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Mth163
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missing qa's
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qa_16,qa_17,Qa-18,qa-19
are there qa_'s for these assignments because I clicked on them and they said path not specified ?????
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There are no more qa's after Assignment 15, just the worksheets and the Queries.
I don't see the qa's on either the Assignments Page or the Brief Assignments Page.
I want to be sure you're viewing the correct page. Can you send me a copy of the Address Box of the page where you see links to these qa's?
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Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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cubes
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Question: `q012. A sculpture is made of small cubes, about the size of playing dice.
Suppose each small cube could be cut into smaller cubes, each having edges 1/3 as long as the edges of the original. Suppose that somehow each of the smaller cubes is then expanded until it is the size of one of the original cubes, forming a larger sculpture. The shape of the sculpture would not change, only its size would be different. Don't worry about how this might be accomplished or even whether this is possible.
If this could be done, how many times higher would the larger sculpture be?
How many times as much volume would the larger sculpture occupy?
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Your solution:
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The same size of the original larger sculpture
The same
Because if you make it small and then expanded it again wouldn’t it be the same size since you are saying that the cube basically gets as big as it was before. But are the 1/3 edges included on how big it was before if not then we could take the orginal size*1/3 to get the new size to get volume
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The figure isn't shrunk in the prcess of subdividing the original cubes. Those cubes are cut into smaller cubes without changing the total volume.
The smaller cubes are now inflated to the size of the original cubes, so the sculpture now gets bigger.
By what factor does its volume change?
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so does the cube get 3 times as big as the original since you have a bunch of cubes cut into 1/3's then each of those small cubes are re-expanded to the size of a original cube???
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See if you can answer the questions below. You should either sketch this or cut up an actual cube to be sure you see what's going on here (you could cut up something like a potato or a piece of fruit).
What we're trying to visualize here is a cube being cut into smaller cubes, then each of the smaller cubes being expanded to the size of the original (of course you have to imagine the expansion).
Suppose you have a cube 3 feet on a side.
Into how many 1-foot cubes can it be cut?
Now if a 1-foot cube is expanded to form a 3-foot cube, by how many times does its volume increase?
If each of the 1-foot cubes into which you divided the original 3-foot cube was expanded back into a 3-foot cube, then how many times as much volume would you have?
So in general if a cube is cut into smaller cubes, each having edges 1/3 as long as the edges of the original, and each of the smaller cubes is then expanded until it is the size of the original cube, by how many times does its volume increase?
What are the implications for this sculpture?
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Mth163
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porportionalty
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Class Notes Precalculus I, 9/29/98
Proportionality and Sugar Piles
Today's quiz problem was to determine a linear force vs. displacement function for a pendulum, if it is known that a 5 in. displacement from a certain reference position results in 3 pound resisting force, tending to pull the pendulum back to the reference point, and also that a 9 in. displacement results of 7 pound resisting force.
We then wish to use this function to determine the force when the displacement is 15 in., and also the displacement when the force is 13 pounds.
To solve the problem it is very helpful to sketch or graph of the force vs. the displacement.
The two known force-and-displacement pairs to give us two points on this graph.
These points are (5 in., 3 lbs) and (9 in., 9 lbs); they are designated by (5, 3) and (9, 7) on the graph below.
We can easily draw the graph of the linear function through these two points.
The slope of this graph is the slope between the points (5 lbs., 3 in.) and (9 lbs., 7in. ); this slope is 1 lb / in..
If (x , y) is a point on this graph, then the slope (y - 7) / (x - 9) must be equal to this slope and we have the slope = slope form of the equation of the straight line: (y - 7) / (x - 3) = 1.
This equation is easily solve for y to obtain y = x - 2.
pc01.jpg
Since y stands for the force, and x for the displacement, we can rewrite the equation y = x - 2 in meaningful terms as
force = displacement - 2,
understanding that force will be in lbs. when the displacement is in inches.
