{\Delta }y/\mathrm{\Delta }x$ of $y$ with respect to $x$ evidentlyrepresents the averagerate of increase of $y$ with respect to $x$ in the interval $\mathrm{\Delta }x$;if $x$ represents time and $y$ distance, then $\mathrm{\Delta }y/\mathrm{\Delta }x$ is theaverage speed over the interval $\mathrm{\Delta }x$ (§7, p. 13);if $y=f(x)$ is thought of as a curve, then $\mathrm{\Delta }y/\mathrm{\Delta }x$ is theslope of a secant or the average rate of rise of the curve in theinterval $\mathrm{\Delta }x$ (§4, p. 6).

The limit obtained in such cases represents the instantaneousrate of increase of one variable with respect to the other, –this may be tho slope of a curve, or the speed of a movingobject, or some other rate, depending upon the nature of theproblem in which it arises.

In general, the limit of the quotient $\mathrm{\Delta }y/\mathrm{\Delta }x$of two infinitesimal differences is called the derivative of$y$ with respect to $x$; it is represented by the symbol $dy/dx$:

$$\frac{dy}{dx}\equiv \mathrm{derivative}\mathrm{of}\mathrm{y}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}=\underset{\mathrm{\Delta }x\doteq 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}.$$

Henceforth we shall use this new symbol $dy/dx$ or otherconvenient abbreviations;but the student must not forget thereal meaning: slope, in the case of curve;speed, in the case ofmotion; some other tangible concept in any new problemwhich we may undertake; in every case the rate of increase of$y$ with respect to $x$.

Any mathematical formulas we obtain will apply in any ofthese cases; we shall use the letters $x$ and $y$, the letters $s$ and$t$, and other suggestive combinations; but the student shouldremember that any formula written in $x$ and $y$ also holds true,for example, with the letters $s$ and $t$, or for any other pair ofletters.

16. Formula for Derivatives. If we are to find the value ofa derivative, as in §§4-7, we must have given one of thevariables $y$ as a function of the other $x$:

(1) $y=f(x)$.

If we think of (1) as a curve, we may, as in §4, take anypoint $P$ whose co"ordinates are $x$ and $y$, and join it by a secant$PQ$ to any other point $Q$, whose co"ordinates are$x+\mathrm{\Delta }x,y+\mathrm{\Delta }y$.Here $x$ aud $y$ represent fixed valuesof $x$ and $y$; this will prove moreconvenient than to use new letterseach time, as we did in’§§4-7.

Since $P$ lies on the curve (1), itsco"ordinates $(x,y)$ satisfy theequation (1), $y=f(x)$. Since $Q$ lies on(1), $x+\mathrm{\Delta }x$ and $y+\mathrm{\Delta }y$ satisfy thesame equation; hence we must have

(2) $y+\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)$.

Subtracting (1) from (2) we get

(3) $\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$;

whence the difference quotient is

(4) ${\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}={\displaystyle \frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}}=\mathrm{𝑎𝑣𝑒𝑟𝑎𝑔𝑒}\mathrm{𝑠𝑙𝑜𝑝𝑒}\mathrm{𝑜𝑣𝑒𝑟}\mathrm{𝑃𝑀}$,

and therefore the derivative is

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\doteq 0}{lim}\frac{\delta y}{\mathrm{\Delta }x}\equiv \underset{\mathrm{\Delta }x\doteq 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}=\mathrm{𝑠𝑙𝑜𝑝𝑒}\mathrm{𝑎𝑡}P$$

This formula is often convenient; we shall apply it at once.

17. Rule for Differentiation.

The process of finding a derivative is called differentiation.To apply formula (5) of §16:

(A) Find $(y+\mathrm{\Delta }y)$ by substituting $(x+\mathrm{\Delta }x)$for $x$ in the given function or equation; this gives$y+\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)$.

(B) Subtract $y$ from $y+\mathrm{\Delta }y$; this gives $\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$.

(C) Divide $\mathrm{\Delta }y$ by $\mathrm{\Delta }x$ to find the difference quotient $\mathrm{\Delta }|J/\mathrm{\Delta }x$; simplify this result.

(D) Find the limit of $\mathrm{\Delta }y/\mathrm{\Delta }x$ as$\mathrm{\Delta }x$ approaches zero; thisresult is the derivative, $dy/dx$.

Example 1. Given $y=f(x)\equiv x^2$, to flnd $dy/dx$.

$(A)f(x+\mathrm{\Delta }x)={(x+\mathrm{\Delta }x)}^2$.

$(B)\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)={(x+\mathrm{\Delta }x)}^2-x^2=2x\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2$.

