course Math 163
I used the simulated data for my flow experiment. I hope I did it correctly. If not, I will re-do it until it is right. I worked it out on paper and then copied it to the computer, so if the numbers don't match it is just an error on my part. It gets difficult sometimes to copy the exact number.
Flow Experiment, Simulated Data________________________________________
The following are simulated depth vs. clock time data for the flow experiment:
clock t 2.8 5.6 8.4 11.2 14 16.8
depth 82.1 76.8 72.7 69.8 66.8 63.7
Clock times are in seconds, depths in cm.
Using a graph and simultaneous linear equations, find a model for this data set and compare your model with the data.
I used simulated data from the website to do my flow experiment. I am a little confused on exactly what to do, so I used the DVD experiment as a guide. I know that here it is asking for a linear equation and the example on the DVD is a quadratic equation. If this is wrong, I will re-do it the correct way after I learn my mistakes.
First, using the quadratic formula y=at^2 + bt + c I chose the points from the graph of clock vs. depth
(2.8, 82.1) (8.4, 72.7) (14, 66.8) to get the following formula
a(2.8)^2 + b(2.8) + c first solving the exponent and working in order of operation
a(7.84) + b(2.8) + c
7.84a + 2.8b + c = 82.1
The next problem:
a(8.4)^2 + b(8.4) + c
a(70.56) + b(8.4) + c
70.56a + 8.4b + c = 72.7
The last one:
a(14)^2 + b(14) + c
a(196) + b(14) + c
196a + 14b + c = 66.8
Next we solve by elimination. Choosing to eliminate “c” I subtracted (2) – (1)
( 70.56a + 8.4b + c = 72.7) – ( 7.84a + 2.8b + c = 82.1) gives us:
62.72a + 5.6b = -9.4
the next step would be to subtract (3) – (2)
(196a + 14b + c = 66.8) – (70.56a + 8.4b + c = 72.7) gives us:
125.44a + 5.6b = -5.9
??On the above, I’m not sure if I put the parenthesis in the right place when subtracting the two since there was an equal sign inside of the parenthesis??
The next step is to eliminate “b” by multiplying the coefficient of b * (1st answer) and multiplying –b * (2nd answer)
5.6(62.72a + 5.6b = -9.4)= 351.232a + 31.36b = -52.64
-5.6(125.44a + 5.6b = -5.9) = -702.464a – 31.36b = 33.04
As you can see the 31.36b when subtracted from the same number cancels “b”
the next step is to solve for “a”
(351.232a = -52.64) – (-702.46a = 33.04) to get
-351.232a = -85.68 dividing both sides by -351.232
a is approximately equal to 0.244
Now we can substitute a for 0.244 in the equation
62.72(0.244) + 5.6b = -9.4 multiplying and then subtracting from both sides you get
5.6b = -9.4-15.30368 dividing both sides by 5.6
b is approx equal to -4.411
Since we have the value of a and b, we can easily solve for c by substituting the values of a and b back into the original quadratic equation
y = at^2 + bt + c
62.72(0.244) + 5.6(-4.411) + c = -9.4
15.304 – 24.7016 + c = -9.4
-9.3976 + c = -9.4
c = -0.0024
y = 0.244t^2 +(-4.411)t + (-0.0024)
The graph is decreasing at a decreasing rate.
"
The first time you do a problem of this nature you should include all the details, as you have done here.
For later solutions, provided you're doing things right (as you are), you could then abbreviate your solution a bit (e.g., 'substituting 14 and 66.8 we get 196 a + 14 b + c).
In any case your work looks great, as I would have expected.