In my first three years at university I was subjected to Calculus. The Calculus they called it. These were the most frustrating years of my life.

The instructor was Korean. He spoke English in a way that I could not understand. Some of the other students in the class were also recent arrivals from other non-English speaking countries (I immigrated from Norway). For them it was even worse. This instructor was a graduate student earning his board and tuition by ruining the academic prospects for others.

Since then I have fortunately discovered the key to the mystery of what calculus is all about.

Calculus was developed to solve real world problems that relate to rate of change.

Two approaches may be used when solving the sorts of problems that calculus was developed to address - the symbolic (abstract, general) approach and the numerical (concrete, specific, computer based) approach.

Symbolic calculus is taught in calculus classes.

This paper uses the numerical approach.

It is my belief that understanding is best achieved through experience with specific concrete applications as a prelude to a study of the underlying concepts. In other words, learning necessarily proceeds from the specific to the general.

As an example, in arithmetic we deal with numerical solutions to specific concrete problems. In algebra we deal with symbolic solutions to abstracted generalizations. And just as it is more efficient to learn arithmetic first and then study algebra, it makes sense to learn about numerical calculus before the symbolic variety.

There are two major divisions of calculus: differential and integral. Differential Calculus has to do with tangents. Integral Calculus has to do with areas.

1. DIFFERENTIAL CALCULUS

   This chapter is organized according to these headings:

   1.1 "Problems leading to use of derivatives"

   1.2 Derivatives of linear functions of the form:

   y = a*x + c

   1.3 Derivatives of polynomial functions

   of the form y = a*x^n + b*x^(n-1) + … c

   1.3.1. First derivatives

   1.3.2. Second and further derivatives

   1.3.3 Derivatives of trigonometric functions

   like y = sin(x)

   1.3.3.1. First derivatives

   1.3.3.2. Second, and further derivatives

   1. Problems Leading to use of Derivatives
      1. acceleration

         From High School Science we know that if something is dropped and allowed to fall freely, it will accelerate at 9.8 m/s/s (metres per second per second). Well, at least that would be true if there was no air resistance! With air resistance the situation becomes very complicated and is beyond the scope of this paper. So for now, let's just go with it as described.

         The velocity at the end of every second will be 9.8 m/s more than it was at the end of the previous second. In other words the velocity changes. Here we are interested in the rate of change.

         velocity = acceleration * time

         and also acceleration = rate of change of velocity

                              = change in velocity/time

         This graph shows velocity as a function of time and acceleration for the first ten seconds.

         [fig 1.1.1a

         

         The velocity is changing. The rate at which it is changing is equivalent to the slope of the velocity-time graph.

         You remember from High School:

         slope is m = (y2 - y1) / (x2 - x1)

         Any two points will do. In this case:

         m = (68.60*m/s - 29.40*m/s) / (7.00*s - 3.00*s)

         Notice how we have kept the units and treat them like numbers.

         = 39.2*m/s / (4*s)

         = 9.8 m/s/s

         The acceleration of a falling object is the rate at which the velocity is changing - found by the slope of the velocity - time graph.

         Another way of saying this: "Acceleration (the rate of change of velocity) is the derivative of the velocity with respect to time."

      2. velocity

         The distance covered by a moving object is found by:
         distance = average velocity * time
         The velocity achieved after a certain time is
         velocity = acceleration * time
         So, during the first five seconds we get
         velocity = 9.8 m/s/s * 5 s
         = 49 m/s
         and the distance fallen:
         distance = average velocity * time
         distance = (0 + 49 m/s)/2 * 5 s
               = 122.5 m
         

         Similar calculations allow us to make this table and graph showing how distance changes with respect to time.

         [fig 1.1.2a]

         

         The rate of change of distance is called speed or velocity. The rate of change is the slope of the graph. By the way, velocity is a vector quantity and therefore has direction associated with it. Speed is not a vector quantity.

         In this case the graph is curved, so there must be different velocities for different times - the object is speeding up. Let's consider the situation at the 5 second mark.

         The velocity we are looking for is the slope of the tangent to the graph at time = 5 s

         How do we know what the value of the tangent is at this point?

         Let's zoom in at time = 5:

         Fig 1.1.2b

         

         As we zoom in, the graph appears straighter. Even though the tangent is drawn onto the graph "by eye", at some magnification we can approximate the value of the tangent by estimating some values from the graph.

