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involved in this process is the following: We know where the ball was at ð sec.
At 5.1 sec, the distance that it has gone all together is 16(5.1)2 = 416.16 ft (see
Eq. 8.1). At 5 sec it had already fallen 400 ft; in the last tenth of a second it
fell 416.16 — 400 = 16.16 ft. Since 16.16 ft in 0.1 sec is the same as 161.6 ft/sec,
that is the speed more or less, but it is not exactly correct. Is that the speed
at 5, or at 5.1, or halfway bebween at 5.05 sec, or when 7s that the speed? Never
mind—the problem was to fñnd the speed ø‡ 5 seconds, and we do not have exactly
that; we have to do a better job. So, we take one-thousandth of a second more
than ð sec, or 5.001 sec, and calculate the total fall as
s = 16(5.001)Ÿ = 16(25.010001) = 400.160016 ft.
In the last 0.001 sec the ball fell 0.160016 ft, and if we divide this number by
0.001 sec we obtain the speed as 160.016 ft/sec. That is closer, very close, but it is
siill not exact. Tt should now be evident what we must do to ñnd the speed exactly.
To perform the mathematics we state the problem a little more abstractly: to
find the velocity at a special time, ứọ, which in the original problem was ð sec.
Now the distance at fọ, which we call sọ, is 16fã, or 400 ft in this case. In order
to ñnd the velocity, we ask, “At the time £o + (a little bit), or fo +, where is
the body?” The new position is 16(fo + e)2 = 16fã + 32foe + 16c?. So it is farther
along than it was before, because before it was only 16/á. Thịis distance we shall
call so + (a little bit more), or sg + # (ïŸ z is the extra bit). Now if we subtract
the distance at ứo from the distance at fọ + c, we get z, the extra distance gone,
as # = 32fo -c-L 16e2. Qur first approximation to the veloeity is
b= h = 39fo + 16c. (8.4)
The true velocity is the value of this ratio, z/c, when e becomes vanishingly small.
In other words, after forming the ratio, we take the limit as e gets smaller and
--- Trang 168 ---
smaller, that is, approaches 0. “The equation reduces to,
9U (at time to) = 32to.
In our problem, #o = ð sec, so the solution is ø = 32 x 5 = 160 ft/sec. A few lines
above, where we took c as 0.1 and 0.001 sec successively, the value we got for 0
was a little more than this, but now we see that the actual velocity is precisely
160 ft/sec.
8-3 Speed as a derivative
The procedure we have just carried out is performed so often in mathematics
that for convenience special notations have been assigned to our quantities
and z. In this notation, the used above becomes A£ and # becomes As. 'This
Af means “an extra bit of £,” and carries an implication that it can be made
smaller. The prefx A is not a multiplier, any more than sỉn Ø means s-i -n - Ø—it
simply defines a tỉme increment, and reminds us of its special character. As has
an analogous meaning for the distance s. Since A is not a factor, it cannot be
cancelled in the ratio As/Af to give s/f, any more than the ratio sin Ø/sin 20
can be reduced to 1/2 by cancellation. In this notation, velocity is equal to the
limit of As/Af when Af gets smailler, or
= lim —. 8.5
_—- 5)
Thịis is really the same as our previous expression (8.3) with e and z, but it has
the advantage of showing that something is changing, and it keeps track of what
is changing.
Incidentally, to a good approximation we have another law, which says that
the change in distance of a moving point is the velocity times the time interval,
or As =0 Af. Thịs statement is true only if the velocity is not changing during
that time interval, and this condition is true only in the limit as Af goes to 0.
Physicists like to write it đs = 0 đf, because by đ£ they mean Af in circumstances
in which it is very small; with this understanding, the expression is valid to a close
approximation. If A£ is too long, the velocity might change during the interval,
and the approximation would become less accurate. Eor a time đÝ, approaching
zero, ds = 0 đf precisely. In this notation we can write (S.5) as
h As ds
= lim -_— =-—.
T— Arso AE — đi
--- Trang 169 ---
The quantity đs/đ£ which we found above is called the “derivative of s with
respect to £” (this language helps to keep track of what was changed), and the
complicated process of ñnding ït is called ñnding a derivative, or diferentiating.
The đs's and đf£s which appear separately are called đjfereniials. To familiarize
you with the words, we say we found the derivative of the funetion 162, or the
derivative (with respect to £) of 16/2 is 32. When we get used to the words, the
ideas are more easily understood. Eor practice, let us fnd the derivative of a more
complicated funetion. We shall consider the formula s = 4£ + B + Œ, which
might describe the motion of a point. The letters 4, Ö, and Œ represent constant
numbers, as in the familiar general form oŸ a quadratic equation. Starting from
the formula for the motion, we wish to ñnd the velocity at any time. To ñnd the
velocity in the more elegant manner, we change # to ý + A# and note that s is
then changed to s-Ƒ some As; then we find the As in terms of A7. That is to say,
s+ As = A(+ At)Ỷ+ B(+ At)+Œ
= Af + Bt+ CƠ +3Af? At+ BAt+3At(A9)? + A(AĐ,
but since
s= Af + Bt + C,
we fñnd that
As=3A/? At+ BAt+3At(A93 + A(A9.
But we do not want As—we want As divided by A¿. We divide the preceding
cquation by A£, getting
As 2 2
Ar E34? + B+3AH(A0) + A(A0).
As Af goes toward 0 the limit of As/Af is đs/đf and is equal to
—=3A4/+D.
'This is the fundamental process of calculus, diferentiating functions. he process
1s even more simple than it appears. Observe that when these expansions contain
any term with a square or a cube or any higher power of A£, such terms may be
dropped at once, since they will go to 0 when the limit is taken. After a little
practice the process gets easier because one knows what to leave out. 'There are
many rules or formulas for differentiating various types of functions. 'Phese can
be memorized, or can be found in tables. A short list is found in Table 8-3.
--- Trang 170 ---
Table 8-3. A Short Table of Derivatives
8, tu, 0, t are arbitrary functions of ; ø, b, c, and m are arbitrary constants
Punction Derivative
=í" —=ni”
S—= Cu đs —=C€C đụ
có dt đt