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