July 1932 Radio News
[Table
of Contents]
These articles are scanned and OCRed from old editions of the Radio & Television News magazine. Here is a list of the Radio & Television
News articles I have already posted. All copyrights are hereby acknowledged.

I
challenge you to find a calculus lesson in a modernday electronics
magazine. In 1932, not all that long after Isaac Newton developed
differential calculus (that's a joke),
Radio News magazine
ran a series of "Mathematics in Radio" articles that included, among
other topics, a few lessons in calculus. Anyone who has taken collegelevel
science or engineering courses knows how indispensible calculus
is in working out many circuit, physics, and chemistry problems.
My appreciation for calculus came when I realized that it actually
allowed me to derive the kinds of standard equations that are commonly
seen in lower level applications. For instance, if you needed to
know the
volume of a sphere, you could look up the familiar Volume = 4/3
π r
^{3}
formula, or you could write the equation
Volume = 
∫ 
2π 
∫ 
π 
∫ 
r 
ρ^{2}
sin
Φ
dρ
dΦ
dθ
= 4/3 π r^{3}. 



0 
0 
0 
Area, mass, center of mass, and length of a spiral line, are other
examples of problems that can be solved by knowing the fundamental
mathematics behind the equations.
Mathematics in Radio
Calculus and its Application in Radio  Part Seventeen
By
J. E. Smith*
In order to apply the formulas for differentiating,
the student should become familiar with the following examples with
reference to the standard forms (1)(5) inclusive (which were shown
in the last lesson
^{+}).
Differentiate the following:
1) y = x
^{4}Here, formula (5) applies where
x = v, and the exponent 4 = u.
Solution:
(A)
, but from the formula (1), dx/dx = 1,
therefore A becomes: dy/dx = 4x
^{3} (answer).
2) y = x
^{6}
3) y = x
^{2}
4) y = x
^{9}
5) y  x
^{5}
6) y = ax
^{3} + bx
^{5}Here, formulas
(3) and (5) apply, where a and b = c of formula (3), and the exponents
3 and 5 respectively = n.
Solution:
= 3ax
^{2} + 5bx
^{4} (answer).
7) y = ax
^{4}
8) y = ax
^{6} + bx
^{2}
9) y = ax
^{7}  x
^{6}
10) y  ax
^{3}  bx
^{3}
11) y = 5x
^{4} + 3x
^{2}  6
The derivative
of a constant is zero; i.e., dc/dx = 0.
Solution:
= 20x
^{3} + 6x (answer).
12) y
= 3cx
^{2}  8dx + 5e
13) y = 2ax
^{2} + 4dx
^{2}  5ax
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5
Figure 6 
The sinusoidal functions are those expressed by the sine and cosine
curves as well as those functions which are closely allied to them,
which are the tangent and cotangent curves so important in geometry.
Electrical and radio theory are very often advanced by the
use of calculus, which takes the liberty of performing the necessary
operations on the sine and cosine functions. Let us apply the theory
which was outlined above in finding the derivative of a sine function.
That is, if a sine function is expressed by y = sin x, it is the
purpose of this analysis to determine its proper derivative.
In
the previous texts it has been shown how to plot the function y
= sin x, and this is shown in Figure 1 for the values of x from
0180 degrees. In order to study the derivative of this function,
reference is made to Figures 2 and 3, where the values of x have
been plotted from 090 degrees. Applying the theory which has been
outlined above, we see that when Δx approaches zero, Δy/Δx
finally approaches a limiting value, and it is remembered that this
limiting value is called the derivative of y with respect to x,
This limit, represented by dy/dx, has been shown above to be equal
to the tangent θ.
With reference to Figure 4, it is
noticed that the angle θ at the point A is somewhat larger than
the values of θ for the points Band C. Finally, at the point D,
the tangent line TD becomes parallel to the abscissa, and it is
noticed that the angle θ has become increasingly smaller until it
has approached the value of zero.
Now, the derivative of
the function y = sin x at the point A is by definition equal to
the tangent of θ. In like manner, the derivatives of the function
at points B, C and D are again equal respectively to the tangents
of the angles θ.
By close analysis of the relative sizes
of these angles, we arrive at the following table:
For point
A, θ equals about 45°
For point B, θ equals about 35°
For point C, θ equals about 17°
For point D, θ equals
about 0°
But the tangent of these angles is by definition
equal to the derivative of the function at its respective points;
thus:
dy/dx at point A is about 1
dy/dx at point
B is about 0.70
dy/dx at point C is about 0.31
dy/dx
at point D is about 0
Plotting these values of dy/dx at
point A, B and C, respectively, we have the graph of Figure 5. But
this dy/dx curve is further noticed to be the same as the function
y = cos x, so it is apparent that:
d (sin x)/ax = cos x
or the derivative of the sin x is equal to the cos x.
Figure 6 shows the functions y = sin x and y = cos x plotted
together.
Formulas for Differentiating Standard
Forms The following formulas are a continuation
of the ones listed above from (15) inclusive.
6.
d/dx (sin v) = cos v (dv/dx)
The derivative of a sine function
is equal to its cosine function, times the derivative of the function.
7. d (cos v) / dx = sin v (dv/dx)
8.
d (tan v) / dx = sec
^{2} v (dv/dx)
9. d (cot
v) / dx = csc
^{2} v (dv/dx)
See v and csc v above
are the abbreviations for secant and cosecant, respectively. The
application of the above differential forms to radio theory is now
quite readily presented.
Answers to Problems:
1) 4x
^{3} 2) 6x
^{5}
3) 2x
4) 9x
^{8} 5)
5x
^{4} 6) 3ax
^{2} + 5bx
^{4}
7) 4ax
^{3} 8) 6ax
^{5} + 2bx
9) 7ax
^{6}  6x
^{5}10) 1  3ax
^{2}
 3bx
^{2}11) 20x
^{3} + 6x
12)
6cx  8d
13) 4ax + 8x  5a
*President National Radio Institute.
+Note  Many of the examples have been taken from the book,
"Elements of the Differential and Integral Calculus,"
by W. A. Granville, published by Ginn and Company, N. Y.
Posted August 22,
2013