#$&*
course Mth 173
2/11 4:02 AM
006. query 6
Question `q Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles Why should this be the proportionality
Your solution
if p=3 was proportionality i confused it with power, and if so it is
y=ax^p
y=ax^3
This is important because of measuring volume, and it can be measured in tiny ""cubes"" which a power of 3 represents.
Confidence Assessment 2
Given Solution
the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares.
If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x Why should this be the proportionality
Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2.
Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the
proportionality for areas is y = k x^2.
By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3.
Self-critique (if necessary)
After the explanation this makes plenty of sense to me, just didn't think about it in the right way.
Self-critique Rating OK
Question `q Explain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and
volume.
Your solution
From what I remember on the class notes, the differential is a measure to the change in y. The derivative is the rate of change (using change in x). and as long the change in y is small, we can use the slope (derivative) times the change in x
so 'y=y'(x)*'x
???I can remember the concept, but when I look at it, I don't quite see it, when I look at it as a right triangle, I see multiplying the the change in x and the slope to be greater than a small leg to the triangle. Does this work because a slope is usually a fraction and
makes x smaller rather than larger? and how will the data change based on the side of the x leg? I think if you need a specific part that confuses me it was this (I am going back and finding it in the class notes):
We note that the derivative at x = 15 is the limiting value of the slopes of the triangles starting from the x = 15 point on the graph, as the run approaches 0. is this similar to the uniform cylinder problem where if we moved the hole up from the bottom of the
cylinder, its ""bottom"" is the hole's location?
Confidence Assessment 2
Given Solution
The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve,
and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y.
The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential
expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative `dx (approx)., or `dy = slope at given point `dx (approx),
or `dy = 3 k x^2 `dx (approx).
The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y.
The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5.
Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k
.01.
}
SPECIFIC EXAMPLE We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k
5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x.
Thus if x changes from 5 to 5.01 we expect that the change will be
change in y = (dydx) `dx =
rate of change change in x (approx) =
.15 .01 = .0015,
so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to
change much over that short increment, so we expect that the approximation is pretty good.
Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change
much over that short interval. But it does change a little, and that's the reason for the discrepancy.
The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.
Self-critique (if necessary)OK
Self-critique RatingOK
Question `q What would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30 (instant response not required)
Your solution
???instant response not required, does that mean you want to see a step by step?
this is a quadratic, the derivative is y=2at+b, which is also the rate of change
???is the derivate the rate of change, or an AVERAGE rate of change, I am thinking the latter.
y'=.04t-3 for this specific function would be the rate of depth change
Confidence Assessment 3
Given Solution
You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate.
Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 30 - 3 = 1.2 - 3 = -1.8.
COMMON ERROR y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change
INSTRUCTOR COMMENT This is the depth, not the rate of depth change.
Self-critique (if necessary)
just reasserting. Derivative is using the change of x to find a rate of change. rate of change always corresponds to changes in x (at least in standard x independent y dependent graph)
differential is the change of y...I believe throughout the course I will understand the importance better, for now I don't see how important it is, but still noting for remembrance
Is this correct?
@&
The derivative of f(x) will be a formula for the derivative function, which we denote as f ' (x).
For example the derivative of f(x) = 2 x^2 + 5 x - 7 is f ' (x) = 4 x + 5.
The derivative, evaluated at a specific x, can be identified with the slope of the graph at the corresponding graph point.
The derivative evaluated at a specific x is the rate at which the function changes with respect to x.
If you multiply the derivative of a function y = f(x) having a smooth graph, with the derivative evaluated at a specific x, by a small change `dx, you get a good approximation of the change in y. The smaller the change in x, the more closely the result will approximate the change in y.
*@
Self-critique Rating OK
Question `qmodeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side Explain how you know this.
Your solution
You noted in the problem think of it in 3d, how many cubes you could put it in. so on one side I could put one more on top to double the height. I now have a rectanglular prism though. In simplest thought processes I can just reshape a cube by adding same sizes
everywhere. now I take two more, and attach the same rectangular prism shape to the side. I have now attached 3 cubes, and I have the height and width proper, but I still have one dimension left, length, so I now a make a replicate of the new rectangular prism I
have to the other side (which should be 4 more) to make the cube, and I have in total added 7 cubes that are 1.5 inches on a side, and recreated another cube, geometrically similar to the previous one. so altogether there are now 7 cubes plus the original, or 8
cubes.
Confidence Assessment 3
Given Solution
You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer.
Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side.
Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups.
COMMON ERROR
It would take 2 quarter-cups.
INSTRUCTOR COMMENT 2 quarter-cups would make two 1.5 inch cubes, which would not be a 3-inch cube but could make a rectangular solid with a square base 1.5 inches on a side and 3 inches high.
Self-critique (if necessary)
I do not understand how to explain what I am seeing so let me just explain how I look at it. I start with one cube, so I add one. I now have two cubes, and solved one dimensions problem. I have two dimensions left. I now have two cubes, so I add two cubes. I
have solved another dimensions problem, I have one dimension left. I have four cubes, so I add four cubes, I now have eight, and no more dimensions left.
I can also reword this into math a little more by saying I fix one dimension, and then use the power I am dealing with, in this case 3 so I have 2 cubes, so 2^3. 2*2*2. 4*2=8 Is there a mathematical law or formula or equation for the way I am thinking.
@&
If go from 1 cube to n cubes, then it will take n cubes to make a single row of n, it will take n^2 cubes to make an n x n array of cubes (i.e., a 2-dimensional array, which would be a square array), and it will take n^3 cubes to make an n x n x n array (i.e., a 3-dimensional array, which would be cubical array).
This is why the areas of geometrically similar figures scale as
A = k * x^2 for 2-dimensional figures
and as
A = k x^3 for 3-dimensional figures.
*@
Self-critique Rating
Question `qWhat value of the parameter a would model this situation How many quarter-cups does this model predict for a cube three inches on a side How does this compare with your previous answer
Your solution
y=ax^3
i said 2^3 would be the way to see this.
a would be 1?
Confidence Assessment
Given Solution
The proportionality would be
y = a x^3,
with y = 1 (representing one quarter-cup) when x = 1.5. So we have
1 = a 1.5^3, so that
a = 1 1.5^3 = .296 approx.
So the model is y = .2963 x^3.
Therefore if x = 3 we have
y = .296 3^3 = 7.992, which is the same as 8 except for roundoff error.
Self-critique (if necessary)
I am not sure how I am looking at it to see it a 2^3 instead of .296* 3^3. This equation just seems to over-complicate it for me. Although I do understand now how to set it up, 1 is the amount, x is the measurement of a dimension, and p equals the amount of
dimensions.
@&
y is the number of quarter-cups, and x is the length of the side of the cube.
y = x^3 would not work, since y has to be 1 when x is 1.5. y = x^3 would give you y = 3.375 when x = 1.5.
That's why the proportionality has to be of the form
y = k x^3
(or if you prefer y = a x^3; it doesn't matter whether you call the proportionality constant k or a or Fred).
You use the correct proportionality, evaluate the proportionality constant, and you get the function that models the situation.
In this case, you could also reason directly without getting the function. 3 inches is double 1.5 inches, so as I believe you clearly see it would require 8 1.5 inch cubes to make a 3 inch cube.
The formula approach gives you the same result.
Using the function y = .2963 x^3, you could for example take its derivative, evaluate the derivative at x = 3, multiply that by .1 and get a pretty good estimate of how many additional quarter-cups you would need to increase the sides of the cube to 3.1 inches. (this is related to your previous comment about not understanding what the differential might be good for)
*@
Self-critique Rating 2
Question `qWhat would be the side measurement of a cube designed to hold 30 quarter-cups of sand What equation did you solve to get this
Your solution
well from looking above y = an amount, in this case quarter-cups of sand.
30=ax^3
Using the above found a
30=.296296x^3
cube root(101.2501012501)=(x^3)
x=about 4.66085
Confidence Assessment
Given Solution
You are given the number of quarter-cups, which corresponds to y. Thus we have
30 = .296 x^3 so that
x^3 = 30 .296 = 101, approx, and
x = 101^(13) = 4.7, approx..
!!!!!!! I am looking at your answer, and for some reason when I copy this document it makes your answer say 101^(13), so I went back to your page, and it shows 1/3, so I do not know what happened there, but it seems quite often my computer does not pick
up certain letters when copying your documents
@&
That can occasionally happen, though since / is an ASCII character I can't think of how or why it does.
*@
Question `qquery problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function was y = .002 x^3. What function would have been obtained using 1/4 cup
!!!!!again my copy did not contain the / symbol, so I got completely confused, and had to go back and check the original document, before I could even figure out what was going on. I manually inserted the / myself after noticing so I could read the problem
better. did you actually type 1 / 2 and 1 / 4 or did you use a program that viewed it as one entity, I do not believe I can see these in plain text .txt extentions, maybe only in .RTF. If this is the case I will switch to it. But I wanted to make sure this was the problem.
