1025

Run Multiple Choice Generator at h:\shares\physics
Run Core Problems by giving the program qzz as your 'course'--enter this when asked for your 3-digit course number. Do the Precalculus Quiz for 1025. Then do the Precalculus Quiz for 1018.
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 is the same as multiply 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.

Practice Test 1

Problem Number 1

Explain what the number e represents.

e is the factor by which money grows in a year if we compound annual interest at 100%, compounding continuously, for 1 year.

This factor is the limiting value of (1 + 1 / n)^ n as n approaches infinity.

If we compound for t years the factor is e^t, so that if our initial principle is P0 our principle after t years is P(t) = P0 * e^t.

If we were to compound for t years at interest rate r the principle function would be P(t) = P0 * e^(rt).

Problem Number 2

If f(x) = x², what are the vertex and the three basic points of the graphs of f(x- .75), f(x) - .35, 5 f(x) and 5 f(x- .75) + .35. Quickly sketch each graph.

At clock times 54.8, 82.2, 109.6 and 137 sec, we observe water depths of -59, -95.4, -116.8 and -123.3 cm.

At what average rate does the depth change during each time interval?

Sketch a graph of depth vs. clock time.

Use a sketch to explain why the slope of this graph between 54.8 and 82.2 sec represents the average rate at which depth changes during this time interval.

This is Major Quiz material. We won't address it today, but be sure you review.

Problem Number 3

We expect that one of the power functions

y = a x^.5,
y = a x^-.5,
y = a x^2 and
y = a x^-2

best fits the following data:

When x takes values

15.448, 21.737, 26.758 and 31.304 ,

respectively, y takes values

5, 10, 15 and 20.

Which power function best fits this data?

As shown in the worksheets, substitute x = 15.45 and y = 5 into the y = a x^.5 and get a. Then substitute, say, the last point x = 31.3 and y = 20. If you get very nearly the same value of a then try the other two points. If they all give you about the same value of a then you have your model.

Otherwise repeat for y = a x^-.5.

You can also use some common sense. It looks like y values increase with increasing x values. This eliminates two of the possibilities.

If you know how the basic power functions behave (e.g., basic points), increasing or decreasing at an increasing or decreasing rate, then you can pretty much spot which function to try first.

Problem Number 4

Find the equation of a line through ( 3, 9) and ( 5, 8) by each of the following methods:

Substitute the coordinates of the points into the form of a straight line and solve the equations for the appropriate parameters.
Use the slope = slope form, complete with explanation and a labeled graph.

If your graph represents the length of a spring, in cm, vs. hanging weight in pounds:

What does the slope of the graph represent?
What is the length when the weight is 9 pounds?
What is the weight when the spring length is 13 cm?

The form of a straight line is y = a x + b. Substituting the points we get the equations

9 = a * 3 + b
8 = a * 5 + b

We can solve to get our values of a and b. We get a = -.5 and b = 10.5 so our model is

y = -.5 x + 10.5

We can then use the slope = slope form. Placing the two points and the general point (x,y) on a graph we find that the slope of the line is (8 - 9) / (5 - 3). The slope from the first point to (x, y) is (y - 9) / (x - 3). Setting the two slopes equal we get

(y - 9) / (x - 3) = (8 - 9) / (5 - 3),

which we can solve for y to get

y = -.5 x + 10.5

Problem Number 5

Explain how to use two simultaneous linear equations, obtained from two given points, to obtain the equation of the line through the two points.

See above.

Problem Number 6

Solve using ratios instead of functional proportionalities:

If a sand pile 2.3 meters high has a mass of 4866.8 kg, then what would we expect to be the mass of a geometrically similar sand pile 4.7 meters high?

The proportionality equation for volume is y = k x^3. Since mass occupies volume we use this proportionality.

This proportionality tells us that y2 / y1 = k x2^3 / ( k x1^3) = (x2 / x1)^3.

So mass2 / mass1 = (height2 / height1) ^ 3 = (4.7 m / 2.3 m)^3 = 8.5 so

mass2 = 8.5 * mass1 = 8.5 * 4867 kg = etc..

If there are 1.7457 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?

Sand grains are exposed at the surface. The proportionality for surface areas is y = k x^2. The same sort of analysis as used above tells us that the ratios are

area2 / area1 = (height2 / height1)^2 so we get

...

Problem Number 7

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

.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.

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?

Problem Number 7

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