         In this case we judge that distance coordinate of the tangent at time = 5.40 s is 142 m, and that the distance coordinate at time = 4.60 s = 103 m.

         Using these estimates, we state that slope of the graph at this point is:

         slope = (142 m - 103 m)/(5.40 s -4.60 s)

               = 48.7 m/s

         This number, though an approximation, is probably a good one.

         The velocity of a falling object at any time is the slope of its distance - time graph at that point.

         Another way of saying this is "Velocity is the derivative distance with respect to time."

      3. Other examples
   2. Derivatives of functions

      In section 1.1 we discussed some problems in which we were interested in the rate of change. This rate of change was found by looking at the slope of a graph, or in the case of a curve, at the slope of a tangent at some point. We called the rate of change a derivative.

      The derivative was found by visual inspection - and the result was approximate.

      In this section we will cover the same material, except that we will use generic algebraic notation without units, with a variety of functions. We will also increase the level of precision of the results.

      1. derivatives of linear functions

         y = a*x + c

          

         Here is the graph of

         y = 0.5*x - 1

         [1.2.1a]

         
         The graph is a straight line. This is a linear function. To find the slope we select any two points and apply:

         m = (y2 - y1) / (x2 - x1)
             = (-2.5 - 0.0) / (-3.0 - 2.0)
             = -2.5 / -5.0
             = 0.5

         The derivative of y with respect to x for the function

         f(x) = 0.5*x + 1
         is 0.5

         This is more formally written:

         d(0.5*x + 1) / d(x) = 0.5

         Sometimes it is written this way:

         f'(x) = 0.5

         In this paper we will use the notation:

         d1f(0.5*x + 1) = 0.5

         Note the following:

         d1f(x + 1) = 1
         d1f(2*x - 3) = 2
         d1f((-5)*x + 15) = (-5)

         In general:

         d1(a*x + c) = a

         Here is the graph of 0.5*x - 1 vs. its derivative:

         [1.2.2b]

         

      2. Derivatives of functions of the form:

         y = a*x^n + b*x^(n-1) + … cx + constant

         Example: y = x^3 / 5 - 2*x^2 + x - 3
         here, a = 0.2
         b = 2.0
         c = 1.0
         constant = -3.0

         1. First derivative

            This the graph of

            y = x^3 / 5 - 2*x^2 + x - 3:

            [1.2.2.1a]

            This graph is not a straight line. The slope at any point is the tangent at that point.

            We will develop a way of calculating the slope value for any value of x for any function.

            Let's zoom in on one point in particular; P5(2.00,-7.40) in this illustration)

            [Fig 1.2b]
            

            Consider the slope at x = 1.

            First, let's find the slope of the line through (0.00,-3.00) and (2.00,-7.40), which is:

            (-3.00 - (-7.40)) / (0.00 - 2.00) = -2.20

            The slope of this line is are obviously different from the slope we want... but it's close! The slope at x = 1 is very approximately -2.20.

            Looking at the lines drawn through the other pairs of points make it apparent that the closer the pairs of points are to the point in question, the more closely such lines through them will approximate the slope we are looking for.

            Now, consider this diagram:

            [1.2.2.1c]

            

            The triangle, or "delta" symbol in this context means "something very small"; so we are considering two points on ether side of P that are some very short distance away. The slope of the line through P1 and P2 ought to be very close to the slope of the tangent at P.

            The slope of the line through P1 and P2 is:

            (1) m = (f(x - dx) - f(x + dx)) / ((x - dx) - (x + dx))

            "dx" refers to "delta x" or "a very small value"

            In this case f(x) = x^3 / 5 - 2*x^2 + x - 3

            Let's see what happens to the slope at x = 1 if we let dx = 0.1:

            In this case f(x - dx) = (1 - 0.1)^3 / 5 - 2*(1 - 0.1)^2 + (1 - 0.1) - 3
            (2)                    = (0.9)^3 / 5 - 2 * (0.9)^2 + 0.9 - 3.0
            (3)                    = 0.146 - 1.620 + 0.900 - 3.000
            (4)                    = -3.574

             

            Similarly    f(x + dx) = -4.054

            and (x - dx) - (x + dx) = -2dx = -0.2  (since dx = 0.1)

            Substituting the values in (2), (3) and (4) in to equation (1), we get:

                                 m = ((-3.575) - (-4.054)) / 2.00
                                   = 0.2395

            If these calculations are re-done, using x = 0.01, we get slope = 0.2400. This is sufficiently precise.

