Avant Brown Want to Play a Game?

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Nyarlathotep's picture
Stare - ...it's hard for me

Stare - ...it's hard for me to visualize a solution...

Don't try to visualize the solution. Just work it out with the method you suggested earlier. It will give you the answer eventually.

This fear of starting a problem that you can't visualize the ending of before you start; is a serious problem I've discovered many students suffer from. Just get your hands dirty and see where the math takes you. Sometimes I call this "mathematical courage". 99% of the stuff I work on, I can't see the solution from the beginning. It is nice when you can do it, but it is not a requirement; acting as if it is a requirement will cause you big problems if you pursue this kind of work.

Stare's picture
Don't try to visualize the

Don't try to visualize the solution. Just work it out with the method you suggested earlier. It will give you the answer eventually.

This fear of starting a problem that you can't visualize the ending of before you start; is a serious problem I've discovered many students suffer from. Just get your hands dirty and see where the math takes you. Sometimes I call this "mathematical courage". 99% of the stuff I work on, I can't see the solution from the beginning. It is nice when you can do it, but it is not a requirement; acting as if it is a requirement will cause you big problems if you pursue this kind of work.

That's excellent advice.

But my worry here, is that the notation you gave namely "(2t^3 + t^2 + 5t +8)ĵ", although resembling a valid position, does not really seem to be a valid position. According to the books I've come across, which is only a few admittedly and other resources like the enotes on acceleration calculation, 2d positions tend to include unit vector pairs at minimum, including something like i and j.

I don't see any position that falls short of this cardinality of unit vectors any where else so far. So I am not confident in progressing, as I don't think it's legal to differentiate what resembles a position, but what really isn't one, as far as I'm seeing online. As far as I've seen there are rules for differentiation herein; measurements of change are occurring wrt important units like the vectors described.

Nyarlathotep's picture
Stare - "(2t^3 + t^2 + 5t +8

Stare - "(2t^3 + t^2 + 5t +8)ĵ", although resembling a position, does not really seem to be a position.

It is a position function. It tells you the position o̶f̶ at every point in time:

  • At time 0, the position is {0,8} if we adopt the convention of putting the coordinate associated with î first.
  • At time 1, the position is {0,16}
  • At time 2, the position is {0,38}
  • At time 3, the position is {0,86}

Clearly it is a position function. You can also see here why the shortcut works, because the particle never changes position in the î direction.

Stare's picture
In the initial problem you

In the initial problem you gave on page 1, you asked for the "î component of acceleration" of the position. What exactly are you asking for? It sounds like the coefficient of i after one would have differentiated a position to attain some velocity, then differentiated again to get the acceleration, then the acceleration/answer would look like this: 82î + 23j (as seen here), from which one would take 82 to be the answer to your question, if " î component of acceleration" is to be interpreted as the "coefficient of î".

If that's the case how would we get the coefficient of î if it is missing? Or did you mean something else by asking for the "î component of acceleration"?

Nyarlathotep's picture
Stare - ...how would we get

Stare - What exactly are you asking for? It sounds like the coefficient of i...how would we get the coefficient of î if it is missing?

Yes that is exactly what I'm asking for. I've already explained this, but I'll try again: the fact that î is "missing" from the position function, means its coefficient is 0.

Are you starting to see how easy this problem really is for someone familiar with the subject? And what a nightmare it is for people who are not? That is by design as well:

The problem is phrased in terms of position functions, acceleration functions, orthonormal unit vectors, and calculus. However I could have phrased it l like this: how much "up" motion (or acceleration) is there in a object that is NOT moving up? Then everyone knows the answer is 0. I wanted to see if the "PhD candidate" knew what the words meant; clearly he does not.

Stare's picture
Yes that is exactly what I'm

Yes that is exactly what I'm asking for. I've already explained this, but I'll try again: the fact that î is "missing" from the position function, means its coefficient is 0.

Are you starting to see how easy this problem really is for someone familiar with the subject? And what a nightmare it is for people who are not? That is by design as well.

The problem is phrased in terms of position functions, acceleration functions, orthonormal unit vectors, and calculus. However I could have phrased it l like this: how much "up" motion (or acceleration) is there in a object that is NOT moving up? Then everyone knows the answer is 0. I just wanted to see if the "PhD candidate" knew what the words meant, he didn't.