Using this form of the equation it is easy to determine what to do define the force when the displacement is 15 inches.
Letting displacement = 15 we obtain
force = 15 - 2 = 13.
It is equally easy to determine what displacement corresponds to a force of 13. (Coincidentally, this is the force we found above for the displacement 15. We will proceed as if we didn't know this.)
When we substitute 13 for the force we obtain the equation
13 = displacement - 2,
which we easily solve for the displacement.
Of course we do obtain displacement = 15.
pc02.jpg
Video File #01
Video File #02
Video File #03
Sand Piles
We will use sand piles to illustrate some of the main ideas of proportionality and to provide an introduction to power functions.
Actually are dealing with sugar piles here, but sand will behave in a similar way.
The first picture below shows a pile of sugar being formed by pouring a container full of sugar in front of the 43-cm mark on a meter stick.
The second picture shows the pile that results when the container has been emptied.
We see that this pile forms an approximate cone with an approximately circular base.
By reading the meter stick we see that the base appears to have a diameter of about 4.5 cm.
We pose following question: when the second container of sugar is added on top of the first, will the diameter of the pile therefore double?
If so, of course, we will end up with a pile whose base has diameter 9 cm (double the 4.5 cm of the first).
Before looking further, decide whether you think this is the case.
If not, decide whether the diameter will more than double or less than double, and be sure that you explain your reasoning to yourself.
Then decide what you think will happen in a third, then a fourth, then a fifth and finally a sixth container are poured on the pile.
pc03.jpg pc06.jpg
The pictures below show the result of pouring the second and the third container of sugar onto the pile.
When we doubled then tripled the amount of sugar in the pile, did the diameter correspondingly double and triple?
It should be clear that the diameter did much less than double and triple.
Rather than having a doubled diameter of 9 cm, the second pile has a diameter of approximately 5.5 cm (you can probably read the diameters to a greater accuracy than the figures that will be given here, and you should do so).
The third pile appears to have a diameter of approximately 7 -- still not double the diameter of the first, though the pile has three times as much sugar as the first.
pc07.jpg pc08.jpg
The fourth and fifth piles have diameters that approach 8 cm, but they still do not reach the double-the-first-diameter measurement of 9 cm.
pc09.jpg pc10.jpg
Except for the last trial, where the sugar was poured carelessly and allowed to 'mound' on the left half of the pile, all these piles seem to have approximately the same slope along their sides.
Another way seeing this is that each pile has the same ratio of height to diameter.
The piles are reasonably close to being perfect cones, though their symmetry isn't perfect and their tops are a bit rounded.
To the extent that the piles form good cones with a consistent height-to-diameter ratio, they are geometrically similar, with each one being a scaled-up version of the ones before it.
If we imagine that the first sugar pile is made up of tiny cubes, and that each subsequent pile is made up by 'inflating' each tiny cube by inflating it to form a larger cube, then to increase the diameter from 4.5 cm to 5.5 cm would require that we increase the length, width and height of each cube by 5.5 cm / 4.5 cm = 1.22 (approx.).
Imagine that we were able to scale up each cube to double its original dimensions.
Then how many times the original volume with the new sugar pile half?
We might be tempted to say that it would double.
However, if we think a little bit, we can see that just doubling the length of a cube, leaving its height and width unchanged, will double its volume -- we would end up with the equivalent of two cubes laid end-to-end.
We would not have a cube with all its dimensions doubled, since the width and height would not be less than the length; this shape would not be a cube at all.
To make this shape into a cube we would have to first double its width, which would give us the equivalent of four cubes laid down, as at four corners of a square.
This figure would have four times the volume of the first, but would still not be a cube until its altitude was doubled.
This would of course require the equivalent of four more small cubes, bringing our total to 8.
It would therefore take the equivalent of 8 small cubes to increase the original cube to twice its original volume.
We can think of the process as three subsequent doublings, each doubling the volume of the preceding figure.
