Look at the sequence 1               3              4              7              11            18           ...

Differences are 2, 1, 3, 4, 7.  No real sense to that?

Jump out of the box and look at other possible patterns.  In this one we add two consecutive numbers to get the next.

Sequence:  13             20            39            76            137         ...

1st diff:             7                19           37             51  ...

2d diff:                     12              18             24         ...

3d diff:                             6                 6                 ?

The third difference is the constant 6.  We can rebuild the sequence:

  13             20            39            76            137             228            355

         7                19           37             61               91            127

                 12              18             24             30             36

                           6                 6              6              6

Since the third difference is a nonzero constant the sequence 13, 20, 39, 76, 137, ... can be generated by a polynomial of degree 3 (a function of the form

a n^3 + b n^2 + c n + d).

The specific function for this sequence is

n^3 + 3n^2 + 3n + 13.

(next to 5 in cell B8 we put

=b8^3 + 3*b8^2 + 3*b8 + 13

and Excel evaluates the expression.  Drag the fill handle down and B8 adjusts for each new row.

In row 466 the command becomes

 =B466^3 + 3*B466^2 + 3*B466 + 13

and since B466 is 463 the computer evaluates

463^3 + 3 * 463^2 + 3 * 463 + 13 = 99897356

).

Evaluate this function for B10, which is B10 = 7.  See if this fits the above patterns of differences.

20    29    37    44    50    55

    9        8     7      6      5

Notes:

• If the first difference of a sequence consists of positive numbers then the graph is increasing.

• If the first difference of a sequence consists of positive increasing numbers then the graph of the sequence is increasing at an increasing rate.

y(t) = .010 (t - 100)^2

The expression y(0) is like y(t) but t has been replaced by 0.

So y(t) = .010 (t - 100)^2 becomes

y(0) = .010 (0 - 100)^2. 

Evaluating this expression we get

y(0) = .010(100)^2 = 100.

y(10) = .010 (10 - 100)^2 = 81.

y(20) = 64

y(30) = 49

y(40) = 36

y(50) = 25

y(60) = 16

y(70) = 9

y(80) = 4

y(90) = 1

y(100) = 0.

Note that:

As a sequence the differences are -19, -17, -15, -13, ... and the second differences are 2, 2, 2, 2 ... .  This means that y is a degree 2 polynomial.

We already knew that. 

.010 ( t - 100)^2 = .010 (t^2 - 200 t + 10000) = .01 t^2 - 20 t + 100.

At t = 0, 10, 20, 30, 40, ... the y values are 100, 81, 64, 49, 25, ... .

The average rates of change of y with respect to t are as follows:

• On the first interval, which runs from t = 0   to t =  10, we see that y changes from 100 to 81. 

The average rate of change of y with respect to t is   change is y / change in t.

 

Change in y is 81 - 100 = -19.  Change in t is 10 - 0 = 10.  So

 

ave rate of change of y with respect to t is

ave rate = -19 / 10 = -1.9.

• On the second interval, which runs from t = 10   to t =  20, we see that y changes from 81 to 64. 

The average rate of change of y with respect to t is   change is y / change in t.

 

Change in y is 64 - 81 = -17.  Change in t is 10 - 0 = 10.  So

 

ave rate of change of y with respect to t is

ave rate = -17 / 10 = -1.7.

 

These average rates would be the slopes on a trapezoidal graph.

Find the average rate for each of the next three intervals and label the corresponding a trapezoidal graph; include labels for the length of lines and for slopes.

temp scale 40% of way to room temperature every 10 minutes; init temp 12 cm, room temp 3 cm.

• 40% of the way to room temperature from the initial 12 cm is 40% of (3 cm - 12 cm) = -9 cm.

• At the end of the first 10 minutes the temperature will have changed by 40% of -9 cm, which is .4 * -9 cm = -3.6 cm.

So at the end of the first 10 minutes the temperature will be

 

initial temp + change in temp = 12 cm + (-3.6 cm) = 8.4 cm.

 

• At the end of the next 10 minutes the temperature will have changed by 40% of the change from 8.4 cm to 3 cm.

The change from 8.4 cm to 3 cm is -5.4 cm so the change is .4 * -5.4 cm = -2.16 cm.

 

• etc.

A graph of temperature vs. clock time will be decreasing at a decreasing rate.

The first differences of the temperature sequence will be negative and the second differences will be positive.

Temperature excesses are temperatures above room temperature.  For example 8.4 cm is 5.4 cm above room temperature; 6.24 cm is 3.24 cm above room temperature.

The temperature excess function is exponential.  It's not a polynomial, it's an exponential function.

Excel does an exponential fit to our data, giving us the function y = 9e^(-0.0511x)

• e is the limiting value of (1 + 1/n)^n, as n approaches infinity.  Approximate value of e is 2.71828; e is an irrational number.

Find the average rate of change of temperature with respect to clock time over the interval from t = 0 to t = 10 and the interval from t = 10 to t = 20.  Label the corresonding trapezoidal approximation graph of temperature excess vs. clock time.

sane/demented 1000, 10%, 20%, init 900 sane

5 trees, 1000 bugs on 2d, 20% transition, 10% at ends

cardboard

find f fn. from T fn.

We know that T = .2 sqrt(L) is the period in seconds of a pendulum whose length in cm is L.

So if a pendulum has length L, what is its frequency in cycles per minute?

Example:  A pendulum of length 25 cm has period T =.2 sqrt(25) sec = .2 * 5 sec = 1 sec.  This pendulum will oscillate 60 times in a minute (1 cycle every second is 60 cycles in a minute).

Another example:  A pendulum of length 100 cm has period T = .2 sqrt(100) sec = .2 * 10 sec = 2 sec.  This pendulum will oscillate 30 times in a minute (1 cycle every 2 seconds is 30 cycles in a 60-second minute).

So how do we find the frequency f in cycles per minute, give the period T?

• Just as we got 30 cycles / minute by dividing 60 by our 2-second period, for period T we get the frequency by dividing 60 by T.

• So the frequency in cycles / minute is

f = 60 / T.

• Since T = .2 sqrt(L) we can get an expression for f in terms of L.  Do this.  Evaluate this function for L = 25 and L = 100 and find the average rate of change of frequency with respect to period between these two lengths.

What is the ave rate at which the period changes with respect to length between length 25 cm and length 100 cm?

• ave rate of change of period with respect to length is change in period / change in length.

change in period is the change from a 1 sec period to a 2 sec period.  The change in period is therefore

 

change in period = 2 sec - 1 sec = 1 sec.

 

change in length is change from 25 cm to 100 cm so

 

change in length = 100 cm - 25 cm = 75 cm.

 

So our average rate of change of period with respect to length is

 

ave rate = change in period / change in length = 1 sec / (75 cm) = .0133... sec / cm.

 

Calculate the average rate of change of period with respect to length between lengths 5 cm and 7 cm, then again between lengths 8 cm and 10 cm.

Does the average rate of change of period with respect to length depend on length?