$(C)\mathrm{\Delta }y/\mathrm{\Delta }x=(2x\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2)+\mathrm{\Delta }x=2x+\mathrm{\Delta }x$.

$(D)dy/dx=\underset{\mathrm{\Delta }x\doteq 0}{lim}\mathrm{\Delta }y/\mathrm{\Delta }x=\underset{\mathrm{\Delta }x\doteq 0}{lim}(2x+\mathrm{\Delta }x)=2x$.

Compare this work and the answer with the work of §4, p. 6.

Example 2. Given $y=f(x)\equiv \theta -12x+7$, to flnd $dy/k$.

$(A)f(x+\mathrm{\Delta }x)={(x+\mathrm{\Delta }x)}^3-12(x+\mathrm{\Delta }x)+7$.

$(B)\mathrm{\Delta }\nu =f(x+\mathrm{\Delta }x)-f(x)=3x^2\mathrm{\Delta }x+3x{\overline{\mathrm{\Delta }x}}^2+{\overline{\mathrm{\Delta }x}}^3-12\mathrm{\Delta }x$.

$(C)\mathrm{\Delta }y/\mathrm{\Delta }x=3x^2+3x\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2-12$.

$(D)dy/dx=\underset{\mathrm{\Delta }x\doteq 0}{lim}\mathrm{\Delta }y/\mathrm{\Delta }x=\underset{\mathrm{\Delta }x\doteq 0}{lim}(3x^2+3x\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2-12)=3x^2-12$

Compare thls work and the answer with the work of Example 3, §6.

Example 3. Given $y=f(x)\equiv 1/x^2$, to flnd $dy/dx$.

$(A)f(x+\mathrm{\Delta }x)=\frac{1}{{(x+\mathrm{\Delta }x)}^2}$ .

$(B)\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)={\displaystyle \frac{1}{{(x+\mathrm{\Delta }x)}^2}}-{\displaystyle \frac{1}{x^2}}=-{\displaystyle \frac{2x\mathrm{\Delta }x+{\overline{\mathrm{\Delta }x}}^2}{x^2{(x+\mathrm{\Delta }x)}^2}}\cdot$

$(C)\mathrm{\Delta }y/\mathrm{\Delta }x=-{\displaystyle \frac{2x+\mathrm{\Delta }x}{x^2{(x+\mathrm{\Delta }x)}^2}}\cdot$

$(D)dy/dx={lim}_{\mathrm{\Delta }x\doteq 0}{\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\doteq 0}{lim}[-{\displaystyle \frac{2x+\mathrm{\Delta }x}{x^2{(x+\mathrm{\Delta }x)}^2}}]=-{\displaystyle \frac{2x}{x^4}}=-{\displaystyle \frac{2}{x^3}}\cdot$

Example 4. Given $y=f(x)\equiv \sqrt{x}$, to flnd $dy/dx$, or $df(x)/d\alpha$

$(A)f(x+\mathrm{\Delta }x)=\sqrt{x+\mathrm{\Delta }x}$.

$(B)\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)=\sqrt{x+\mathrm{\Delta }x}-\sqrt{x}$.

(C) ${\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}={\displaystyle \frac{\sqrt{x+\mathrm{\Delta }x}-\sqrt{x}}{\mathrm{\Delta }x}}={\displaystyle \frac{\sqrt{x+\mathrm{\Delta }x}-\sqrt{x}}{\mathrm{\Delta }x}}\cdot {\displaystyle \frac{\sqrt{x+\mathrm{\Delta }x}+\sqrt{x}}{\sqrt{x+\mathrm{\Delta }x}+\sqrt{x}}}$

$$=\frac{1}{\sqrt{x+\mathrm{\Delta }x}+\sqrt{x}}$$

$(D){\displaystyle \frac{dy}{dx}}=\underset{\mathrm{\Delta }x\doteq 0}{lim}{\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\doteq 0}{lim}{\displaystyle \frac{1}{\sqrt{x+\mathrm{\Delta }x}+\sqrt{}\overline{x}}}={\displaystyle \frac{1}{2\sqrt{x}}}$.

(Compare Ex. 11, p. 10.)

Example 6. Given $y=f(x)\equiv x^7$, to flnd $df(x)/dx$.

$(A)f(x+\mathrm{\Delta }x)={(x+\mathrm{\Delta }x)}^7=x^7+7x^0\mathrm{\Delta }x+$( $\mathrm{terms}$ with a factor ${\overline{\mathrm{\Delta }x}}^2$).

$(B)\mathrm{\Delta }\nu =f(x+\mathrm{\Delta }x)-f(x)=7x^6\mathrm{\Delta }x+$( $\mathrm{terms}$ with a factor $\overline{\mathrm{\Delta }}{x}^2$).