Can you check this out for me?
Your solution
half cups for what. what are we trying to reach? at any rate using have cups it would take half as long for him to reach a desired amount. therefore his amount would be halved. he would have to double the function value. if it is .002 for his half cups, then double
it. .004
Confidence Assessment
Given Solution
In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y.
He would have obtained half as many half-cups as the actual number of quarter-cups.
To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3.
Self-critique (if necessary)
??? you stated To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. how does this double y. this just looks like doubling a, not y.
Self-critique Rating
@&
If you evaluate .002 x^3 for any value of x, and then evaluate .004 x^3 for the same value of x, the second result will be double the first.
Since the result is the value of y, then, replacing .002 x^3 by .004 x^3 will double the value of y.
*@
Question `qquery problem 4. number of swings vs. length data. Which function fits best
Your solution
I didn't actually test that so I will hypothesis. When doing TRUE oscillations, it should only be going in two directions (which looks like one, but as it goes left and right, its height also gets higher or lower, so i will see it as two anyway). But we were decreasing
the ""speed"" so the division alternate is 1/2 and it is declining, so it would be negative 1/2 or -.5. Of course I had free form swinging, so it was a little off and had 3 dimensions, forward and back, so assuming I kept the same kinetic energy constantly it would
maybe have been closer to a number between -.5 and -.333, closer to -.5. I do remember however using -.5 my a's stayed relatively the same (of course with some fluctuation, but I believe that to be testing error from me having to use energy to keep the ball
moving for an entire minute).
Confidence Assessment 2
Given Solution
For each function you can substitute the x and y data for each point, then solve for the constant a. If the values of a are not relatively constant, then the function is not constant.
For example, you can try the function y = a x^-2.
You can plug in every (x, y) pair and solve of the resulting equations for a.
Try this for the data and you will find that y = a x^-2 does not give you consistent a values—every (x, y) pair you plug in will give you a very different value of a.
If you know the shapes of the basic graphs, you can compare them with a graph of the data and get a pretty good indication of which functions to try.
For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate.
The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. So you could try these functions, in any reasonable order
You might start with the y = a x^-.3 function. The values of the parameter a will probably vary quite a bit from data point to data point, and if so this function won't be a good candidate.
You could then try the y = a x^-.4 function. The values of the parameter a will likely still vary significantly, but probably not as much as for the y = a x^-.3 function.
You could then try the y = a x^-.5 function. The values of the parameter a will likely still vary a bit, but probably not by a whole lot. If you have good data this function will probably be your best candidate.
You could then try the y = a x^-.6 function. The values of the parameter a will likely vary more than they did for the y = a x^-.5 function, confirming that the y = a x^-.5 function is the best candidate so far.
The y = a x^-.7 function will probably result in quite a bit of variation in a, and is not likely to be the best function.
STUDENT COMMENT
this concept is a little fuzzy to me. Im not quite sure what you mean when you say that ax^-.5 results in a nearly constant value.
INSTRUCTOR RESPONSE If the proportionality is y = a x^-2, then if we solve for a we get a = y x^2.
If you evaluate a = y x^2 for each of your data points, then if y = a x^-2 is a good model you should get about the same value of a for each point.
For example if x values 2, 4, 7, 10 give you respective y values .5, .3, .2, .1, then your a values would be
a = .5 2^2 = 2
a = .3 4^2 = 4.8
a = .2 7^2 = 9.8
a = .1 10^2 = 10
These values appear to be increasing. So the data don't appear to be consistent with the form y = a x^-2.
Another proportionality might yield relatively constant values for a. If so, then your data would be consistent with that proportionality. For example y = a x^-1, or y = a x^-.5 might give you good results.
It is also possible that the data aren't consistent with any power-function proportionality.
Check this out with the data for this problem, and see what you find.
Self-critique (if necessary)
???Can we not figure out what to use relative to the dimensions and if it is increasing or decreasing, if everything else stays the same (or close to the same, with fluctuation)
@&
The pendulum moves in space, but the pendulum is not a geometric figure so spatial dimensions aren't relevant to the nature of the proportionality.
The proportionality actually arises because of the (very near) linearity of the net force on the pendulum with respect to its position as measured relative to equilibrium.