I'm thinking that the way you just phrased it, seems like an entirely different question altogether.

You asked for the "î component of acceleration", which seems to imply by default, that an acceleration would have already been calculated, and hence why the "î component of acceleration" or "î coefficient" was asked for. (Which would imply that unit vector i should have been in the position, from what I understand?)

Can you point me to any resource that shows what happens if the î component is missing? I don't mean an example question, I simply mean the mathematical implication of having the î component omitted?

Stare's picture
use the product rule on it,

BTW, the answer to the question is just 0.

Can be calculated with some simple first semester calculus.

Or it can be gotten in an even easier way with high-school trigonometry: by just realizing the answer contains (ĵ · î) which is just cos(π/2) which is 0.

Why did you multiply j by i above?

What rule gets you from the item you proposed as a valid position, namely: ""(2t^3 + t^2 + 5t +8)ĵ"" to multiplying j by i?

Nyarlathotep's picture
Stare - Why did you multiply

Stare - Why did you multiply j by i above?...What rule gets you from the item you proposed as a position, namely: ""(2t^3 + t^2 + 5t +8)ĵ"" to multiplying j by i?

Vectors are not multiplied like scalars (the multiplication that most people know how to do is only for scalars), I took the dot product. Which tells us the magnitude in the î direction, of a vector in the ĵ direction; which is 0.

In imprecise English: I calculated how much "East" is in an arrow that points "North"; which is 0.

Stare's picture
What I'm wondering really, is

What I'm wondering really, is given your position ""(2t^3 + t^2 + 5t +8)ĵ", if the actual position is "(2t^3 + t^2 + 5t +8)ĵ ± (undefined-value-associated with i)", where i is omitted, how would we attempt to calculate the derivative "d/dt" of an undefined value, in order to identify the "i component of acceleration" from an acceleration calculation as you requested?

Nyarlathotep's picture
Stare - "is given your

Stare - "is given your position ""(2t^3 + t^2 + 5t +8)ĵ", if the actual position is "(2t^3 + t^2 + 5t +8)ĵ ± (undefined-value-associated with i)", where i is omitted"

AGAIN: the position function could be written as (2t^3 + t^2 + 5t +8)ĵ + 0î + 0k̂; but the simplest way to write it is how I wrote it in the problem. The coefficient of an unlisted object is always 0; not undefined as you suggested.

Nyarlathotep's picture
Another simple way to put it

Another simple way to explain it might be:

If I asked a person how many pets they have they might respond:

  • 1 cat
  • 3 dogs
  • 1 horse
  • so 5 pets total

But that is equivalent to:

  • 1 cat
  • 3 dogs
  • 1 horse
  • 0 birds
  • 0 elephants
  • so 5 pets total

The lesson is: unlisted objects have a coefficient of 0. They are 0 because they don't contribute anything and can therefore be safely omitted.

Tin-Man's picture
@Nyar

@Nyar

Holy crap, dude, even I can understand that. And the most I have ever had is Pre-Cal. Hate to say it, but this Miss Stare is starting to sound a bit fishy, just as you mentioned a bit earlier in this thread. And if I am wrong, Miss Stare, then I do sincerely apologize. However, something ain't addin' up here. (Pardon the pun.)

Stare's picture
If I asked a person how many

If I asked a person how many pets they have they might respond:

1 cat
3 dogs
1 horse
so 5 pets total
But that is equivalent to:

1 cat
3 dogs
1 horse
0 birds
0 elephants
so 5 pets total
The lesson is: unlisted objects have a coefficient of 0. They are 0 because they don't contribute anything and can therefore be safely omitted.

Alright, I guess it's time to make things clear.

Note: Forgive me for some of the all-caps below, it's just used to emphasize some key points, to describe the huge issue I see with the problem you proposed on page 1, if I am not mistaken.