$(C)\mathrm{\Delta }y/\mathrm{\Delta }x=7x^6+$( $\mathrm{t}ϵ\mathrm{ms}$ with a factor $\overline{\mathrm{\Delta }}{x}^2$).

$(D)dy/dx=\underset{\mathrm{\Delta }x\doteq 0}{lim}\mathrm{\Delta }y/\mathrm{\Delta }x=\underset{Deltax\doteq 0}{lim}$[ $7x^6+($ terms with a factor $\mathrm{\Delta }x)$] $=7x^6$.

EXERCISES VI.–FORMAL DIFFERENTIATION

1. Find the derivative of $y=x^3$ with respect to $x$. [Compare Ex. 8(c), p. 11.] Write the equation of the tangent at the point $(2,8)$ to thecurve $y=x^3$.

2. Find the derivatives of the following functions with respect to $x$:

(a)$x^2-3x+4$. (b) $x^3-6x+7$. (c) $x^4+5$.

(d) $x^4+3x^2-2$. (e) $x^3+2x^2-4$. (f) $x^4-3x^3+6x$.

(g) ${\displaystyle \frac{1}{x^2}}$. (h) ${\displaystyle \frac{1}{x+1}}$.(i) ${\displaystyle \frac{1}{2x-3}}$.

(j) $\sqrt{x+1}$. (k) ${\displaystyle \frac{x}{x+1}}$. (l) ${\displaystyle \frac{2x+3}{x-2}}$

3. Find the equation of the tangent and the equation of the normalto the curve $y=1/x$ at the point where $x=2$. (See Ex. 8, p. 11.)

4. Find the values of x for which the curve $y=x^3-16x+1$ risesand those for which it falls; find the highest point (maximum) and thelowest point (minimum). Draw the graph accurately.

5. Draw accurate graphs for the following curves:

(a)$y=x^3-18x+3$. (c) $y=x^4-32x$.

(b)$y=x^3+3x^2$. (d)$y=x^4-18x^2$.

6. Determine the speed of a body which moves so that

$$s=16t^2+10t+5.$$

[A body thrown down from a height with initial speed 10 ft. per secondmoves in this way approximately, if s is measured downward from amark 5 ft. above the starting point.]

7. If a body moves so that its horizontal and its vertical distancesfrom a point are, respectively, $x=10t,y=-16t^2+10t$, flnd itshorizontal speed and its vertical speed. Show that the path is

$$y=-16x^2/100+x,$$

and that the slope of this path is the ratio of the vertioal speed to thehorizontal speed. [These equations represent, approximately, the motionof an object thrown upward at an angle of ${45}^{\mathrm{o}}$ with a speed $10\sqrt{}\overline{2}.$]

8. A stone is dropped into still water. The circumference $\mathrm{c}$ of thegrowing circular waves thus made, as a function of the radius $r$, is $c=2\pi r$.

Show that $d\mathrm{c}/dr=2\pi ,i.e$. that the circumference changes$2\pi$ times as fast as the radius.

Let $A$ be the area of the circle. Show that $dA/dr=2\pi r$; i.e. therate at which the area is changing compared to the radius is numericallyequal to the circumference.

9. Determine the rates of change of the following variables:(a) The surface of a sphere compared with its radius, as the sphereexpands.(b) The volume of a cube compaoed with its edge, as the cube enlarges.(c) The volume of a right circular cone compared with the radius ofits base (the height being flxed), as the base spreads out.

10. If a man 6 ft. tall is at a distance $x$ from the base of an arc light10 ft. high, and if the length of his shadow is $s$, show that $s/6=x/4$, or$s=3x/2$. Find the rate $(ds/dx)$ at which the length $s$ of his shadowincreases as compared with hls distance $x$ from the lamp base.

11. The specific heat of a substance (e.g. water) is the amount ofheat required to raise the temperature of a unit volume of that substance $1^{\text{o}}$. (Centigrade). This amount is known to change for the same substance for different temperatures. The average speciflc heat between twotemperatures is the ratio of the quantity of heat $\mathrm{\Delta }H$ consumed inraising the temperature divided by the change $\mathrm{\Delta }t$ in the temperature; show that the actual specific heat at a given temperature is $dH/dt$.

12. The coefflcient of expansion of a solid substance is the amount abar of that substance 1 ft. long will expand when the temperature changes$1^{\mathrm{o}}$. Express the average coefflcient of expansion, and show thatthe coefficient of expansion at any given temperature is $dl/dt$, if the baris precisely 1 ft. long at that temperature. (See also Ex. 12, p. 145.)