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Run Multiple Choice Generator at h:\shares\physics
Run Core Problems by giving the program 163 as your 'course'--enter this when asked for your 3-digit course number.
Do choice #3, then do choice #9.
If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution

Compounding $1000 initial principle continuously, for 1 year, at 100% annual interest:

If we compound just once we end up with $2000, which is obvious.
If we compound twice then we get to add 50%, then add another 50% to the resulting amount. Adding 50% is the same as multiply by 1.5 So we multiply the initial principle by 1.5, then multiply the result by 1.5. We end up with $1000 * 1.5 * 1.5 = $1000 * 1.5^2 = $2250.
If we compound, say, 100 times we get to add 1%, then add another 1% to the resulting amount, etc., 100 times. Adding 1% is the same as multiply by 1.01 So we multiply the initial principle by 1.01, then multiply the result by 1.01, etc., 100 times. We end up with $1000 * 1.01^100 = $2704, a factor of 2.704.
If we compound n times then we get to add 100% / n, a total of n times.
Adding 100% / n = 1.00 / n or just 1/n is the same as multiplying by 1 + 1 / n. So we multiply the initial principle by 1+1/n, then multiply the result by 1+1/n, etc., n times.

We end up with $1000 * (1 + 1/n) ^ n.
As n -> infinity compounding approaches continuous and (1 + 1/n) ^ n approaches e. e is approximately 2.71828.

Give an equation whose solution answers the question 'how long would it take to double principle, compounding 100 times per year at 100%?

Obvious that at 100% it takes less than a year.

After .7 year

we will have compounded .7 * 100 = 70 times.

After .9 year we will have compounded .9 * 100 = 90 times.

After t years we will have compounded t * 100 = 100 t.

So after t years we will have

$1000 ( 1.01) ^ ( 100 t).

If our principle doubles, we will have $2000 so we know that

$2000 = $1000 ( 1.01) ^ (100 t).

Problem Number 7

Show that the slopes of the function y = .9 t^2 + -30 t + -75 change at a constant rate.

To show that the slopes change at a constant rate, pick a few equally spaced points and determine the average slopes for the corresponding intervals.

The vertex of this graph is at t = 15 or so (easily found as t_vertex = - b / ( 2 a) ). So let's use t values 0, 5, 10, 15. You could use any equally-spaced set of t values, as long as your slope calculations have a reasonable number of significant figures.

The corresponding y values are easily found using a calculator, or alternatively using the DERIVE commands f(t):= .9 t^2 - 30 t - 75 and vector(f(t),t,0,15,5), the latter telling us to calculate f(t) from t = 0 to 15 by step 5. We get y values -75, -202.5, -285, -322.5.

The slopes over the corresponding intervals are calculated by slope = rise / run = change in y / change in x. The DERIVE command vector( (f(t+5) - f(t) ) / 5, t, 0, 10, 5) also accomplishes this.

Either way we get slopes -25.5, -16.5, -7.5.

Treating these slopes as a sequence we find that the first differences are always 9, plausibly demonstrating that the slope always changes by 9 units when the t-interval is 5.

Problem Number 8

Problem: Obtain a quadratic depth vs. clock time model if depths of 67.75163 cm, 49.07069 cm and 38.95717 cm are observed at clock times t = 15.01528, 30.03055 and 45.04583 seconds.

Problem: The quadratic depth vs. clock time model corresponding to depths of 67.75163 cm, 49.07069 cm and 38.95717 cm at clock times t = 15.01528, 30.03055 and 45.04583 seconds is depth(t) = .019 t² + -2.1 t + 95. Use the model to determine the clock time at which depth is 54.61883 cm.

This is a typical question from the Major Quiz.

Problem Number 4

If y = -.4 t^2 + -6 t + 66, what symbolic expression stands for the slope between the graph points for which t = x and t = x+h?

Slope is rise / run.

The rise is the change in the y function, between t = x and t = x + h, and the run is x + h - x = h.

Regarding y as a function y(t) of t, the y values at the two t values are

y(x + h) =

-.4 ( x + h ) ^2 - 6 ( x + h ) + 66 and

y(x) = -.4 x^2 - 6 x + 66.

The rise, which is the change in y, is

rise = `dy = -.4 ( x + h ) ^2 - 6 ( x + h ) + 66 - ( -.4 x^2 - 6 x + 66 ).

We simplify by expanding the square and using the distributive law. Many terms cancel out in the process and we end up with

rise = `dy = -.8 x h - 6 h - 0.4 h^2.

Again, the run is h so we get

slope = rise / run =

`dy / `dt =

( -.8 x h - 6 h - 0.4 h^2 ) / h =

-.8 x - 6 - .4 h.

Problem Number 7

If a(n) = a(n-1) + 9, with a(0) = -9, then what is the value of a( 310)?