There are logically correct questions that no human being can comprehend. Can anyone write one down?
How do we get a graph of flow rate in cm^3 / sec vs. clock time in seconds from our depth vs. clock time information?
Find the volume of the cylinder that done went missin. Divide the volume of the water that ain't there no more in the container by the time it took it to done go missin and you find the average rate at which water flowed out.
How do we find the area of a trapezoid?
We multiply width be average height. Think of changing it into a rectangle of equal area (you get a triangle of area equal to that that done went missin).
y vs. x means the same thing it means when you make a table with x values in the first column and y values in the second. You graph y vs. x by putting second-column values on the y or vertical axis and first-column values on the x or horizontal axis.
Label your axes and your graphs
Given rate information, length change: is 50 cm supposed to change by 5 cm or what ???
For your depth vs. clock time data what is the volume change corresponding to each of your intervals? What is the average rate at which volume changes on each interval?
See above.
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Assuming that the water exiting your container was filling a tube whose
cross-sectional area is .27 cm^2, on each of your time intervals at what average
rate would the depth of water in the tube be rising?
Divide the volume change by the .27 cm^2 cross-sectional area.
That gives you the change in height of water in the tube (see previous notes).
Divide the change in the height in the tube by the time required.
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Give your data for flow range vs. clock time. Copy your trapezoidal
approximation graph of flow range vs. clock time, including labeling, onto your
paper.
Be sure to label the graph. It ain't a trapezoidal approximation graph if it ain't completely labeled.
c&b
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Between two pendulum lengths, one 17 cm greater than the other, the rate at
which period changes with respect to length is increases from .0125 to .0111 sec
/ cm. By how much does the period of the pendulum change between the two
lengths? On your paper make a sketch of the situation, and a sketch of a graph
of rate of change of period with respect to length vs. length.
The rate changes from .0125 sec/cm to .0111 sec/cm. Length changes by 17 cm.
Change in period is
change in period = rate of change of period with respect to length * change in length.
We know change in length. We need the ave rate of change of period with respect to length.
We have the .0125 sec/cm and the .0111 sec/cm.
Approximate average rate is (.0125 sec/cm + .0111 sec/cm) / 2 = .0118 sec / cm.
Now change in period = ave rate * change in length = .0118 sec / cm * 17 cm = .2 sec approx.
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Find the first three differences of the sequence 1, 2, 3, 5, 8, 13, 21, 34, 55,
89. What does the first difference tell you about the graph of the sequence?
What does the second difference tell you?
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Make a sequence of the ratios of the sequence 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
The first member of the sequence is 2/1 = 2, the second is 3/2 = 1.5, the third
is 5/3=1.666.., etc.. Find the first and second differences of this new sequence
of ratios. What do the first and second differences tell you about the graph of
the sequence of ratios?
Sequence of ratios is 2, 1.5, 1.666.., 1.60, 1.625, etc.
First difference of sequence of ratios is
-.5, +.1666..., -.0666.., .025, ...
This first-difference sequence goes from negative to positive to negative to positive ... . The original sequence is therefore alternately increasing and decreasing.
Magnitudes of first differences are decreasing. So the up-down behavior is getting less and less drastic.
The graph is sorta tryin' to smooth out.
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Sketch and label a trapezoidal approximation graph of velocity vs. clock time,
with velocities 1, 2, 4, 7, 11 cm/s at clock times 0, 3, 6, 9 and 12 seconds.
Calculate the slopes of each trapezoid and the area of each. Label slopes as
usual; label areas as circled numbers in the middle of each trapezoid. Interpret
the meaning of the slope of a trapezoid. Interpret the meaning of the area of a
trapezoid.
To get areas we multiply ave ht by width.
approx. ave. rate of change of position with respect to clock time
and
change in clock time
which gives us
change in position.
To get slopes we divide rise by run
A quadratic function y of clock time is of the form
Suppose our depth vs. clock time information tells us that depths are 20.0, 17.0, 14.5, 12.5, 11.0, 10.0 and 9.5 cm at clock times t = 0, 5, 10, 15, 20, 25 and 30 seconds.
Pick three data points (t, y) and plug the t and y coordinates into the form of a y vs. t quadratic function. You will get three equations.
25a + 5b + c = 17
225 a + 15 b + c = 13
625 a + 25 b + c = 11
What are your three equations? What are the unknowns in these three equations?
The unknowns are a, b and c.
What do you get when you subtract the first equation from the second?
200a + 10b= -4
What do you get when you subtract the first equation from the third?
600 a + 20 b = -6.
What system of equations do you now have, and what variable has been eliminated?
c has been eliminated to give us
200a + 10b= -4
600 a + 20 b = -6.
Solve this system of two equations by elimination. What is your result?
Starting with
200 a + 10 b = -4
600 a + 20 b = -6
multiply the first equation by -2 to get
-400 a - 20 b = 8
600 a + 20 b = -6.
We get
200 a = 2 so
a = .01.
Now we substitute .01 for a in one of the equations. Plugging into 200 a + 10 b = -4 we get
200 * .01 + 10 b = -4. We easily solve for b to get
b = -.6
Substitute your results back into the first of your three original equations and solve for the remaining unknown. What do you get?
The first equation is 25 a + 5 b + c = 17. Substituting a = .01 and b = -.6 we get
25 * .01 + 5 * (-.6) + c = 17. We easily solve for c to get
c = 19.75.
Substitute your values of a, b and c into the form y = a t^2 + b t + c. For the clock time corresponding to your first data point find the value of y predicted by this equation.
Our solution is a = .01, b = -.6, c = 19.75. Substituting into y = a t^2 + b t + c gives you
y = .01 t^2 - .6 t + 19.75.
This function gives us depth y as a function of clock time t. If we plug in a value of clock time t we'll get the corresponding depth y.
Note that for the entire data set Excel gives us the function
y = 0.01t^2 - 0.65 t + 20
Find the values of y for the other clock times, according to the equation.
The other clock times were t = 0, 10, 20 and 30.
Plugging into y = .01 t^2 - .6 t + 19.75 we get y values
19.75, 14.75, 11.75, 10.75.
These compare pretty well to the values
20.0, 14.5, 11.0, 10.0.
By how much does each value differ from the observed value?
Sketch the function on the same set of coordinate axes as the graph of the original data. How close are the two graphs? How well does the function model the behavior of the actual system?
According to the function at what instant would the depth be 15 cm?
According to the function what is the average rate of change of depth between t = 15 sec and t = 15.1 sec?