            In a similar way we can calculate the slope at many points on the graph - and plot the results:

            [1.2.2.2.1d]

            

            Note that

            d1f(x^3/5 - 2*x^2 + x - 3) = 5x^2 - 4*x + 1

            (a parabola)

         2. 2nd and Higher Order Derivatives

            If you can find the derivative of a function, you should be able to find the derivative of the derivative too. In other words you can find the slopes of the parabola which represents the slopes of the original function at various values for x.

            And if you can do that you can find the derivative of the derivative of the derivative as well. And so on. These are called second derivatives, third derivatives, and so on.

            The calculations and graphs in this paper were all done with a mathematics utility program called MathPad.

            Here is a MathPad document which, if considered carefully in this context, should be self-explanatory:

   It may be that you find the following MathPad calculations puzzling. For a more detailed
   explanation of what is going on here, click on this link:

   -- function defined --

   f(x) = x^3 / 5 - 2*x^2 + x - 3

   -- derivatives defined ---

   dx=0.01
   d1f(x)=(f(x+dx) - f(x-dx)) / ((x+dx)-(x-dx))
   d2f(x)=(d1f(x+dx) - d1f(x-dx)) / ((x+dx)-(x-dx))
   d3f(x)=(d2f(x+dx) - d2f(x-dx)) / ((x+dx)-(x-xd))
   d4f(x)=(d3f(x+dx) - d3f(x-dx)) / ((x+dx)-(x-dx))
   d5f(x)=(d4f(x+dx) - d4f(x-dx)) / ((x+dx)-(x-dx))
   d6f(x)=(d5f(x+dx) - d5f(x-dx)) / ((x+dx)-(x-dx))

   -- make a table of values --

   Nmin = -5;Nmax=10
   Nsteps= 11

   table N, f(N), d1f(N), d2f(N), d3f(N)

   -5.00 -83.00 36.00 -10.00 1.20
   -3.50 -39.58 22.35  -8.20 1.20
   -2.00 -14.60 11.40  -6.40 1.20
   -0.50  -4.03  3.15  -4.60 1.20
    1.00  -3.80 -2.40  -2.80 1.20
    2.50  -9.88 -5.25  -1.00 1.20
    4.00 -18.20 -5.40   0.80 1.20
    5.50 -24.73 -2.85   2.60 1.20
    7.00 -25.40  2.40   4.40 1.20
    8.50 -16.17 10.35   6.20 1.20
    10.00  7.00 21.00   8.00 1.20

   -- plot graphs --

   Xmin=-5; Xmax=15
   Ymin=-40; Ymax=70
   Xdiv=5; Ydiv = 10
   Xhowxgrid=0; Yshowgrid=0

   plot d1f(X)
   plot d2f(X)
   plot d3f(X)

   And here are the resulting graphs:

   [1.2.2.2a]

   

    

   It is recommended that the reader evaluate and plot each of the functions

   y = x^3/5 - 2*x^2 + x - 3

   y = 3*x^2/5 - 4*x + 1

   y = 6*x/5 - 4

   1. The derivatives of sin(x)

This is the graph of sin(x) where x is in radians:

[1.2.2.3a]

The slope of the graph has been shown at several points.

The numerical value of the slope anywhere along the curve can be calculated using:

d1f(sin(x)) = (sin(x + dx) - sin(x - dx))/ ((x + dx) - (x - dx))

This table, also produced by MathPad shows the numerical values of the slopes and also the slopes of the graph of the slopes, in other words, the first and second derivatives of sin(x)

x in rads                    x              sin(x)     d1f(sin(x))       d2f(sin(x))

      0   0.0000    0.0000   1.0000     0.0000
   pi/4   0.7854    0.7071   0.7071    -0.7071
   pi/2   1.5708    1.0000   0.0000    -1.0000
pi*3*/4   2.3562    0.7071  -0.7071    -0.7071
     pi   0.0000   -1.0000   0.0000     0.0000
 5*pi/4   3.9270   -0.7071   -0.7071     0.7071
 3*pi/2   4.7124   -1.0000    0.0000     1.0000
 7*pi/4   5.4978   -0.7071    0.7071     0.7071
    2*pi  6.2832    0.0000    1.0000     0.0000

Here we see the graphs of sin(x), d1f(sin(x)) and d2(sin(x)):

[1.1.1.3b]

This completes our minicourse in differential calculus. Next Chaper - Curve Fitting in progress.

Continue with Integral Calculus