If the average rate of depth change in a cylinder is -.3 cm / second at the instant t = 8 seconds, and .1 cm/second at t = 14 seconds, then what is your best estimate of how much depth changes between these two clock times?
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I don't know a formula to use in calculating this problem. I understand what it is asking, just not how to respond to it!

If water is flowing into a cylinder of radius 4 cm at the rate of 100 cm^3 / second, then at what rate is depth changing with respect to clock time?
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I don't understand what to do with the radius in this problem, or even why it is good to know it.

** Remember the concept that the volume of a cylinder is cs area * altitude, and the formula (one of three formulas we can't just reason out) A = pi r^2.  **

For the depth vs. clock time situation of the preceding problem (uniform cylinder of radius 4 cm are depths 25, 18, 13, 10 cm at clock times 0, 5, 10 and 15 seconds, and if you didn't have all that written down get in the habit because you won't get a reminder next time) how many cm^3 of water flowed out between the 25 cm depth and the 18 cm depth? How much water flowed out between t = 5 seconds and t = 10 seconds? At what average rate, in cm^3 / second, was water flowing out between t = 10 s and t = 15 s? At what average rate, in cm^3/second, was water flowing out during the first time interval?
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I don't know how to get how many cm^3 from cm for how much water flowed out between 25 cm and 18cm.

** The volume flowing out is the volume of a cylinder of height 25 cm - 18 cm = 7 cm and radius 4 cm. **

Between 5 and 10 seconds, 5 cm flowed out of the cylinder.

** 5 cm isn't a volume.  Volume is flowing out of the cylinder, not depth. **

 

I also don't know how to get the average rate in cm^3/seconds between 10 and 15 seconds, which is 3 cm or during the first interval, which is 7 cm.

 

 

When I divided 1 by 43, I got 0.023255814...

The graph goes up overall, but at some points, it is scattered in different places. So no...my graph is not evenly distributed.

If the water flowing out of the container between t = 5 sec and t = 10 sec flowed into a tube whose cross-sectional area is .1 cm^2 how long would the water column in the tube be? How many cm of the tube would therefore fill, per second?
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I don't understand this question. I do understand that you have water flowing out a the same container as in the previous problems. So this means that the raduis of this container is 4 cm. I also understand that you're trying to figure out how much water would flow into the tube whose cross-secional area is .1 cm^2 between t=5 sec and t=10 sec. To solve this question you would have to find the volume of the tube. I don't know how to find the volume of the tube without having the radius and I'm not sure of how to find the radius.

 

At what average rate does the period of a pendulum change with respect to length between lengths of 9 cm and 16 cm? At what average rate does it change between lengths of 16 cm and 25 cm? What is your best estimate, based only on these results, of how much the period changes between lengths of 15 cm and 17 cm? How accurate do you think this estimate will be (e.g., within .03 seconds, or within .0007 seconds, or whatever you think; explain your thinking)?
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The avg rate that the period of a pendulum changes between 9 and 16 cm is about .02857 s/cm

The avg. rate that the period changes between 16 and 25 cm is .02222 s/cm

I think the period changes between lengths 15 and 17 cm by about .0250 s/cm. It should be accurate to about .0001 because the the change in distance is not very much, which allows it to be very accurate.

What do the numbers .8, 1, 1.2, 1.4 represent on the trapezoidal graph just sketched?

On trapezoidal graphs we use plain numbers, no squares or circles or anything, to represent the lengths of lines.  The given numbers represent the lengths of vertical lines.  The vertical direction represents period in sec, so these numbers represent periods in seconds. 

 

 

What information does the graph give us relevant to the calculation of the average rate of change of the period with respect to length between lengths 16 cm nd 25 cm?

First, what does ave rate of change of period with respect to length mean?

To find this rate we need to find difference in period and difference in length.

The graph tells us that at 16 cm the period is .8 s and at 25 cm the period is 1.0 sec.

The change in period is therefore 1 s - .8 s = .2 s and change in length is 25 cm - 16 cm = 9 cm.

So we have

This means that

We label slopes with boxes around them.

 

What are the changes in the periods for the other two intervals shown?

Periods go from .8 to 1 to 1.2 to 1.4 s.  The change in period for any interval is .2 sec.

What are the changes in the lengths for the other two intervals shown?

The change on the second interval is 36 cm - 25 cm = 11 cm; on the third interval it is 49 cm - 36 cm = 13 cm.

What are the rates of change of periods with respect to the lengths for the other two intervals shown?

The rates are .2 sec / (11 cm) = .0182 sec/cm and .2 sec / 13 cm = .0154 sec / cm, both approximate.

How are these rates labeled on the trapezoidal graph?

The rates are labeled with boxes next to the slope segment.

What aspect of a trapezoid is the rate?

The rates are represented by the slopes.

Why?

Slope is rise / run and rise is change in the period, run is change in the length.  So slope represents change in period / change in length.

 

Suppose that the rate at which the depth of water in a container changes is -.2 cm/s * sqrt(y), where y is the depth in cm. If the depth is initially 100 cm then at what rate is the depth changing at that instant? If this rate holds for the next 10 seconds then how much does the depth change during this 10 seconds? What will then be the depth at that instant and at what rate will the depth be changing? Is there any inconsistency in these results?
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RESPONSE -->
Depth at 100 cm - rate of change = -2 cm/s

The depth will go down -20 cm/s.

