class 0217

040217 Practice Major Quiz

DERIVE exercises: Go to any computer in 307 or the Learning Lab and click on the DERIVE icon. Follow the instructions on the DERIVE exercises.

To use the multiple-choice program:

Download multiple choice generating program from bb.vccs.edu (log in, if necessary you can use vhstudent and college as username and password), Supervised Study Current Semester under Course Documents, about the sixth item listed under that category.
If you are running on a different computer (e.g., at home) run the program q_a_prelim, about the second item under Course Documents. As soon as the first form appears close the program. Alternatively you can create a folder on your computer named c:\vhmthphy, but be sure you do it right.
Still under Course Documents click on Downloads, about the fifth item.
Click on the name of your course.
Copy the source file 163_c.txt to the c:\vhmthphy folder on your computer.
Run the multiple choice generating program. If you are on a different computer (e.g., at home) and it won't run then under Course Documents run the program vhrun60.exe, which will install the files needed to run vb programs.

Practice Major Quiz

Problem Number 1

Problem: Obtain a quadratic depth vs. clock time model if depths of 74.44157 cm, 67.70271 cm and 63.78343 cm are observed at clock times t = 6.855136, 13.71027 and 20.56541 seconds.

Problem: The quadratic depth vs. clock time model corresponding to depths of 74.44157 cm, 67.70271 cm and 63.78343 cm at clock times t = 6.855136, 13.71027 and 20.56541 seconds is

depth(t) = .03 t² + -1.6 t + 84.

Use the model to determine the clock time at which depth is 74.18898 cm.

Problem Number 2

Problem: Sketch a graph of the basic exponential function y = 2^x. If this function is stretched vertically by factor -1.41 then shifted 2.46 units vertically, what is the algebraic form of the function obtained? Sketch a graph of this new function, and show that the graph is different than that obtained if the vertical shift is performed before the vertical stretch.

Problem: Water depths of 67.536, 50.544, 39.024 and 32.976 cm are observed at clock times t = 12, 24, 36 and 48 seconds. What is the average rate of depth change during each of the three time intervals? Predict what the next average rate would be, and use this result to predict the depth at t = 60 seconds.

Problem Number 3

Give the full statement of the Quadratic Formula.

Problem Number 4

Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = .25 x^2 over the same domain.

By what factor do we vertically stretch the first graph to obtain the second?
What are the three basic points of the first graph (i.e., the vertex and the points 1 unit to the right and left of the vertex)?
What are the three basic points of the second graph?

Sketch the second graph shifted 1.75 units in the x direction and 1.75 units in the y direction.

What are the three basic points of this graph?

If f(x) = x^2, then what are

f(x- 1.75), f(x) + .75, .25 f(x) and .25 f(x- 1.75) + .75?

Problem Number 5

At clock times 49.4, 74.10001, 98.8 and 123.5 sec, we observe water depths of -57, -93.2, -117.2 and -129 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 49.4 and 74.10001 sec represents the average rate at which depth changes during this time interval.

If f(x) = x², what are the vertex and the three basic points of the graphs of f(x- 1.25), f(x) - .85, 5 f(x) and 5 f(x- 1.25) + .85. Quickly sketch each graph.

Problem Number 1

Problem: Obtain a quadratic depth vs. clock time model if depths of 74.44157 cm, 67.70271 cm and 63.78343 cm are observed at clock times t = 6.855136, 13.71027 and 20.56541 seconds.

This means plug the coordinates into the form y = a t^2 + b t + c and get three simultaneous equations in a, b and c. Then solve the equations to get a, b and c. Finally plug a, b and c into the form to get your function.

The function will look something like y = .02 t^2 - 3 t + 80, but with different numbers.

Problem: The quadratic depth vs. clock time model corresponding to depths of 74.44157 cm, 67.70271 cm and 63.78343 cm at clock times t = 6.855136, 13.71027 and 20.56541 seconds is

depth(t) = .03 t² + -1.6 t + 84.

Use the model to determine the clock time at which depth is 74.18898 cm.

To get the clock time at which depth is 74.2 cm (you can round off to 3 significant figures) reason as follows:

74.2 cm is the depth.
depth(t) stands for depth. So we gotta plug 74.2 cm in for depth(t) (not for the clock time t)
We get the equation 74.2 = .03 t^2 - 1.6 t + 84.
We solve this equation for t using the quadratic formula. Be sure to use the quadratic formula in its full statement.

Problem Number 2

The graph is easily sketched using its basic points (0, 1) and (1, 2) and knowing the basic shape of the exponential function (asymptotic to the negative x axis, increasing at an increasing rate, etc.).

The vertical stretch takes the basic points to (0, -1.41) and (1, -2.82). The asymptote will still be the negative x axis. y values have been multiplied by -1.41 so the function is now y = -1.41 * 2^x.

Vertically shifting 2.46 units moves every point 2.46 units in the vertical direction. The basic points become (0, -1.41 + 2.46) = (0, 1.05) and (1, -2.82 + 2.46) = (1, -.36). The horizontal asymptote rises from the negative x axis (y = 0) to the horizontal line y = 2.46. The equation of the function is now y = -1.41 * 2^x + 2.46.

If the vertical shift is performed first

The basic points (0, 1) and (1, 2) will change to (0, 3.46) and (1, 4.46) while the horizontal asymptote will become y = 2.46.

The equation of the function at this point will be y = 2^x + 2.46.

The vertical stretch will then move all ponts -1.41 times as far from the y axis. The basic points (0, 3.46) and (1, 4.46) will be transformed to (0, -1.41 * 3.46) = (0, -4.88) and (1, -1.41*4.46) = (-1, - 6.29). The horizontal asymptote y = 2.46 will be transformed to y = -1.41 * 2.46 = -3.47.

