course Mth 173
7/1 1
Time and Date Stamps (logged): 12:18:01 07-01-2009 °±Ÿ°•Ÿ¯°¯¶Ÿ¯°Ÿ±¯¯¸ Applied Calculus I Major Quiz
________________________________________
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
Problem: The quadratic depth vs. clock time model corresponding to depths of 41.26766 cm, 20.8432 cm and 8.726604 cm at clock times t = 12.1801, 24.3602 and 36.276-739-2590. 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.s is depth(t) = .028 t2 + -2.7 t + 70.
t=12.1801 cm = 41.26766
t= 24.3602 cm = 20.8432
t= 36.376 cm = 8.726604
• Show the system of equations we would solve to get this model.
y= .028 (12.1801)^2 + -2.7(12.1801) + 70 = 41.26766
y= .028 (24.3602)^2 + -2.7(24.3602) + 70 = 20.8432
y= .028 (36.376)^2 + -2.7(36.376) + 70 = 8.726604
• Then use the model to determine whether the depth will ever reach zero.
• According to the depth function what is the average rate at which depth changes between clock times t = 34.4 and t = 134.4 seconds?
( 212.89 - 10.25) / (134.4 – 34.4) = 2.03
• What is the instantaneous rate of depth change at each of these clock times?
y` = .056t – 2.7
(4.8264 +.7736) / (134.4 – 34.4) = .056
• How does the average of the instantaneous rates compare with the average rate, and why should the comparison be as it is?
.
.
.
.
.
.
.
.
.
.
Problem Number 2
Problem: At what average rate does the exponential principle function P(t) = 13 * 1.16 ^ t change between clock times 2.7 and 2.7001?
2.7 = 19.40796155
2.7001 = 19.40824961
AVERAGE RATE= 19.40810558
These are the values of the function, not the average rate of change between the given clock times.
The average rate is (change in value) / (change in clock time).
• Write the same expression for the principle function P(t) = P0 * (1 + r) ^ t.
• Using the laws of exponents simplify your expression as much as possible.
P(t) = 13 * (1 +.16) ^ t
Problem: If it takes 9 million level tablespoons of sand to build a sandpile 4 meters in diameter, then what function gives the number of tablespoons of sand required to build a geometrically similar sandpile as a function of diameter in meters?
• How many tablespoons what therefore take to increase the diameter of this sandpile to 4.00001 meters?
• At what average rate is the number of tablespoons of sand changing with respect to diameter between these two diameters?
.y = ax^2
(9) = a(4)^2
a = .5625
so far so good with the basic procedure.
You would substitute a in your original equation y = a x^2 and conclude that
y = .5625 x^2.
Then you could substitute x= 4.00001 meters for x and find your new value of y.
(9) = a(4.00001) ^2
when x = 4.00001, the value of y will no longer be 9. a would be the unchanging quantity.
It's a that is constant, not y.
a =.56249
You were fine up to a point.
However the correct proportionality for volume is y = k x^3, not y = k x^2.
You also need to find the average rate of change.
.
.
.
.
.
.
.
.
Problem Number 3
The depth vs. clock time function y = .029 t2 + -1.5 t + 88 indicates the depth y of water in a certain uniform cylinder at clock time t.
• At what average rate does depth changes between clock times t = 6 and t = 12?
y = .029 t2 + -1.5 t + 88
t=6 => 80.044 cm
t=12 => 74.176 cm
74.176-80.044/ 12-6 = -.978
good
• What clock time lies midway between t = 6 and t = 12, at what rate is depth changing at this instant?
t = 9
y` = .058t -1.5 = -.978
good
What is the function that represents the rate r of depth change at clock time t?
y =at^2 + bt + c
y` = 2at + b
• Evaluate this function at the clock time halfway between t = 6 and t = 12.
y` = .058t -1.5 = -.978
If the rate of depth change is given by dy/dt = .021 t + -2.6 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 6 and t = 12? .
.021(6) + -2.6 = -2.474
.021(12) + -2.6 = -2.348
-2.348 – 2.474/ 12-6 = .021
-2.474 and -2.348 are the instantaneous rates of depth change at t = 6 and t = 12.
-2.348 – 2.474/ 12-6 = .021 represents the change in the rate, divided by the change in clock time. By the definition of average rate, (change in rate) / (change in clock time) is the average rate of change of the rate with respect to clock time.
To get change in depth, you would multiply the average rate of depth change by the change in clock time.
You have two instantaneous rates of depth change. What therefore is your best estimate of the average rate of depth change? What therefore is the change in depth?
• Give the function that represents the depth. What would this specific function be if at clock time t = 0 the depth is 150?
.
.
.
.
.
.
.
.
Problem Number 4
Sketch and completely label a trapezoidal approximation graph for the function y = 2 x/ 6, for x = 0 to 2.7 by increments of .9.
x | y
0 1
.9 1.1
1.8 1.2
2.7 1.4
.
.
.
.
.
.
.
.
.
Problem Number 5
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
I DIDN’T UNDERSTAND ALL OF THIS PART. WHICH NOTES ARE THESE?
• If a concrete sphere 3.1 meters in diameter has a mass of 905646.3 kg, then what would we expect to be the mass of a concrete sphere 8.1 meters high?
• If it requires 2.0181 liters of paint to cover the first sphere, how many liters will be required to cover the second?
35. x( 3x^3 – 375)
3. -14.5
These notes are in the Modeling Project on proportionalities.
The proportionalities here are the y = k x^3 proportionality for volumes and the y = k x^2 proportionality for areas.
See my notes. You might well have additional questions, and I'll be glad to answer them if you do.