This yields the equation
F_net = - k x,
where x is the position and F_net the net force on the pendulum's mass.
F_net being the net force is equal to the product of the pendulum mass and its acceleration.
For reasons you won't understand until you're at least halfway through this course, the acceleration of the pendulum is x '', the second derivative of its position function with respect to time.
So the equation becomes
m x '' = - k x.
As you will understand in a month or so, this equation is satisified if x = A cos(omega t + phi), where A and phi are arbitrary constants and omega = sqrt(k / m).
Now, k happens to be inversely proportional to the length of the pendulum, and the period of the pendulum is inversely proportional to omega. So since omega = sqrt(k / m), the period is proportional to the square root of the length.
I don't intend that you completely understand this. I just want you to be aware that this proportionality arises from the physics of the situation and not from geometrical dimension.
I should add that k is also proportional to m, so that omega, and therefore the period, are independent of the mass of the pendulum.
*@
Self-critique Rating 1
Question `qproblem 7. time per swing model. For your data what expression represents the number of swings per minute
Your solution
I used 2,4,6,8,10,12, and 14 inches, with free form swinging, and my hand trying to keep it at a steady rate.
y=162.829x^3
Confidence Assessment
Given Solution
The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55.
The model is pretty close to
# per minute frequency = 55 x^-.5.
As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 1^.5 = 53; for y = 40 and x = 2 we would
get a = 40 2^.5 = 56; for y = 34 and x = 3 we get a = 33 3^.5 = 55; for y = 26 and x = 4 we get a = 26 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54.
The value of a for accurate data turns out to be about 55.
Self-critique (if necessary)
I can't compare since I used inches, however I do understand.
Self-critique Rating
Question `qIf the time per swing in seconds is y, then what expression represents the number of swings per minute
Your solution
if y=ax^3 would be the equation for swings in seconds then 60/divided by that value of y (60/y) would be the swings per minute.
Confidence Assessment 3
Given Solution
To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 y, where f is frequency in swings per minute.
COMMON ERROR y 60
INSTRUCTOR COMMENT That would give more swings per minute for a greater y. But greater y implies a longer time for a swing, which would imply fewer swings per minute. This is not consistent with your answer.
Self-critique (if necessary)OK
Self-critique RatingOK
Question `qIf the time per swing is a x ^ .5, for the value determined previously for the parameter a, then what expression represents the number of swings per minute How does this expression compare with the function you obtained for the number of swings
per minute vs. length
Your solution
i am not sure how to answer this. I just note that as the string gets longer, the time per swing should grow, so the power function would have a positive exponent. but the number of swings will decline as the string gets longer, so if viewing it in that aspect the
power function would have a negative exponent. I see them as using the same ""dimensions"" so the exponent's absolute value would stay the same.
as for the expression. it would just be a modified version of the original.
if 60 seconds is a minute, and you have the time per swing, then however many times that goes into 60, thats the amount of swings per minute.
swings per minute= 60/(ax^-.5). This seems the same as what you asked in the exact previous question.
comparing to the swings per minute vs length. the function was y=ax^-.5 the difference just being in a. and the modifier of 60 to know it in per minute while using seconds.
Confidence Assessment
Given Solution
Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1.
Since the number of swings per minute is 60(time per swing), you have f = 60 (a x^.5), where f is frequency in swings minute.
Simplifying this gives f = (60 a) x^(.-5).
60a is just a constant, so the above expression is of form f = k x^-.5, consistent with earlier statements.
60 a = 60 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5.
Self-critique (if necessary)
How are you able to put 60/a together. to me it would be 60/y value which is the entire (ax^-.5) how can you split a away from x and pair it to 60 first?
@&
x / (y z) is the same as (x / y) * (1 / z), because of you multiply numerators and denominators in the latter expression you get x / (y z).
So 60 / (a x^.5) is the same as (60 / a) * 1 / x^.5, and since 1 / x^.5 is x^-.5, this is the same as (60 / a) * x^-.5.
The reason we write it like this is to put it in the form of
(proportionality constant) * power of x,
which is the standard form of a power function.
The proportionality constant is 60 / a, and the power is -.5.
*@
I am not understanding this problem.
@&
It does appear that your system is somehow eliminating the slashes that indicate division. The solution as displayed above lacks the slashes.
The correct solution is shown below. Let me know if it doesn't display correctly:
Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1.
Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute.
Simplifying this gives f = (60 / a) * x^(-.5).
60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements.
60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5.