We 'stretch' the cube along its length, obtaining twice the volume of the original.
We then 'stretch' the resulting figure along its width, doubling the volume of the previous figure and obtaining four times the original volume.
We finally 'stretch' this figure in the vertical direction, doubling its height and thereby doubling the volume again, obtaining eight times the original volume.
By doubling the dimensions of each tiny cube, we will therefore have increased the volume by 2 * 2 * 2, or 8, times.
This is why doubling the amount of sugar was not sufficient to double the diameter of the pile.
To double the diameter, while maintaining a geometrically similar shape, we would have to double width, length and height.
This would require 8 containers full of sugar.
If we wish to increase the diameter by the factor 1.22 (remember, this is the diameter ratio between the 1-container pile and the 2-container pile), we could imagine that we take each tiny cube and increase its length by the factor 1.22, obtaining 1.22 times the volume, then increase the width of the resulting figure by the same factor, obtaining 1.22 * 1.22 times the original volume, and finally increase the height by factor 1.22 to obtain 1.22 * 1.22 * 1.22 = 1.22 ^ 3 times the original volume.
1.22 ^3 = 1.8, which is nearly 2, so the 1.22 factor is close to the factor that would double the volume. T
The actual factor for doubling the volume (take a calculator and check for yourself) is between 1.25 and 1.26.
Since our diameters 4.5 cm and 5.5 cm are only approximate, it is very possible that the actual ratio is, between 1.25 and 1.26. It is also possible that the shape of the pile changed a little bit, so that the second pile is not as nearly geometrically similar to the first as we might have thought, and the tiny cubes would not have been stretched in such a way to give us perfect cubes.
A situation like this, in which we expect that one quantity, like the volume, changes in proportion to the cube of another quantity, can be represented by the proportionality equation y = k x^3.
The number k represents a 'constant', a number which is the same no matter what the values of y and x are.
We will see that if we have a pair of x and y values, we can easily calculate k and obtain a specific rule for determining values of y from values of x (or, with a little simple algebra, determining values of x from values of y).
By making reasonably careful diameter measurements on a series of sugar piles made in class, we obtain the y = volume vs. x = diameter data shown below.
Diameters in class were not measured in cm, so these data cannot be precisely compared to those in the pictures above.
We can test our hypothesis that the volume y should be proportional to the cube of the diameter x by assuming that there is some constant k such that y = k x^3.
We can then put different values of x and y into this proportionality and determine if k is the same, or nearly the same, for all of our data points.
When we do this we obtain the k values shown in red.
These values are not perfectly constant, but they are fairly close to being constant.
pc04.jpg
The calculation of one of the k values is shown below.
(The right-hand part of the figure has been cut off; the value of k, shown as approximately .004, is in fact .0048).
The process is very straightforward -- we just substitute x = 8.6 and y = 3, corresponding to the data point, into y = k x^3 and solve for k.
pc05.jpg
Video File #04
If we plot our data using DERIVE and fit a cubic proportionality to our data (using FIT([x, kx^3], #1), we obtain y = .0046 x^3.
Derive gives us a value of k very close to that determined above.
The curve fits the data reasonably well.
However, it should be noted that a y = k x^4 fit works just about as well.
With our small but significant inconsistency in the values of k, which in the above table seem to increase from .004 to .005, this is not surprising.
In any case our results are not conclusive. We would have to repeat the experiment very carefully, and measure all dimensions of each sugarpile to determine whether the shape of the sugarpile does in fact change as it grows.
If careful measurements show that the shape remains constant, and if we measure the diameter accurately, we should obtain a consistent value for k.
However, if geometric similarity is violated in any significant way, we will not have this expectation.
Video File #05
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Question: `q013. Of the six proportionality equations y = k x, y = k x^2, y = k x^3, y = k / x, y = k / x^2, y = k / x^3, which one would apply to the surface area of the sculpture in the preceding?
