Q-000

course PHY 231

6/30 20:46

ph1 query 0Most queries in this course will ask you questions about class notes, readings,

text problems and experiments. Since the first two assignments have been lab-

related, the first two queries are related to the those exercises. While the

remaining queries in this course are in question-answer format, the first two

will be in the form of open-ended questions. Interpret these questions and

answer them as best you can.

Different first-semester courses address the issues of experimental precision,

experimental error, reporting of results and analysis in different ways and at

different levels. One purpose of these initial lab exercises is to familiarize

your instructor with your work and you with the instructor 's expectations.

Comment on your experience with the three lab exercises you encountered in this

assignment or in recent assignments.

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Question: This question, related to the use of the TIMER program in an

experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling

down a straight wooden incline about 50 cm long. If the computer timer

indicates that on five trials the times of an object down an incline are

2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you

think the discrepancies could be explained by each of the following:

· The lack of precision of the TIMER program.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Not as much as the error of the operator.

#$&*

· The uncertain precision of human triggering (uncertainty associated

with an actual human finger on a computer mouse)

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

The discrepancies are explained better by this factor than the precision of the

timer.

#$&*

· Actual differences in the time required for the object to travel the

same distance.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Not very much.

#$&*

· Differences in positioning the object prior to release.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

They affect the outcome slightly, maybe by a tenth of a second or so.

#$&*

· Human uncertainty in observing exactly when the object reached the

end of the incline.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Combined with the muscle response delay this could be a significant factor.

#$&*

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Question: How much uncertainty do you think each of the following would

actually contribute to the uncertainty in timing a number of trials for the

ball-down-an-incline lab?

· The lack of precision of the TIMER program.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

10% of the uncertainty

#$&*

· The uncertain precision of human triggering (uncertainty associated

with an actual human finger on a computer mouse)

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

30% of the uncertainty

#$&*

· Actual differences in the time required for the object to travel the

same distance.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

10% of the uncertainty

#$&*

· Differences in positioning the object prior to release.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

25% of the uncertainty

#$&*

· Human uncertainty in observing exactly when the object reached the

end of the incline.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

25% of the uncertainty

#$&*

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Question: What, if anything, could you do about the uncertainty due to each of

the following? Address each specifically.

· The lack of precision of the TIMER program.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Create a more precise timer.

#$&*

· The uncertain precision of human triggering (uncertainty associated

with an actual human finger on a computer mouse)

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Create a device that when activated activates the timer and releases

#$&*

· Actual differences in the time required for the object to travel the

same distance.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Make sure that the object performs the same on each trial.

#$&*

· Differences in positioning the object prior to release.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Rig a device which had a static starting point.

#$&*

· Human uncertainty in observing exactly when the object reached the

end of the incline.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Not much other than by an automated system.

#$&*

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Question: If, as in the object-down-an-incline experiment, you know the

distance an object rolls down an incline and the time required, explain how you

will use this information to find the object 's average speed on the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Use Avg Velocity = Distance/Time

confidence rating #$&* 3

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Question: If an object travels 40 centimeters down an incline in 5 seconds then

what is its average velocity on the incline? Explain how your answer is

connected to your experience.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

R = D/T = 40cm/5sec = 8cm/sec

The average velocity is the overall distance over the overall time.

confidence rating #$&* 3

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Question: If the same object requires 3 second to reach the halfway point, what

is its average velocity on the first half of the incline and what is its average

velocity on the second half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A) R = D/T = 20cm/3sec = 6.67cm/sec for the first half.

B) It is not possible to calculate without the time for the 2nd half.

(Assuming that it takes 2 seconds for the 2nd half the rate would be 10cm/sec)

confidence rating #$&* 1

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Question: `qAccording to the results of your introductory pendulum experiment,

do you think doubling the length of the pendulum will result in half the

frequency (frequency can be thought of as the number of cycles per minute), more

than half or less than half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The frequency will be less than half.

confidence rating #$&* 2

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Question: `qNote that for a graph of y vs. x, a point on the x axis has y

coordinate zero and a point on the y axis has x coordinate zero. In your own

words explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This is so because the y axis lies on the x = 0 coordinate and vice versa. It's

a basic nature of the cartesian coordinate system.

