#$&*
course Math 163
This is a practice test that I would like your feedback on. Submitted 10-21-10 @10:25pm
Time and Date Stamps (logged): 20:54:20 10-20-2010 ąŻ´łąŻ°ŻąŻąŻ°Ż Precalculus I Test 1
________________________________________
Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
________________________________________
Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).
Write on ONE SIDE of paper only
If a distance student be sure to email instructor after taking the test in order to request results.
Signed by Attendant, with Current Date and Time: ______________________
If picture ID has been matched with student and name as given above, Attendant please sign here: _________
Instructions:
Test is to be taken without reference to text or outside notes.
Graphing Calculator is allowed, as is blank paper or testing center paper.
No time limit but test is to be taken in one sitting.
Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..
Directions for Student:
Completely document your work.
Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.
Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.
Please write on one side of paper only, and staple test pages together.
Test Problems:
. . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
Problem Number 1
What values of A and b would we use to express Q(t) = 93 * 2^( -.41 t) in the form Q(t) = A b^t?
.Q(t) = 93 * 2^(-.41t)
Q(t) = 93 * (2^-.41)^t
Q(t) = 93 * .7526^t
A = 93 and b = .7526
.
.
.
.
.
.
.
.
.
Problem Number 2
Use the slope = slope formulation to find the linear function streamRange(t) for the range of the water stream flowing from the side of a uniform cylinder, if the stream range is 50 centimeters at clock time t = 76 seconds, and if the stream range changes by -4 centimeters over a period of 6 seconds. Use your function to find the clock time at which the stream range first falls to 17 centimeters.
.To find the clock time at which the stream range first falls to 17 cm, you first locate your three points. The first point is given which is (76, 50) and then we are given data to determine the second point. The steam ranges changes by -4 cm so you subtract 4 from your first point of 50 and the change takes 6 seconds so you add 6 to your first x point of 76. Therefore you second set of points is (82, 46). Next you are given your thirdThe first point is given which is (76, 50) and then we are given data to determine the second point. The steam ranges changes by -4 cm so you subtract 4 from your first point of 50 and the change takes 6 seconds so you add 6 to your first x point of 76. Therefore you second set of points is (82, 46). Next you are given your third y point of 17 cm and ask to find your third x point. Now you have the points (76,50), (82,46) and (x, 17). To find x you use the formula (y-y1)/(x-x1) = (y2-y)/(x2-x1) so therefore, we substitute the data into this equation. (x1y1) = (76,50), (x2y2) = (82,46), and (xy) = (?, 17)
(17-50)/(x-76) = (46-50)/(82 -76)
-33/x-76 = -4/6 Now multiplying each side by (x-76) will isolate the x to one side
-33/x-76(x-76) = -4/6(x-76) = -33 = -4/6(x-76) Now you divide each side by -4/6
-33 / -4/6 = -4/6(x-76)/-4/6
49.5 = x-76 Now you add 76 to both sides to find the value of x.
125.5 = x Therefore the clock time would be 125.5 when the stream range falls to 17 cm.
.
Well done, though you did have one sign error.
Note that the straight-line model gives you y as a function of x.
Your equation
(17-50)/(x-76) = (46-50)/(82 -76)
would generally be
(y-50)/(x-76) = (46-50)/(82 -76),
which would simplify to y = -2/3 x + 76/3.
This is your function model, where y is given as a function of x.
Solving for x you would get
x = -3/2 y + 152
For y = 17 this gives you
x = -3/2 * 17 + 152 = 125.5.
The question did ask that you use the function to solve the equation. As you can see this solution uses reasoning identical to yours, but keeps y as a variable until the end.
.
.
.
.
.
.
.
Problem Number 3
Solve for x, using the laws of exponents and showing each step:
8 ( 3 x / 3) ^ (- 3/ 4) = 21
x ^ 8 / 3 = 21
.8(3x/3)^(-3/4) = 21 First multiply each side both side by 1/8 to get rid of the 8
8(3x/3)^(-3/4) * 1/8 = 21 * 1/8 = (3x/3)^(-3/4) = 2.625 Next raise both sides to the -4/3 power = 3x/3 = 2.625^-4/3 - since 3/3 = 1 then we now have
X = 2.625^-4/3
X= .0070
X^8/3 = 21 If you raise both sides by 3/8 then you can isolate the x
X^8/3 ^(3/8) = 21^3/8
X = 21^3/8
X = 1157.6
The expression is 21 ^ (3/8).
Your expression 21*3 / 8 does mean 'raise 21 to the power 3 then divide by 8', but this calculation doesn't follow from taking the 3/8 power of both sides.
.
.
.
.
.
.
.
.
.
Problem Number 4
If a(n) = a(n-1) + b, with a(1) = 5, then if a( 380) = 0, what is the value of b?
.a(n) = a(n-1) + b with a(1) = 5
When n = 1
a(1) =a(1-1) + b
a(1) = a(0) + b
5=b
That would be correct if you knew a(0), and its value happened to be 0.
But if b = 5, it would then follow that a(380) = 1900, not 0.
a(2) = a(1) + b
a(3) = a(2) + b = a(1) + b + b = a(1) + 2 b
Then, reasoning similarly
a(4) = a(1) + 3 b
a(5) = a(1) = 4 b
etc..
until finally a(380) = a(1) + 379 b.
Since a(380) = 0, it follows that
a(1) + 379 b = 0
so that
b = -a(1) / 379 = -5/379.
.
.
.
.
.
.
.
.
.
Problem Number 5
Given the depth vs. clock time function y = f(t) = .02 t2 + -2.09 t + 69, with depth in cm when clock time is in seconds, find the clock time t such that f(t) = 21.39875 cm and find f(t) when t = 26 cm. Find the clock time when water depth is 40.39875 cm. Using the same function determine the depth at clock time t = 10 sec.