All y values will have been multiplied by -1.41 so the equation is now y = -1.41 * (2^x + 2.46) = -1.41 * 2^x - 3.47.

The key word here is 'each'. We need to calculate the average rate over each of the three intervals, meaning we will get three average rates.

To get the average rate over an interval we find the change in depth and the change in clock time over the interval and use these values to calculate average rate = change in depth / change in clock time.

The average rate will change from one interval to the next. Writing down the three average rates we can then identify a pattern. For example if the average rates change by the same amount as we move from the first interval to the second, then from the second to the third, we will predict that the average rate will change by the same amount as we move to the next interval and we can use this pattern to predict the next average rate.

Having predicted the next average rate of depth change over the next interval we can predict the change in the depth over that interval. Since ave rate = change in depth / change in clock time, it follows that change in depth = average rate * change in clock time. The change in clock time for the new interval is again 12 seconds so it will be easy to make the prediction.

clock time (sec)	depth (cm)	rise = change in depth	slope = change in depth / change in clock time	change in slope
12	67.5
24	50.5	-17	-1.417
36	39	-11.5	-0.958	0.458
48	33	-6	-0.5	0.458
60	?	?	?	?

The next slope will be .458 greater than the previous, which makes the next slope -0.5 + .458 = -.042.

The next rise will therefore be

rise = slope * run = -.042 * 12 = -.504

The next depth (the t = 60 sec depth) will therefore be

depth = previous depth + rise = 33 + (-.504) = 32.496.

Problem Number 3

Give the full statement of the Quadratic Formula.

Use the full statement, including the conditions and the 'if and only if' logic.

Problem Number 4

Sketch a graph of y = x^2, from x = -3 to x = 3. Then sketch a graph of y = .25 x^2 over the same domain.

By what factor do we vertically stretch the first graph to obtain the second?
What are the three basic points of the first graph (i.e., the vertex and the points 1 unit to the right and left of the vertex)?
What are the three basic points of the second graph?

We vertically stretch the y = x^2 graph by factor .25 to get the graph of y = .25 x^2. The basic points (-1, 1), (0, 0) and (1, 1) become (-1, .25), (0, 0) and (1, .25).

Sketch the second graph shifted 1.75 units in the x direction and 1.75 units in the y direction.

What are the three basic points of this graph?

The three basic points (-1, .25), (0, 0) and (1, .25) will

shift 1.75 units horizontally to give us (.75, .25), (1.75, 0) and (2.75, .25), then
shift vertically 1.75 units to give us (.75, 2), (1.75, 1.75) and (2.75, 2).

If f(x) = x^2, then what are

f(x- 1.75), f(x) + .75, .25 f(x) and .25 f(x- 1.75) + .75?

We get

f(x-1.75) = (x-1.75)^2
f(x) + .75 = x^2 + .75
.25 f(x) = .25 x^2 and
.25 f(x-1.75) + .75 = .25 (x-1.75)^2 + .75.

Problem Number 5

At clock times 49.4, 74.10001, 98.8 and 123.5 sec, we observe water depths of -57, -93.2, -117.2 and -129 cm.

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

This is done in the same manner as #2.

Sketch a graph of depth vs. clock time.

Use a sketch to explain why the slope of this graph between 49.4 and 74.10001 sec represents the average rate at which depth changes during this time interval.

This requires a sketch indicating the graph points (49.4 sec , -57 cm) and (74.1 sec, -93.2 cm). The rise and run from one point to the other should also be indicated. The explanation goes something like this:

The rise from the first point to the second is -36.2 cm, indicating a change in depth of -36.2 cm.
The run from the first point to the second is +24.7 sec, indicating a change in clock time of +24.7 sec.
The slope is rise / run, which represents change in depth / change in clock time = -36.2 cm / (24.7 sec) = -1.4 cm/sec, very approximately.
Since change in depth / change in clock time during an interval is average rate of change of depth with respect to clock time during that interval, the slope therefore represents the average rate at which depth changes during this time interval.

If f(x) = x², what are the vertex and the three basic points of the graphs of f(x- 1.25), f(x) - .85, 5 f(x) and 5 f(x- 1.25) + .85. Quickly sketch each graph.

The three basic points of the y = x^2 graph are (-1, 1), (0, 0) and (1, 1).

The graph of f(x-1.25) is shifted 1.25 units in the x direction, transforming the basic points to (.25, 1), (1.25, 0) and (2.25, 1). Note that the function is y = (x-1.25)^2.
The graph of f(x) - .85 is shifted -.85 units in the y direction, transforming the basic points of the f(x) = x^2 graph to (-1, .15), (0, -.85) and (1, .15). Note that the function is y = x^2 - .85.
The graph of 5 f(x) is vertically stretched by factor 5, transforming the basic points of the f(x) = x^2 graph to (-1, 5), (0, 0) and (1, 5). Note that the function is y = 5 x^2
The graph of 5 f(x-1.25) - .85 is first shifted 1.25 units in the x direction, transforming the basic points to (.25, 1), (1.25, 0) and (2.25, 1).

The graph is then vertically stretched by factor 5, transforming the basic points (.25, 1), (1.25, 0) and (2.25, 1) to (.25, 5), (1.25, 0) and (2.25, 5).

The graph is finally shifted -.85 units in the x direction, transforming the basic points (.25, 5), (1.25, 0) and (2.25, 5) to (.25, 4.15), (1.25, -.85) and (2.25, 4.15).

The function y = x^2 transforms to y = (x-1.75)^2, then to y = 5 ( x-1.75)^2 and finally to y = 5 ( x - 1.75)^2 - .85.