When a pendulum moves in simple harmonic motion with period .8 seconds, the angular frequency of the corresponding circular model is 2 `pi / .8 = 2.5 `pi.

• If the amplitude of the motion of the object is 20 mm, how long is a take the object to go from position 5 mm to position 0 mm? How long from 20 mm to 5 mm?

A model of this motion is as given below. We can answer the questions by determining the clock time t at which the position is 20, 5 and 0.

Since we are using a sine function, when t = 0 we have x = 0.

To find the clock time when x = 5, we solve the equation indicated in the figure below.

• We see that the x = 5 position is attained at .03 sec..

If we solve a similar equation for the x = 20 position we see that this position is attained at t = .2 sec..

It therefore takes .03 sec to go from the 5 mm position to the 0 mm position, and .17 sec to go from the 20 mm position to the 5 mm position.

Video Clip #01

The two figures below depict the numbers in Pascal's triangle and the naming scheme for these numbers.

• We note that, for example, C(4,2) is the number of ways are getting 2 heads on 4 coin flips.
• In general, C(n,r) is the number of ways of getting r heads on n coin flips.

C(4,2) is also the number of ways to choose 2 people out of 4 for a committee, or in general the number of ways to choose 2 objects out of n, where the order of the choice does not matter.

We can apply this idea to the problem of finding the cube of a binomial like (x + y).

• When we multiply (x+y)(x+y)(x+y), we get our terms by choosing either x or y from each binomial. When we have made every possible choice then we have all the terms of the product.

• We could for example pick x from the first and for the second binomial and y from the last, obtaining x^2 y.
• We will also obtain x^2 y if we choose x from the first and last binomial and y from the second; we will obtain x^2 y again if we choose x from the second and third binomials and y from the first.

• For each of these choices we have chosen one of the three binomials from which to pick y.
• As a result we have obtained x^2 y a total of C(3,1) = 3 times.

• To obtain x y^2 we must pick two of the three binomials from which to select y.
• This time we are choosing 2 of the three binomials from which to pick y. We thus obtain x y^2 in C(3,2) = 3 ways.

• The entire process is depicted in the figure below.

Video Clip #02

To compute (x + y) ^ 4

• we obtain x^4 by choosing y from 0 of the 4 binomials, which we can do in C(4,0) = 1 way
• we obtain x^3 y by choosing y from 1 of the 4 binomials, which we can do in C(4,1) = 4 ways
• we obtain x^2 y^2 by choosing y from 4 of the 4 binomials, which we can do in C(4,2) = 6 ways
• we obtain x y^3 by choosing y from 3 of the 4 binomials, which we can do in C(4,3) = 4 ways
• we obtain x^4 by choosing y from all 4 of the 4 binomials, which we can do in C(4,4) = 1 way.

We thus obtain the expansion in the last line below.

To expand (3a - 5b)^3 we use the expansion of (x+y)^3, letting x = 3a and y = 5b, as shown below.

Video Clip #03

We can write a formula for C(n, r) as depicted below.

• If we substitute n = 10 and r = 4 into this formula, we obtain the expression indicated below.
• We see that all the the first 4 factors of the 10! cancel, and that the denominator then contains 4!.

Note, as in the lower right-hand corner of the figure,  that if we are picking 4 objects out of 10, there are 10 choices for the first, 9 for the second, 8 for the third, and 7 for the fourth so that there are 10 * 9 * 8 * 7 ways to choose the four objects; however according to this calculation we can choose the same 4 objects in two or more different orders, and we are counting every one of these orders.

• In order to get the number of ways to choose the objects without regard for order, we therefore have to divide this result by the 4! ways in which four objects can be ordered.

Again we note at least 3 different ways of writing C(n, r).

The bottom half of the figure above states to Binomial Theorem, or Binomial Formula, for finding the nth power of a binomial.

Video Clip #04