????I don't quite understand the question.

your answer is good

confidence rating #$&* 0

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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on

the vertical axis and length on the horizontal), what would it mean for the

graph to intersect the vertical axis (i.e., what would it mean, in terms of the

pendulum and its behavior, if the line or curve representing frequency vs.

length goes through the vertical axis)? What would this tell you about the

length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

It would be the highest possible frequency of the pendulum since the length of

the pendulum was zero.

confidence rating #$&* 2

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Question: `qOn a graph of frequency vs. pendulum length, what would it mean for

the graph to intersect the horizontal axis (i.e., what would it mean, in terms

of the pendulum and its behavior, if the line or curve representing frequency

vs. length goes through the horizontal axis)? What would this tell you about

the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

It would mean the frequency reached zero.

confidence rating #$&* 2

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Question: `qIf a ball rolls down between two points with an average velocity of

6 cm / sec, and if it takes 5 sec between the points, then how far apart are the

points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

D = R*T = 6cm*5sec/1sec = 30cm/sec

confidence rating #$&* 3

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.............................................

Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it

will move 30 cm.

The formal calculation goes like this:

We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement

and `dt is the time interval.

It follows by algebraic rearrangement that `ds = vAve * `dt.

We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

`ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

vAve = `ds / `dt. We multiply both sides of the equation by `dt:

vAve * `dt = `ds / `dt * `dt. We simplify to obtain

vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

???Why do you use the terms ""ds"" and ""dt"". Is that the similar as dy/dx from

calculus? Or does it mean ""delta""?

`d means 'Delta', as explained in the very first problem of the Introductory Problem Sets.

Without all the qualifying details, we can roughly say that dy/dx is the limiting value of `dy / `dx, as `dx goes to zero. This idea, the definition of the derivative, should be familiar to you from calculus and will be used extensively in this course.

confidence rating #$&* 2?

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Question: `qYou were asked to read the text and some of the problems at the end

of the section. Tell your instructor about something in the text you understood

up to a point but didn't understand fully. Explain what you did understand, and

ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The part about multiplying vectors was new to me. It seems not too difficult

but I will probably have more questions as I do more problems throughout the

course on this subject.

I'll welcome your questions.

#$&*

STUDENT QUESTION

I understand that we cannot measure to exact precision, but when we are dealing

with estimated uncertainty, do we always

just increment our lowest unit by one and that is our uncertainty? Is there a

standard that is used to figure out this?

INSTRUCTOR RESPONSE

The standard answer is that we assume an uncertainty of +- 1 of our smallest

unit of precision. However, depending on how well we can 'see' that smallest

unit, we can get pretty close to +- 1/2 of a unit.

A more sophisticated answer can be given in terms of the statistics of the

normal distribution, but in this course we're not going to go into a whole lot

of depth with that. A calculus background would be just about required to

understand the analysis well enough to apply it meaningfully.

STUDENT QUESTION

I fully understand how to calculate uncertainty, but what if the uncertainty

isn’t given? For example,

problem 6 asks us for the uncertainty of 1.67. Do we just use .01 as the

uncertainty?

INSTRUCTOR RESPONSE

Depending on the nature of the instrument and the observation, +- .01 might be

necessary, but we could go to +-.005 if can regard 1.67 as an accurate roundoff.

Without very good reason, though, +-.01 would be the safer assumption.

STUDENT QUESTION

I had trouble grasping the uncertainty. I understand the bit about significant

figures, but I’m not sure how that applies

to the uncertainty. Is it just the last digit of the significant figure that

could be wrong?

INSTRUCTOR RESPONSE

Any measurement is uncertain to some degree.

On some of the initial videos, despite the fact that the ruler was marked in

inches and subdivided to eighths of an inch, the resolution of the image was

poor and it wasn't possible to observe its position within eighths of an inch.

Had the videos been very sharp (and taken from a distance sufficient to remove

the effects of parallax), it might have been possible to make a good estimate of

position to within a sixteenth of an inch or better.

So for the videos, the uncertainty in position was probably at least +- 1/4

inch, very possibly +- 1/2 inch. But had we used a better camera, we might well

have been able to observe positions to within +-1/16 inch.

The video camera is one instrument, and each camera (and each setup) introduces

its own unique uncertainties into the process of observation.