.Y = f(t) = .02t^2 Ź 2.09t + 69
21.39875 = .02^2 -2.09t +69 = .02t^2 2.09t +69 -21.39875= 02t^2 2.09t +47.6
2.09 + sqrt-2.09^2 4(.02*47.6) / 2(.02)
2.09 + sqrt 4.3681 3.808 / .04 = 2.09 +/- sqrt.5601/.04
(2.09 + .56)/.04 = 66.25
0r (2.09-.56)/.04 = 38.25 This is the answer because with the given data as 66.25 is not realistic
.02t^2-2.09t + 69 with t = 26 sec = 13.52 54.34 + 69 = 28.18 cm
40.39875 = .02t^2-2.09t + 69 = = .02t^2-2.09t + 28.60125
2.09 + sqrt-2.09^2 4(.02*28.6) / 2(.02)
2.09 + sqrt 4.3681 2.288 / .04 = 2.09 + sqrt.2.0801/.04
2.09 + 2.0801/.04 = 4.1701/.04 = 104.2525 sec
Or 2.09 - 2.0801/.04 = .0099/.04 = .2475 sec - This is the answer because with the given data as 104.25 sec is not realistic
.02t^2-2.09t + 69 with t = 10 sec = 2 20.9 + 69 = 50.1 cm
.
.
.
.
.
.
.
.
.
Problem Number 6
At clock time t = 13 years the population of a certain organism in an ecosystem is 68, in thousands, while at clock time a years the population is 38 thousand. Plot the corresponding points on a graph of population vs. clock time and determine the slope of the straight line segment connecting these points. Explain why this slope represents the rate at which the population changes over this time interval.
For the exponential function y = f(t) = 102 * 1.018t, determine the average rate of change of y with respect to t, between clock times t = 28 and t = 29.
In plotting the points on a graph for the first problem, the data given is (13, 68) and (a, 38). To find the slope I subtracted y2-y1 divided by x2-x1, therefore, I had the following:
38-68 / a-13 = -30 / a-13 This is a negative slope because I assumed the population decreasd.
Good, but you haven't explained the meaning of the slope.
The rise is the change in population, the run is change in clock time, so slope = rise / run = (change in population) / (change in clock time), which is the average rate of change of population with respect to clock time.
f(t) = 102 * 1.018^t substituting t = 28 in this function you have
f(t) = 102 * 1.018^28 = 102 * 1.647926297 = 168.09
f(t) = 102 * 1.018^29 = 102 * 1.67758897 = 171.11
171.11 168.09 = 3.02
.
.
.
.
.
.
Problem Number 7
A population starts at 4000 and grows at the rate of 7.5 percent per year.
What will be the population 1, 2, 3 and 10 years later?
What function P(t) gives population as a function of number t of years?
What are the growth rate and growth factor of this function?
.The growth rate of 7.5 percent per year = (1+7.5) = 8.5
7.5% is .075. So the function would be 4000 * 1.075 ^ t.
Your function would correspond to a 750% increase every year.
4000 * 8.5^t
1 yr = 34,000
2 yrs = 4000*8.5^2 = 289,000
3yrs = 4000*8.5^3 = 2,456,5000
10yrs = 4000*8.5^10 = 7.874976174
The P(t) function for population for t number years is
P(t)= PO* (1+r)^t = 4000 * (1+7.5) ^t = 4000 * 8.5^t
The growth rate is r = 7.5 so the growth factor is 1 + 7.5 = 8.5
.
.
.
.
.
.
.
.
.
Problem Number 8
Problem: The quadratic function y = .017 t2 + -1.7 t + 83 models depth vs. clock time. Find the vertex of the graph and the graph points 1 unit to the right and the left of the vertex. Find the zeros of the function, and indicate all these points on a graph of the function.
Problem: What function do we get if we vertically stretch the basic y = x 2 parabola vertically by a factor of .25, then shift it -1.7 units vertically? Sketch a graph showing the effect of the vertical stretch on at least four selected points on the entire graph, then do the same for the vertical shift.
To find the x vertex using this function you use b/2a = 1.7 / .034 = 50 and now that you know the value of x you substitute 50 for the value of t to find To find the x vertex using this function you use b/2a = 1.7 / .034 = 50 and now that you know the value of x you substitute 50 for the value of t to find the value of y vertex. Substituting 50 in the function y = .017(50^2) 1.7(50) +83 = 40.5. Therefore the vertex is (50, 40.5). Graphing points 1 unit to the right and left of the vertex, you have 49 and 51 substituting these values into the function the y points are 40.517.
That would work, but the y values 1 unit right and left will be a = .017 units higher, where a is the coefficient of t^2.
The point of this model is for you to be able to quickly locate the vertex and get some idea of how 'spread out' the parabola will be.
Therefore, those points are (49, 40.517) and (51, 50.517). Finding the zeros of the function you set y = 0 the result is 83 and it never passes through the x coordinate therefore there is no other zeros.
The function you get if you vertically stretch the basic y = x^2 vertically by a factor of .25 then shift it -1.7 unit vertically is y = .25x^2 1.7
The four selected points for the graph y = x^2: (1,1), (2,4), (3,9), (4,16) and the points for the stretch and shift are y = .25x^2 1.7: (1,-1.45), (2,-.7), (3, .55), (4, 2.3)
The basic points of the original graph are (-1, 1), (0, 0) and (1, 1). It is these points from which you should begin the series of transformations, which are direct and can be done without resorting to a calculator. The point is to be able to visualize and construct the transformations without resorting to technology or involved arithmetic, so that you fully understand the transformation process.
You're in good shape. See my notes and you should be in even better shape. Let me know if you have questions.