  • It is possible to calculate motion in 1D or 1 dimension, where our position vector of coordinates would simply consist of 1 TYPE OF direction or 1 unit vector, like ĵ, OR any letter of your choosing, typically representing a change in the x (OR whichever coordinate) ONLY. An example is seen here, with valid position functions, in 1D.
  • It is possible to calculate motion in 2D or 2 dimensions, where we could include something like î and ĵ or any two other unit vector labels, to represent our 2D position vector, where each unit vector would pertain to an axis, typically x and y axis ONLY. An example is seen in the enotes url we previously discussed.
  • It is possible to calculte motion in 3D, or 3 dimensions, using something like the unit vectors î, ĵ, and k̂, ONLY.
  • etc.

    Conclusion:

    1. This is why I asked why you multiplied î by ĵ above, your position from page 1 indicated a 1D position, which is not 2D, so there would be no rule to suddenly graduate that 1D position, in order to apply the property gained when a zero valued component unit vector like unit vector î, occupies shared vector space with some non-zero unit vector like ĵ, that is, your multiplication would perhaps apply only if you had actually written 0î in your problem. Since you didn't write 0î, i.e. since your position vector encompassed only 1D, there doesn't seem to be any rule that would enable your 1D position to become 2D, so you could benefit from that dot product involving shared unit vectors.

      In simpler words, unlike dealing with coefficients in maths, the dimensions we are dealing with wrt your problem are not arbitrary, where you can simply invoke unstipulated items to exist as zero valued items, as you suggested above.

    2. What this means based on evidence, is that you asked for "î component of acceleration" on page 1, which would imply a calculation on a 2D position vector, not the 1D one you proposed in your problem.
    3. This means your answer zero would seem to be wrong, from what I can garner.
    4. Crucially, what you asked for, seems to be in conflict with the space or position you gave us to work with; i.e. you asked for data related to "î" regarding a dimension that was not present in your position or problem space "(2t^3 + t^2 + 5t +8)ĵ". Correct me if I am wrong.
    Meepwned's picture
    Math is my worst subject and

    Math is my worst subject and I barely passed algebra, but wouldn't the y axis of anything 1D always be 0 because that axis wouldn't be represented?

    Tin-Man's picture
    Stare is starting to sound

    Stare is starting to sound less and less like a fourteen year old girl with each post.

    Randomhero1982's picture
    Everyone is a 14 year old

    Everyone is a 14 year old girl online lol

    Tin-Man's picture
    @Random Re: "Everyone is a

    @Random Re: "Everyone is a 14 year old girl online lol"

    Not true. I'm only 13 and a half.

    Stare's picture
    Math is my worst subject and

    Math is my worst subject and I barely passed algebra, but wouldn't the y axis of anything 1D always be 0 because that axis wouldn't be represented?

    Not at all, as far as physics/motion calculations seems to go. The y-dimension would have to be specified to exist in the problem, or be given in the problem.

    In math without any physics-like considerations, we can arbitrarily define coefficients to exist as zero-valued items, even if the associated variable is not stipulated, because they don't refer to specific structures with distinct properties.

    However, from what I can see of physics or motion calculations, the math becomes specific; i.e. the quantities associated with the unit vector labels, like î, are very specific values, that describe very specific regimes. This means in the absence of labels, like î, we can't just assume that such an î based dimension exists in the problem, since it was not stipulated.

    In simpler words, we can't just presume the y dimension exists in our particular problem space. Y=0 implies that dimension exists. Y not specified at all, implies the dimension doesn't exist at all in that space. Notably, 0 is not equal to being undefined.

    I gave a link in my previous response about coefficients in math and physics: https://whatis.techtarget.com/definition/coefficient

    Nyarlathotep's picture
    Meepwned - ...wouldn't the y

    Meepwned - ...wouldn't the y axis of anything 1D always be 0 because that axis wouldn't be represented?

    A little garbled, but yes, that is correct.

    Nyarlathotep's picture
    @Stare

    @Stare

    This recent post of yours contains a large number of false claims, not questions.

    Recently you repeatedly suggested assigning a scalar value to a vector, this is a neophyte mistake, the mistake of someone who doesn't really know what a vector is. Someone who is need of instruction (if they actually want to learn the subject). I'm more than willing to help provide some.

    What I'm not willing to do is argue the finer points of vector and vector spaces with a neophyte. If that is what you are here for, you are wasting everyone's time.