The same can be said of any setup and any instrument or combination of

instruments.

STUDENT QUESTION: I understood the portion discussing the nature of science and

felt familiar with much of the measurement. What I did not fully understand was

how do you know when to write an answer using the powers of 10 or to leave it

alone? Several of the tables had values in powers of 10 for metric prefixes such

as centi and mili.

INSTRUCTOR RESPONSE

Whether you use scientific notation or not depends a lot on the context of the

situation.

As a rule of thumb, I would recommend going to scientific notation for numbers

greater than a million (10^6) and less than a millionth (10^-6). When numbers

outside this range are involved in an analysis it's a good idea to put

everything into scientific notation.

And when you know that scientific notation is or is not expected by your

audience, write your numbers accordingly.

QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics

students)

I don’t fully understand the dot product rule

INSTRUCTOR RESPONSE

The dot product of vectors A = a_1 i + a_2 j + a_3 k and B = b_1 i + b_2 j + b_3

k is a_1 * b_1 + a_2 * b_2 + a_3 * b_3. The dot product is simply a number.

The magnitude of A is | A | = sqrt( a_1 ^ 2 + a^2 ^ 2 + a_3 ^ 2); the magnitude

of B is found in a similar manner.

The dot product is equal to | A | * | B | * cos(theta), where theta is the angle

between the two vectors.

If you have the coefficients of the i, j and k vectors, it is easy to calculate

the dot product, and it's easy to calculate the magnitudes of the two vectors.

Setting the two expressions for the dot product equal to one another, we can

easily solve for cos(theta), which we can then use to find theta.

More importantly for physics, we can find the projection of one vector on

another. The projection of A on B is just the component of A in the direction

of B, equal to | A | cos(theta). The projection of one vector on another is

important in a number of situations (e.g., the projection of the force vector on

the displacement, multiplied by the displacement, is the work done by the force

on the interval corresponding to the displacement).

Dot products are a standard precalculus concept. Check the documents at the

links below for an introduction to vectors and dot products. You are welcome to

complete these documents, in whole or in part, and submit your work. If you

aren't familiar with dot products, it is recommended you do so.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_09.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_10.htm

confidence rating #$&*

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Question: `qTell your instructor about something in the problems you understand

up to a point but don't fully understand. Explain what you did understand, and

ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Looking ahead in the assignments I was having problems with percent error with

velocity.

This is address in a number of documents in the early part of the course. Let me know if you are still having trouble with the idea after completing the first few assignments.

#$&*

SOME COMMON QUESTIONS:

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QUESTION: I didn’t understand how to calculate uncertainty for a number such as

1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had

a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when

no estimated uncertainty is given.

INSTRUCTOR RESPONSE:

The +- number is the uncertainty in the measurement.

The percent uncertainty is the uncertainty, expressed as a percent of the number

being observed.

So the question in this case is simply, 'what percent of 1.34 is 0.5?'.

0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.

.037 is the same as 3.7%.

I recommend understanding the principles of ratio, proportion and percent as

opposed to using a formula. These principles are part of the standard school

curriculum, though it does not appear that these concepts have been well

mastered by the majority of students who have completed the curriculum. However

most students who have the prerequisites for this course do fine with these

ideas, after a little review. It will in the long run save you time to do so.

There are numerous Web resources available for understanding these concepts.

You should check out these resources and let me know if you have questions.

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QUESTION: I understood the main points of changing the different units, but I’m

not sure when in the problem I should change the number to 10 raised to a

certain power. In example 1-8 I did not understand why they changed 70 beats/min

to 2 x 10^9 s.

2 * 10^9 is about the number of seconds in 70 years.
70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats /
minute to beats in a lifetime, there was a step where it was necessary to
multiply by 2 * 10^9 seconds.
The example actually used 80 beats / min as a basis for the solution. This was
converted to beats / second by the calculation
80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats /
second.
This was then multiplied by 2 * 10^9 seconds to get the number of beats in a
lifetime:
2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.
In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually
calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2
* 10^9 seconds in one step
80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.
In your instructor's opinion the unit 'beats' should have been left in the
result; the text expressed the result simply as 3 * 10^9, apparently ignoring
the fact that the unit 'beats' was included in the quantities on the left-hand
side.
Also the text identified this number as 3 trillion. In the British terminology
this would be correct; in American terminology this number would be 3 billion,
not 3 trillion.