Depth at that instant 80 cm.

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Suppose the depth vs. clock time results for a uniform cylinder of radius 4 cm are depths 25, 18, 13, 10 cm at clock times 0, 5, 10 and 15 seconds. Find the average rate of depth change with respect to clock time for each of these intervals and sketch and label a trapezoidal approximation graph representing depth vs. clock time.
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What does the slope -1.4 on the depth vs. clock time graph tells us and why?

Slope is equal to rise over run, or change in depth / change in clock time. 

It means the average rate of change of depth with respect to clock time is -1.4.  The units are units of rise / units of run, or in this case cm / sec.

For the depth vs. clock time situation of the preceding problem (uniform cylinder of radius 4 cm are depths 25, 18, 13, 10 cm at clock times 0, 5, 10 and 15 seconds, and if you didn't have all that written down get in the habit because you won't get a reminder next time) how many cm^3 of water flowed out between the 25 cm depth and the 18 cm depth? How much water flowed out between t = 5 seconds and t = 10 seconds? At what average rate, in cm^3 / second, was water flowing out between t = 10 s and t = 15 s? At what average rate, in cm^3/second, was water flowing out during the first time interval?

Between the 25 cm and 18 cm depth we lost a 7-cm-high cylinder of water whose radius is 4 cm.  Its volume is therefore

Vol = area of cs * height,

with area of the cs being pi r^2 = pi * (4 cm)^2 = 16 pi cm^2, or approximately 50 cm^2.  The volume is therefore

volume between 25 cm and 18 cm = 50 cm^2 * 7 cm = 350 cm^3, approx.

Between t = 5 and t = 10 sec depth changes by 13 cm - 18 cm = -5 cm so the container loses a cylinder of altitude 5 cm, cs area 50 cm^2 so the amount flowing out it 50 cm^2 * 5 cm = 250 cm^3.

Similar reasoning shows that during the third interval the loss is 150 cm^3.

The average rate at which the contents of the cylinder changes during the first time interval is

ave rate of change of vol with respect to clock time = change in volume / change in clock time = -250 cm^3 / (5 sec) = -50 cm^3 / s.

So the rate at which water is leaving is 50 cm^3 / s.

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Sketch a trapezoidal approximation graph depicting the volume of water in the container vs. clock time and label the slopes. What is the meaning of the slopes?
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If the water flowing out of the container between t = 5 sec and t = 10 sec flowed into a tube whose cross-sectional area is .1 cm^2 how long would the water column in the tube be? How many cm of the tube would therefore fill, per second?

Hints: 


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What did you get when you divided 1 by 43 in your homework? Describe the graph you got when you paired the digits and plotted them. Is your graph evenly distributed?
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At what average rate does the period of a pendulum change with respect to length between lengths of 9 cm and 16 cm? At what average rate does it change between lengths of 16 cm and 25 cm? What is your best estimate, based only on these results, of how much the period changes between lengths of 15 cm and 17 cm? How accurate do you think this estimate will be (e.g., within .03 seconds, or within .0007 seconds, or whatever you think; explain your thinking)?
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If the rate of depth change in a cylinder is -.3 cm / second at the instant t = 8 seconds, and -.1 cm/second at t = 14 seconds, then what is your best estimate of how much depth changes between these two clock times?

What could we do to make the problem easier to understand? One answer is 'draw a picture'.  Another is 'draw a graph'.

From the picture, suggestions:

Good idea but we aren't given the depths, we're given the rates.

Be sure you identify what sort of information you've been given.

How do we do this?  We can note that we are given two rates of change and we want an average rate of change.  We can average the two rates to get an approximate average rate.

The two rates are -.3 cm/s and -.1 cm/s.  The average of these rates is -.2 cm/s.

Change in clock time is 6 sec; at -.2 cm/s the change is therefore -.2 cm/s * 6 s = -1.2 cm.

A graph of | rate | vs. clock time gives us a trapezoid with altitudes .3 and .1; average altitude is .2, representing .2 cm/s and width is 6, representing 6 seconds.

 

 

 

 

 

Hint:  You are given two rates.  How do you use two rates to find the average rate?  Once you find the average rate how do you find the change in depth over the given time interval?

 
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Suppose that the rate at which the depth of water in a container changes is -.2 cm/s * sqrt(y), where y is the depth in cm. If the depth is initially 100 cm then at what rate is the depth changing at that instant? If this rate holds for the next 10 seconds then how much does the depth change during this 10 seconds? What will then be the depth at that instant and at what rate will the depth be changing? Is there any inconsistency in these results?
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If water is flowing into a cylinder of radius 4 cm at the rate of 100 cm^3 / second, then at what rate is depth changing with respect to clock time?
 

 

Choose 4 questions and write up the sort of response you think people would need if these questions were in q_a_ format.

You will receive bonus credit on the test in inverse proportion to the number of people who give good solutions to each of the problems to which you give a good solution.