1027
Compounding $1000 initial principle continuously, for 1 year, at 100% annual interest:
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)?