    I set up the 1st derivative for you to solve, but I see you haven't even attempted it yet. There is no 2nd orthonormal vector in that part of the problem, so your misguided complains don't even apply there. Are you planning on trying? It's OK to get it wrong, I'll walk you though any mistake. But you aren't going to learn anything if you don't try.

    If you want me to review those false statements, I will; but I'd like to get some sign from you that you want to learn, not troll.

    Stare's picture
    @Stare

    @Stare

    This recent post of yours contains a large number of false claims, not questions.

    Recently you repeatedly suggested assigning a scalar value to a vector, this is a neophyte mistake, the mistake of someone who doesn't really know what a vector is. Someone who is need of instruction (if they actually want to learn the subject). I'm more than willing to help provide some.

    What I'm not willing to do is argue the finer points of vector and vector spaces with a neophyte. If that is what you are here for, you are wasting everyone's time.

    I set up the 1st derivative for you to solve, but I see you haven't even attempted it yet. There is no 2nd orthonormal vector in that part of the problem, so your misguided complains don't even apply there. Are you planning on trying? It's OK to get it wrong, I'll walk you though any mistake. But you aren't going to learn anything if you don't try.

    If you want me to review those false statements, I will; but I'd like to get some sign from you that you want to learn, not troll.

    Which post are you talking about, where I made a neophyte mistake? This is possible since I really am a newbie here, but I can't find the mistake you refer to.

    The only post I could find about related to scalars, is where I asked if i should be equal to 1 other dimension. I don't see how that makes it seems that I would think i to be a scalar.

    Could you please explain why that would sound like I'm talking about scalars? (Ps, I know what vectors are, there just collections of numbers. A vector can almost look like a scalar, with "one entry" or 1 item. )

    Stare's picture
    Meepwned - ...wouldn't the y

    Meepwned - ...wouldn't the y axis of anything 1D always be 0 because that axis wouldn't be represented?

    A little garbled, but yes, that is correct.

    Yes, I did agree to him, but with one extremely crucial condition, 1D suggests that a y dimension exists unless I am mistaken.

    I don't see anywhere so far in physics literature, where one can unspecifiy a dimension in a problem, then simply presume said problem includes other dimensions.

    This means to me, that if Meepwned proposed a problem with 1D being on an x axis, there would be no reason to presume his proposed problem also included a y axis as well. Please correct me if I'm wrong.

    Stare's picture
    @Stare

    @Stare

    This recent post of yours contains a large number of false claims, not questions.

    Recently you repeatedly suggested assigning a scalar value to a vector, this is a neophyte mistake, the mistake of someone who doesn't really know what a vector is. Someone who is need of instruction (if they actually want to learn the subject). I'm more than willing to help provide some.

    What I'm not willing to do is argue the finer points of vector and vector spaces with a neophyte. If that is what you are here for, you are wasting everyone's time.

    I set up the 1st derivative for you to solve, but I see you haven't even attempted it yet. There is no 2nd orthonormal vector in that part of the problem, so your misguided complains don't even apply there. Are you planning on trying? It's OK to get it wrong, I'll walk you though any mistake. But you aren't going to learn anything if you don't try.

    If you want me to review those false statements, I will; but I'd like to get some sign from you that you want to learn, not troll.

    Please correct me if I'm wrong.

    Ps: Differentiations to begin, are very simple, it just takes one from the power of a variable, before which one would have multiplied the coefficient etc. I didn't differentiate the term for a simple reason, what you asked for didn't seem to agree with the dimension of your position.

    As far as I understand, your position is 1D namely associated with "ĵ", but you asked for an item related to another dimension, namely associated with "î".

    Is it legitimate to introduce a new dimension to your position, in order to try to solve what you asked for as I worried about in the recent post of mine you linked?

    How many dimensions does your position refer to? Does what you ask for "the î component of acceleration" refer to a 2nd dimension that was not stated in your problem position? Do you agree that the position you gave was 1D as seen in these examples?

    In simpler words, should I have ignored the difference between 1D and 2D kinematics, where what you asked for seems to concern 2D, while the position you gave was 1D?

    I think the answers to these questions could clear up my confusion. Thanks.

    Nyarlathotep's picture
    Stare - As far as I

    Stare - As far as I understand, your position is 1D namely associated with "ĵ", but you asked for an item related to another dimension, namely associated with "î".