COMMENT:

I thought that these problems were pretty basic and felt that I understood them

well. However, when I got to questions 14 (determine your own mass in kg) and 15

(determining how many meters away the Sun is from the Earth), I did not

understand how to complete these. I know my weight in pounds, but how can that

be converted to mass in kilograms? I can look up how to convert miles to meters,

but is this something I should already know?

INSTRUCTOR RESPONSE:

Both of these questions could be answered knowing that an object with a mass of

1 kg has a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes

that you know how many feet in a mile, and that the Sun is 93 million miles

away. All these things should be common knowledge, but it doesn't appear to be

so.

For my own weight I would reason as follows:

I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass

than I will pounds of weight, so it's reasonable to conclude that my mass is 170

/ 2.2 kg, or about 78 kg.

More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx..

(technical point: this isn't really right because pounds and kilograms don't

measure the same thing--pounds measure force and kg measure mass--but we'll

worry about that later in the course).

Converting 93 million miles to kilometers:

93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) *

(1 meter / (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.

QUESTION

What proved to be most tricky in the problems portion was the scientific

notation. I am somewhat familiar with this from

past math classes, but had trouble when dealing with using the powers of 10. I

had trouble dealing with which way to move my decimal according to the problems

that were written as 10^-3 versus 10^3. Which way do you move the decimal when

dealing with negative or positive powers of 10?

INSTRUCTOR RESPONSE

Using your numbers, 10^3 means 10 * 10 * 10 = 1000.

When you multiply a number by 1000 you move the decimal accordingly. For example

3.5 * 1000 = 3500.

10^-3 means 1 / 10^3 = 1 / (10 * 10 * 10) = 1 / 1000.

When you multiply by 10^-3 you are therefore multiplying by 1 / 1000, which is

the same as dividing by 1000, or multiplying by .001.

For example 3.5 * 10^-3 = 3.5 * .001 = .0035.

As another example 5 700 000 * 10^-3 would be 5 700 000 * (1 / 1000) = 5 700.

From these examples you should be able to infer how the decimal point moves.

You can also search the Web under 'laws of exponents', 'arithmetic in scientific

notation', and other keywords.

There isn't a single site I can recommend, and if I did find a good one its URL

might change by the time you try to locate it. In any case it's best to let you

judge the available materials yourself.

When searching under 'arithmetic in scientific notation' using Google, the

following appear as additional suggested search phrases:

scientific notation

exponents

scientific notation metric prefixes

significant digits

multiply with scientific notation

scientific notation decimal

scientific notation lessons

addition and subtraction with scientific notation

scientific notation metric system

'scientific notation lessons' might be a good place to look.

QUESTIONS AND RESPONSES

1)In the text question five asks for the percent uncertainty of a measurement

given 1.57 m^2

I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or

approximately one. ??????Am I correct in how I

calculate this??????? Can I asuume that if the number given was 1.579 then we

would calculate it by .001/1.57 = .1 % approximately or am I incorrect?????

You're on the right track.

There are two ways to look at this.

1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies

between 1.565 and 1.575.

This means that the quantity is within .005 of 1.57.

.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the

same as 0.3%, of 1.57.

Another way to look at it:

1.57 could be interpreted to mean a number between 1.56 and 1.58. The

uncertainty would then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.

2)In the text question number 11 the book asks what is the percent uncertainty

in the volume of a sphere whose radius is

r=2.86 plus or minus .09.

I know that the Volume of a sphere is 4/3 pi r^3, so I calculated the volume to

be 4/3 pi (2.86)^3 = 97.99 and to get the

percent uncertainty I tried to divide 0.09/97.99 * 100 =.091846, but the book

answer is 9% ??????I am not sure what i am doing wrong here?????????????????

Again there are two ways to approach this.

I believe the book tells you that the uncertainty in the square of a number is

double the uncertainty in the number, and the uncertainty in the cube of the

number is trip the uncertainty in the number.

An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or

about 3%. As you state, you cube the radius to find the volume. When 2.86 is

cubed, the resulting number has three times the uncertainty, or about 9%.

Another approach:

Calculate the volume for r = 2.86.

