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Precalculus I Test 1________________________________________
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Test Problems:
Problem Number 1
Problem: Obtain a quadratic depth vs. clock time model if depths of 46.35971 cm, 41.86877 cm and 39.52721 cm are observed clock times t = 6.429146, 12.85829 and 19.28744 seconds.
Problem: The quadratic depth vs. clock time model corresponding to depths of 46.35971 cm, 41.86877 cm and 39.52721 cm at clock times t = 6.429146, 12.85829 and 19.28744 seconds is depth(t) = .026 t2 + -1.2 t + 53. Use the model to determine the average rate which depth changes between clock times t = 17.7 and t = 17.701 seconds.
The three points:
(6.429,46.359)
(12.858,41.869)
(19.287,39.527)
The three equations:
46.359=a(6.429)^2+b(6.429)+c
41.869=a(12.858)^2+b(12.858)+c
39.527=a(19.287)^2+b(19.287)+c
Which simplifies to:
46.359=41.332a+6.429b+c
41.869=165.328a+12.858b+c
39.527=371.988a+19.287b+c
Subtract the 1st equation from the second and you get:
41.869=165.328a+12.858b+c 46.359=41.332a+6.429b+c
which equals 4.49=123.996a+6.429b, because the cs cancel out
Then subtract the 2nd equation from the 3rd:
39.527=371.988a+19.287b+c 41.869=165.328a+12.858b+c, which equals
-2.342=206.66a+6.429b, multiply this equation by 1 and get
2.342=-206.66a-6.429b
Eliminate b by:
-4.49=123.996a+6.429b plus 2.342=-206.66a-6.429b and you get 2.148=-82.664a, which equals .02598=a or a=.026
The you solve for b by substituting a in and you get:
-4.49=123.996 (.026)+6.429b
-4.49=3.223896+6.429b
-7.713896=6.42b
-1.2=b
Then you substitute in a and b to get c:
46.359=41.332 (.026)+6.429 (-1.2) +c
46.359=1.07463-7.7148+c
46.359=-6.64+c
52.999=c or c=53
The model is depth (t)=.026t^2+-1.2t+53
Average rate:
Y=.026(17.7)^2-1.2 (17.7) +53
Y=.026 (313.29)-1.2 (17.7)+53
Y=8.14554-21.24+53
Y=39.91 (approx.)
Y=.026 (17.701)^2-1.2 (17.701)+53
Y=8.14646-21.2412+53
Y=39.91 (approx.)
Dy/dt in cm/sec, find slope by y2-y1/x2-x1:
39.90526-39.90554/17.701-17.7=0/.001=0
this doesnt seem right to me, where did I go wrong?
You didnt use enough significant figures in your numerator.
However, note that 39.90526-39.90554 = -.00028, not 0. This would give you an average rate of -.28, not 0.
Problem Number 2
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 9.788, 13.62, 17.14 and 19.399 , respectively, y takes values 6, 12, 18 and 24. Which power function best fits this data?
First, I made a table consisting of the x and y values
Y=ax^.5
6=a(9.788)^.5
6=3.129a
a=1.92
12=a(13.62)^.5
12=3.691a
a=3.251
18=a(17.14)^.5
18=4.14a
a=4.348
Y=ax^-.5
And I got the following:
For 6, a=18.77
For 12, a=44.28
For 18, a=74.53 and
For 24, a=405.73
Y=ax^2
For 6, a=.063
For 12, a=.065
For 18, a=.061
For 24, a=.064
Y=ax^-2
For 6, a=600
For 12, a=2222.222222
For 18, a=5294.1
For 24, a=8000
So, y=ax^2 best fits the data because all of the answers are close together.
Problem Number 3
If Q(t) = Q0 * .879 t, estimate to 2 significant figures the time necessary for the original quantity to dwindle to .01 of its original value.
.879*.01=.00879, this is .01 of the value .879
So the time necessary for the quantity to dwindle to .01 of its original value, would be .01. I got you e-mail Mr. Smith, thats o.k, Technology is great when it works, but those things happen. So on this, I did what you said and continued to narrow my search for the right exponent. I found this,
.879*36.7=.00879
Do you mean .879 ^ 36.7 = .00879?
That would be correct. However .879 Q0 is the value when t = 1, not when t = 0. The value .00879 Q0 would occur at t = 36.7. So the time required to fall to .01 of the initial value .879 Q0 would be the time from t = 1 to t = 36.7, which is 35.7.
Easier to start with the t = 0 value, which is Q0 * .879^0 = Q0. Then you just have to fall to .01 Q0. Using the same methods you find that the time required is 35.6, in agreement with the preceding.
Problem Number 4
Explain why the key points on a graph of a linear function y = mx + b are (0, b) and (1, b+m).
The basic points of a linear function are the y-intercept (0,b) which is the point where x=0 and the point one unit to its right, which is x=1. Slope is rise over run=m, so the run is 1 and the rise (or y coordinate) is b+m. The y-intercept is when x=0 and the x-intercept is when y=0.