*@
Self-critique Rating
Question `qquery problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds
Your solution
I used 2,4,6,8,10,12, and 14 inches.
for time per swing.
y= ax^.5
1.78333=a2^.5
1.78333=1.41421a
a=1.26100
1.15=a6^.5
1.15=2.4495a
a=.469485534
.8=a12^.5
.8=3.464101615a
a=.2309401
avg of a is about .65
@&
A 2-inch pendulum wouldn't require 1.78 seconds per swing. Now would a 6-inch pendulum require 1.15 seconds per swing. The respective times per swing would be, very approximately, .45 seconds and .8 seconds.
Using these approximations you would find that a would come out in the neighborhood of .3 for both calculations.
So while your procedure is fine, your data do not fit the model.and your value of a doesn't work for your data.
*@
I believe I had a deviation because of human error, but not knowing exactly how much I should adjust to it, I will leave it at that amount.
.1=.65x^.5
.15385=x^.5
x=.0236698225 inches
100=.54x^.5
156.25=x^.5
x= 24414.0625 inches
Confidence Assessment 2
Given Solution
You would use your own model here.
This solution uses T = 1.1 x^.5. You can adapt the solution to your own model.
According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x
For T = .1 we get
.1 = 1.2 x^.5 which gives us
x ^ .5 = .1 1.2 so that
x^.5 = .083 and after squaring both sides we get
x = .083^2 = .0069 approx., representing .0069 feet.
We also solve for T = 100
100 = 1.2 x^.5, obtaining
x^.5 = 100 1.2 = 83, approx., so that
x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long.
Self-critique (if necessary) OK
Self-critique Rating OK
Question `qquery problem 9. length ratio x2 x1.
. If one pendulum has length x1 and the other has length x2, then the ratio of their lengths is x2 / x1. What expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums?
Self-critique (if necessary)
I don't understand this question.
@&
If y1 and y2 stand for the times per swing, then if a = .3 (which is a very approximate value) we use the function
y = .3 x^.5
and find that
y1 = .3 * x1^.5
and
y2 = .3 x^.5.
*@
Self-critique Rating
Question `qWhat expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums
Your solution
???What is the purpose of knowing this ratio?
it would be Frequency 1= 60/y(x1) and Frequency 2 = 60/y(x2)
for my equation it would be y=.65x^.5
Confidence Assessment
Given Solution
The solution is to be in terms of x1 and x2.
If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 (1.1 `sqrt(L)) to get 60 (1.1 `sqrt(x1) ) and 60 (1.1 `sqrt(x2)).
Alternative form is f = 55 L^-.5. Substituting would give you 55 x1^-.5 and 55 x2^-.5.
If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5
Self-critique (if necessary)
I do not understand where you are getting 55L^-.5 when 60 is the seconds equivalent to a minute. This may just be to my document not copying your / to making me confused. at any rate I would like a repeat explanation on finding the frequency equation.
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If the time per cycle is 1.1 L^.5 then the number of cycles in a minute is 60 / (1.1 L^.5), which is equal to about 55 L^-.5 (55 being 60 / 1.1).
A copy of the solution, which should show the slashes:
The solution is to be in terms of x1 and x2.
If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)).
Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5.
If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5
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Self-critique Rating
Question `qWhat expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums
Your solution
Frequency1/Frequency2
this is 60/(ax1^.5) and 60/(ax2^.5)
you would divide the two from each other.
Everything is the same except x1 and x2 values
so that should just reduce to a
x1^.5/x2^.5
they are both raised to .5
so (x1/x2)^.5
???what rule is that
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This is one of the rules of exponents:
(ab)^c = a^c * b^c
(a / b)^c = a^c / b^c
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??? would 60/60 make it 1/(x1/x2)^.5
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That would be the ratio of y1 to y2.
Usually you would find the ratio of the second to the first, but as long as you specify that you are finding the ratio of the first to the second your answer would be valid.
1 / (x1/x2)^.5 is the same as
(x2 / x1)^.5.
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Confidence Assessment 1
Given Solution
We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5.
With these expressions we would get
f2 f1 = a x2^-.5 (a x1^-.5) =
x2^-.5 x1^-.5 =
(x2 x1)^-.5 =
1 (x2 x1)^.5 =
(x1 x2)^.5.
Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get
f2 f1 = 55 x2^-.5 (55 x1^-.5) =
x2^-.5 x1^-.5 =
(x2 x1)^-.5 =
1 (x2 x1)^.5 =
(x1 x2)^.5.