Then calculate the volume for r = 2.86 - .09 = 2.77.

You will find that the resulting volumes differ by about 9%.

You could just as well have calculated the volume for r = 2.86 + .09 = 2.95.

Again you would find that the volume differs from the r = 2.86 volume by about

9%.

STUDENT QUESTION

When reading the section about the scientific notation some of the answers were

written in powers of 10 and some were just

written regularly. How do I know when to turn my answer into a power of 10 or to

leave my answer as is?

INSTRUCTOR RESPONSE

Good question.

Convenience and readability are the main factors. It's a lot less typing or

writing to use 438 000 000 000 000 000 000 than 4.38 * 10^20, and it's easier

for the reader to understand what 10^20 means than to count up all the zeros.

For readability any number greater than 100 000 or less than .001 should

probably be written in scientific notation.

When scientific notation is first used in a calculation or result, it should be

used with all numbers in that step, and in every subsequent step of the

solution.

QUESTION

In my problems (I am working from the University Physics text- exercise 1.14)

they are asking for the ratio of length to

width of a rectangle based on the fact that both of the measurements have

uncertainty. ?????Is there anything special you

have to do when adding or multiplying numbers with uncertainty?????? I know that

there are rules with significant figures,

but I don’t understand if the same is true for uncertain measurements.

INSTRUCTOR RESPONSE:

For example:

If there is a 5% uncertainty in length and no significant uncertainty in width,

then area will be uncertain by 5%.

If there is a 5% uncertainty in length and a 3% uncertainty in width, then it is

possible for the area result to be as much as 1.05 * 1.03 = 1.08 times the

actual area, or as little as .95 * .97 = .92 times the actual area. Thus the

area is uncertain by about 8%.

This generalizes. The percent uncertainty in the product or quotient of two

quantities is equal to the sum of the percent uncertainties in the individual

quantities (assuming the uncertainties are small compared to the quantities

themselves).

(optional addition for University Physics students): The argument is a little

abstract for this level, but the proof that it must be so, and the degree to

which it actually is so, can be understood in terms of the product rule (fg) ' =

f ' g + g ' f. However we won't go into those details at this point.

QUESTIONs RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics

students)

I understand everything but the part on measuring the individual i j k vectors

by using cosine.

INSTRUCTOR RESPONSE

It's not completely clear what you are asking, but I suspect it has to do with

direction cosines.

The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x

axis, the y axis and the z axis.

Let's consider first the x axis.

The direction of the x axis is the same as the direction of the unit vector i.

The projection of A on the x direction is just a_1. This is obvious, but it can

also be found by projecting the A vector on the i vector.

This projection is just | A | cos(alpha), where alpha is the angle between A and

the x direction.

Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i

+ a_3 k dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.

It's also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it

follows that A dot i = | A | cos(alpha), so that

cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

Making the convention that alpha is the angle made by the vector with the x

direction, we say that cos(alpha) is the direction cosine of the vector with the

x axis.

If beta and gamma are, respectively, the angles with the y and z axes, reasoning

similar to the above tells us that

cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and

cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

cos(alpha), cos(beta) and cos(gamma) are called the 'direction cosines of the

vector A' with respect to the three coordinate axes.

Recall that alpha, beta and gamma are the angles made the the vector with the

three respective coordinate axes.

If we know the direction cosines and the magnitude of the vector, we can among

other things find its projection on any of the coordinate axes.

STUDENT QUESTION (University Physics)

Chapter 1 wasn’t bad of course I had to read in detail the vector section there

is little confusion on what is meant by

antiparallel. Does that mean that you wouldn’t displace anything if the

magnitude was equal only the direction was different?

Also when handwritten vectors are written above the say A the arrow is only in

one direction (to the right) not the direction

traveled?

INSTRUCTOR RESPONSE

I don't have that reference handy, but my understanding of the word

'antiparallel' is two vectors, one of which is in the direction exactly opposite

the other.

If two vectors are antiparallel, then their dot product would equal negative of

the product of their magnitudes:

The angle theta between antiparallel vectors v and w would be 180 degrees, so v

dot w = | v | * | w | * cos(180 deg) = - | v | * | w | .

Please feel free to include additional comments or questions:

&#Good work. See my notes and let me know if you have questions. &#

#$&*