Use the appropriate equation to answer the following:
If the sculpture originally had a exposed surface of are 4 square meters, what would be the surface area of the larger sculpture (per the conditions of the preceding problem)?
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Your solution:
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y = k / x^3
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4= k / 1/27
27*4= k / 1/27*27
k=108
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Good application of the general procedure, but you chose the wrong proportionality.
If the proportionality is y = k / x^3, then a larger value of x will result in a larger value of x^3, and hence in a smaller value of y. Since a bigger dimension x would imply a larger surface area, this proportionality doesn't apply.
You know that when you double the size of a square, it takes 4 squares of the original dimension to cover it.
So which of the proportionalities would quadruple the value of y when x is doubled?
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confidence rating #$&*:
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y = k x^2
When you double the square of 2 you get 4
4= k 2^2
so when x doubles 4 quads
...I still feel like i am doing something wrong
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First let me address your use of fractions (and order of operations) when you solved the equation
4= k / 1/3^3
for k.
In the first place you meant
4 = k / (1/3)^3.
k divided by 1/3 would not be k / 1 / 3, which by order of operations tells you to divide k by 1 then divide the result by 3.
k divided by 1/3 would be expressed as k / (1/3).
Of course in your equation you are cubing the 1/3. I didn't include the exponent above so we could focus on just the order of operations. Including the cube you would have to write this as
k / (1/3)^3.
This is then equal to k / (1/27), so your equation becomes
4 = k / (1/27).
Multiplying 27 by both sides of this equation doesn't help, because k / (1/27) * 27 means to divide k by 1/27, then multiply the result by 27. But k divided by 1/27 is k multiplied by 27/1, so
k / (1/27) * 27 = k * 27 * 27 = 729 k, not just k.
To solve the equation
4 = k / (1/27)
you would multiply both sides by 1/27, not by 27.
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See my notes directly above.
First be sure you understand the fractions and order of operations addressed in my first note.
Then I suggest you solve the problem in three different ways, as indicated by my second note.
If you can do this successfully you'll be well on your way to a solid mastery of the extremely important concept of proportionality.
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Now you can use the correct proportionality
y = k x^2.
There are many ways to set this up.
You set this up by saying that
y = 4 m^2 when x = 1/3
and find y when x = 1. This would be along the lines of your original thinking, but you have to specify both values of x. I'm not sure you saw that if x for the first figure is 1/3, then x for the second has to be 1.
You could alternatively set this up as
y = 4 m^2 when x = 1
and find y when x = 3. You would get a different value of k, but the final result for the volume would be the same.
I recommend that you set this up in both ways and verify that is doesn't matter whether x goes from 1 to 3, or from 1/3 to 1. You get the same result both ways.
Another way to solve the problem is to recall that
y = k x^2
implies that
y2 / y1 = (x2 / x1)^2.
For this question, what is the ratio x2 / x1, and what therefore is the ratio y2 / y1? What do you therefore conclude is the area of the larger figure?
This reasoning will lead to the same result as before.
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Mth163
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k does not matter
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Question: `q004. If y = k / x and if y = 4 when x = 2, what is the value of y when x = 8?
What is the ratio of the new value of y to the original?
What is the ratio of the new value of x to the original?
If y = k / x and if (x2 / x1) = 3, then what is the value of (y2 / y1)?
In general how is the ratio y2 / y1 related to the ratio x2 / x1?
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Your solution:
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4=k/2
k=8
4=k/8
.k=32
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From the given information you have concluded that the value of k is 8.
That value does not change.
What is the proportionality between x and y, if k has the unchanging value 8?
What therefore is the value of y when x = 8?
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I now realize that the k does not matter so when x=8,y=8
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k matters. Your proportionality is
y = 8 / x
so that when x = 8, y = 8 / 8 = 1.
However in this problem we are looking at the ratios, and when finding the ratio y2 / y1 the k divides out.
It's not that k can be disregarded or doesn't matter. The point is that k divides out.
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Good, but check my note directly above.
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