This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step.
Self-critique (if necessary)
i note that you use ax^-5 for your frequency, is this because of corespondence of length of string making less swings, and how at the end your equation is ^.5 not ^-.5. is this because of that 60/60 making it 1/(x1/x2)^-.5
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We will usually simplify the result so that the exponent is positive.
Per previous solutions, since time per swing is of the form a x^.5, frequency is of the form a x^-.5. For example period 1.1 x^.5 leads to frequency (in cycles/min) 55 x^-.5.
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I do notice I messed up frequency with time per swing, and changed the value of exponent of negative.
Self-critique Rating 3
Question `qquery problem Challenge Problem for Calculus-Bound Students how much would the frequency change between lengths of 2.4 and 2.6 feet
Your solution
well following the (x1/x2)^.5 would we not just need to plug the numbers in?
(2.4/2.6)^.5
the frequency ratio is .9607689228 from 2.4 to 2.6, and that ratio...is the number of swings per foot....i am not sure. and since this is decimal values with feet, and I need less than that I am not sure how to convert it. thought maybe cross multipling decimals/10 to x/12 I guess ill trash that idea
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Your idea can lead to a correct solution, but you would need one of the two frequencies.
If the ratio is .96, then this means that the second frequency is .96 of the first. This means that the change in frequency is -.04 times the frequency of the first.
The frequency for 2.4 feet is 55 / sqrt(2.4) = 35, approximately.
-.04 * 35 = -1.4.
This agrees well with the change you calculated below.
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following the formula you gave f = 55 L^-.5
f=35.50234734 at 2.4
f=34.10955201 at 2.6
the frequency changed 1.392795328
thinking about it...why is my frequency positive.....as the length of string grew it should be slower, So i guess I shall just say the frequency should be a negative to that value.
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Frequency changes from about 35.5 cycles / minute to about 34.1 cycles / minute.
That's a decrease, so the change in frequency is negative.
The change in a quantity is the second quantity minus the first. This change comes out positive if the second quantity is greater, negative if the second quantity is less.
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Confidence Assessment
Given Solution
STUDENT SOLUTION Note that we are using frequency in cycles minute.
I worked to get the frequency at 2.4 and 2.6
y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178.
subtracted to get -1.40949 difference between 2.4 and 2.6.
This, along with the change in length of .2, gives average rate -1.409 cyclesmin (.2 ft) = -7.045 (cyclesmin)ft , based on the behavior between 2.4 ft and 2.6 ft.
This average rate would predict a change of -7.045 (cyclesmin)ft 1 ft = -7045 cyclesmin for the 1-foot increase between 2 ft and 3 ft.
The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cyclesmin.
The answers are different because the equation is not linear and the difference between 2.4 and 2.6 does not take into account the change in the rate of frequency change between 2 and 2.4 and 2.6 and 3
for 4.4 and 4.6
y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5)
y = 26.5341 y = 25.6508
Dividing difference in y by change in x we get -2.9165 cyclesmin ft, compared to the actual change -2.938 obtained from the model.
The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points.
Self-critique (if necessary)
I believe I understand this...just unsure if I could perform it on my own, I got a little confused at first, especially after you where talking about frequency ratio's.
I
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Self-critique Rating
Question `q query problem 1.2.24 5th; 1.2.19 4th formula for exponential function through left (1,6) and (2,18)
Your solution
foormulas for exponential function.
I know y =a*b^x
and that there is one with e...that has a certain value (makes me think of natural logarithm)
6=a * b ^1 (we could simplify this specific instance further, couldn't we?)
18=a * b^2
18= a * b^2
6=a *b^1 tried subtracting at first, that didnt work, noticed exponents might mean need to divide
3= 1b
b=3
18= a * 3^2
18=9a
a=2
so formula is y= 2 * 3^x
Confidence Assessment
Given Solution
An exponential function has one of several forms, including y = A b^x and y = A e^(kx).
Using y = A b^t and substituting the t and y coordinates of the two points gives us
6 = A b^1
18 = A b^2.
Dividing the second equation by the first we get
3 = b^(2-1) or b = 3.
Substituting this into the first equation we get
6 = A 3^1 so
A = 2.
Thus the model is y = 2 3^t .
Self-critique (if necessary)OK
Self-critique RatingOK"
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#*&!
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Good work and good questions. I've inserted a number of responses. Let me know if you have additional questions, and I'll be glad to clarify further.
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