    Yes, that is exactly right, which means the answer depends on the "relationship" between ĵ and î; which is ĵ ∙ î; which depends on cos(π/2); which is 0.
    ---------------------------------------------------------

    Stare - Differentiations to begin, are very simple...I didn't differentiate the term for a simple reason, what you asked for didn't seem to agree with the dimension of your position.

    You have gone back to the same disease, you seem unwilling to do any work on the problem because you can't see the finish line; even though you agree this first step is easy. Finish (or at least try) the first step I gave you.
    ---------------------------------------------------------
    You have a lot of false assumptions about working with dimensions. You need to abandon them if you want to learn anything. It is a very common trick to express motion in the standard basis (î, ĵ, k̂) as one dimensional motion. It is one of the first things you will learn how to do in an introductory physics class because it greatly reduces the difficulty of a problem.

    Stare's picture
    You have a lot of false

    You have a lot of false assumptions about working with dimensions. You need to abandon them if you want to learn anything. It is a very common trick to express motion in the standard basis (î, ĵ, k̂) as one dimensional motion. It is one of the first things you will learn how to do in an introductory physics class because it greatly reduces the difficulty of a problem.

    Hold on, no where did I say one can't work with a 3D motion concerning X dimensions, in terms of a 1D motion concerning X-n, that is, if you start with a 3D motion, depending on the task, one can "constrain", but still stipulate some dimensions such that the initial 3D structure "looks like" a 1D structure.

    However, your question didn't start at X dimensions, it started at X-n dimensions namely 1D, so structures involving 2D would not be there to support a scenario, where we can benefit from shared vector space found in 2D structure, from what I can see.

  • If we started out at 3D, we wouldn't need to generate any extra dimensions, to act as 2D or 1D. However, if we started out at 1D, we would need to generalize our model to 3D, which is non-trivial, if I am not mistaken.
  • Now, stating how each dimension is represented in terms of different sets of vectors is "trivial", but generalizing from one dimension to another higher one is a different problem altogether, as far as what I see in physics so far.
  • Please correct me if I'm wrong.

    Stare's picture
    You have gone back to the

    You have gone back to the same disease, you seem unwilling to do any work on the problem because you can't see the finish line; even though you agree this first step is easy. Finish (or at least try) the first step I gave you.

    Differentiating your position does not yield any i term, which is related to what your problem asks for; from what I can see differentiation in this regard does not yield new dimensions.

    Can we benefit from shared vector space operations, between j and a supposed i, when i was unspecified to begin with?

    Nyarlathotep's picture
    Nyarlathotep - If you want me

    Nyarlathotep - I set up the 1st derivative for you to solve, but I see you haven't even attempted it yet...If you want me to review those false statements, I will; but I'd like to get some sign from you that you want to learn, not troll.

    Do you want to troll or learn? Last chance. Don't blow it.

    Stare's picture
    Do you want to troll or learn

    Do you want to troll or learn? Last chance. Don't blow it.

    I wish to learn.

    Crucial question: I know that one can observe relationships between vectors in shared vector space, for eg, between two unit vector data in 2D i.e. between i and j or k and i for eg.

    However, can we still benefit from that type of relationship, if we start with 1D? Is there anything for dimension 1 unit vector to relate to, if there's not shared space between two sets of unit vector, i.e. only 1D exists in problem? Is going from 1D to 2D not a separate physics question altogether?

    Nyarlathotep's picture
    Stare - I wish to learn.

    Stare - I wish to learn.

    Then calculate the velocity function (or at least attempt it) in your next post.

    The answers to your questions will become obvious if you do the work. If you continue to try to troll by just stringing together random mathematical terms you have googled, you won't learn shit.

    arakish's picture
    You know, I am glad I ain't

    You know, I am glad I ain't got to work on mathematical equations anymore. Except maybe for some algebraic rearrangement. I just know how to find them, use them, and program them into a script/program. Mostly that is what I do. I just program the equations in with appropriate text boxes to enter the variables, then let the program solve it for me. To hell with plugging it into a calculator. Give me a program where I can enter the variables and hit <Enter> to calculate.

    rmfr

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