Problem Number 5
Find the first 4 terms of the sequence defined by a(n) = a(n-1) + -4/n, a(0) = -2.
a(0)=-2
a(1)=a(-2-1)+-4/-2=1
To get the formula for a(1) you have to use n = 1, which gives you
a(1) = a(1 1) 4 / 1, or
a(1) = a(0) 4 / 1 so that
a(1) = -2 4 = -6.
a(2)=a(-1-1)+-4/-1=2
To get the formula for a(2) you have to use n = 2, which gives you
a(2) = a(2 1) 4 / 2, or
a(2) = a(1) 4 / 2 so that
a(2) = -6 2 = -8.
a(3)=a(2-1)+-4/2=-1
To get the formula for a(3) you have to use n = 1, which gives you
a(3) = a(3 1) 4 / 3, or
a(3) = a(2) 4 / 3 so that
a(3) = -8 4/3 = -9 1/3 or -28/3.
Continue the process to find a(4).
a(4)=a(-1-1)+-4/-1=2
I can see that this is linear because it repeats itself.
Problem Number 6
Solve using ratios instead of functional proportionalities:
If a sand pile 2 meters high has a mass of 4160 kg, then what would we expect to be the mass of a geometrically similar sand pile 8.2 meters high?
If there are 1.24 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?
M=kh^3
So, 4160=k(2)^3
4160=k(8)
If you divide these, you would get k=520, this cant be right, because a sand pile that is 8.2 m high would have more volume that one that is 2m high.
k isnt the volume of the pile, its the proportionality constant.
Since k = 520 you have M = 520 * h^3. Substitute h = 8.2 and you get the considerably greater mass of the second pile.
So then I thought you may cube each side, but that gives you 7.2. So then I tried to cross multiply and got:
4160/x=2/8.2
2x=34,112
x=17.056 which seems more rational. But then I realized that I may have done the above wrong. Maybe I should solve for M, mass and put in 4160 for k and 2 for h. So I got,
M=kh^3
M=(4160)(2)^3
M=(4160)(8)
M=33,280kg
Could you please let me know the correct way to use this equation?
Youve got the right ideas. Hopefully my note will clarify this method.
Your equation 4160/x=2/8.2 is also a right idea, but mass is proportional to the cube of height so the equation should read
(4160/x)=(2/8.2)^3. Solve this for x and you will get the same result you got using M = 520 h^3.
This solution method is actually the one requested, which uses ratios. The previous solution uses functional proportionalities.
The second part of the problem.
N=kh^2, so
N=(33,280)(8.2)^2
N=(33,280)(551.368)
N=18,349,527.04
But this is not rational either, because if the first one had 1.24 billion, then this is not enough.
The proportionality is correct. Correct your procedure using functional proportionalities, as before, and you should get a reasonable answer. Then use ratios, using the squares of the ratios.
Problem Number 7
Given the depth vs. clock time function y = f(t) = .02 t2 + -2.07 t + 63, with depth in cm when clock time is in seconds, find the clock time when water depth is 31.43875 cm. Using the same function determine the depth at clock time t = 10 sec. Find t such that f(t) = 59.43875. Find f(t) when t = 24.
I tried to find the clock time when water depth was 31.43875
31.43875=.02t^2-2.07t+63 I wanted to factor this, but realized I could not, then I tried to use the quadratic formula, but ended up with a negative under the square root sign.
Factoring is a good idea but with the given numbers its very unlikely you will find the factors.
In any case you have to first put the equation into the form a t^2 + b t + c = 0; do this by subtracting 31.4 from both sides. It is still possible that you will get a negative discriminant, in which case you would say that there is no solution and the depth cannot reach 31.4. However it is also possible that you will get two solutions, in which case you have to use common sense to determine which one actually applies to the depth vs. t situation.
For t=10 sec.
Y=.02(10)^2+-2.07(10)+63
Y=.02(100)+-2.07(10)+63
Y=2-20.7+63
Y=44.3
For t=24 sec.
Y=.02(24)^2-2.07(24)+63
Y=.02(576)-2.07(24)+63
Y=11.52-49.68+63
Y=24.84
With t=59.43875
The problem does not say that t = 59.4. The problem says that f(t) = 59.4, so your equation should be 59.4 = =.02t^2-2.07t+63.
Y=.02(59.43875)^2-2.07(59.43875)+63
Y=.02(3532.965)-123.03821+63
Y=70.6593-60.03821
Y=10.6
Given this information, I know that for depth 31.43875, that time must lie somewhere between t=10 and t=24. I tried t=15 and got a depth of 36.45, so I tried t=17 and got a depth of 33.59.
For t=18, y=32.22
For t=19, y=30.89, so I know that for depth 31.43875, t must be between 18 and 19 seconds. So I tried 18.5 and got a depth of 31.55. This is a very long, hard way to do. Id love to know how to solve this an easier way. Help!
Problem Number 8
The graph below depicts the length of a spring vs. the weight suspended from it. Construct a good straight-line model for this data and use your model to determine the length of the spring when weight 38 grams is suspended. Determine also the weight required to achieve length 17 cm. The x axis is marked in units of 8 grams and the y axis in units of 17 cm, both starting at (0,0).
Interpret both the x and the y intercepts of this graph.
For whatever reason, I could not access the graph. I selected all, copied and pasted this test into Word, but the graph would not even show up on the website from which I printed it. I tried to right click and Show Picture, but it still would not come up.
You could put your own scale and straight line on the graph and give it a try; or you could simply select to omit the problem, in which case it would not be counted against you.
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