Alright, once again my (insert many derogatory terms here) math teacher has given us a ton of work and not bothered to teach us a damn thing about any of the things we're doing.

I'm somewhat familiar with vectors, but right now we're doing projections of vectors and how to find how much work is done and things of that nature.

The problem I'm having trouble with is this:


So far this is what I have done and the formulas used (an asterisk stands for taking the Dot Product, and < > stand for component form of vectors):

proj_PQv = [(v _* PQ)] / (||PQ||²)] (PQ)

W = ||proj_PQF|| ||PQ||


PQ = <-3-1, 5-3>
PQ = <-4,2>

(v _* PQ) = <2,3> _* <-4,2> = -2

||PQ|| = sqrt(-4² + 2²)
||PQ|| = sqrt(16 + 4)
||PQ|| = sqrt(20)

proj_PQv = (-2 / 20)<-4,2>
proj_PQv = (-1 / 10) <-4,2>
proj_PQv = <(4/10), (-2/10)>
proj_PQv = <(2/5), (-1/5)>

W = ||<(2/5), (-1/5)>|| ||<-4, 2>||
W = [sqrt( {2/5}² + {-1/5}²)] [sqrt( -4² + 2² )]
W = [sqrt( {4/25} + {1/25} )] [sqrt( 16 +4)]
W = [sqrt( 5/25 )] [sqrt( 20 )]
W = sqrt(100/25)
W = sqrt(4)
W = 2


So can someone with a math background or something tell me if I did that right? I'm just kind of guessing at what to do since my teacher pretty much didn't show us how to do any of it.

Hmm in Physics work can be negative, and it's -2 in this case, but by using this method to calculate work it'll always be positive. If that's the method your teacher told you it's fine.

Do you have to use projection? If you don't it's much easier just doing the internal product between the force and displacement (PQ vector)

I have no idea what I have to use

So far we've only learned Dot Product and Cross product (still barely know how to do that one).

She pretty much said "Ok here's formula A, here's formula B, now time for the next section. Oh and you have a test on that next Wednesday and two assignments on it."

She tends to show more than teach unless you go in there after class and get tutored (which I do every day), and it doesn't help much. I have the highest grade in the entire class (87/B ), and the class is comprised of the top 16 people of the Junior and Senior classes. It's pretty much a guaranteed F for the year if you don't spend 50% of your day working on that class. I have to spend an hour after school, and skip two classes a day just to get the work done. It's ridiculous.

May I inquire what grade you're in? Or are you in university.

I'll represent vectors in bold.

W = F?S, where F= force, S= displacement

F = 2i+3j
S = PQ = Q-P
=> (-3i+5j) - (1i+3j) = -4i+2j

Since, the dot product is |F||S|cosΘ ...(where, |F| = magnitude of force)

W = (2i+3j)?(-4i+2j)
=> (2)(-4)i?i+(2)(2)i?j+(3)(-4)j?i+(3)(2)j?j
=> (-8) |i|² + 0 + 0 + (6) |j|^2
=> -8 + 6 = -2 Joules

Note that the dot product of any two of i, j, k, is 0, since the vectors are mutually perpendicular.

It says in his profile info. Pre-calc/trig is usually taken by juniors in high school.

That's it, but careful because when referring to vectors' magnitude use ||F||, not |F|. | is for modulus, so it only applies to numbers. Not that many people care about that, but mathematics teachers usually do.

^^^yep. I'm a Junior this year.

Thanks! I'll ask my teacher about this today when I go in there for tutoring. I don't know if she wanted us to do them all as positives or if she didn't care how we got them.

By the nine I am screwed!

I will attempt to take a hopefully more intuititive, possibly geometric, but mathematical perspective.

First, we define "work:"

• We call the real number that is the product of the distance and the corresponding vector between two points the work done by the vector. That is, in Physics, we define work W to be the dot product of a force F displacing an object through a displacement of s as W = F * s (* denotes "dot product").

Note that we need the "component" of the force that acts exactly along the displacement, i.e. along the line segment connecting the two points.

Thus, we want to find the work done by v = 2i + 3j (I recommend using the "unit vectors" i, j as vectors. For example, if we are considering 2-space, or 2-dimension, i = [1, 0], j = [0, 1], so v = [2, 3], which is what we had before) along the line segment connecting P(1, 3) and Q(-3, 5).

It helps if you draw this out: draw P and Q in some arbitrary manner (you don't need to think about their placement on the R² (the xy plane)), but it would help if they were not vertical and that they have some distance between them. Next, draw the vector v extending from P; it would help if you drew it above the line segment, going towards Q, but this is arbitrary, as well.

Next, draw a line segment connecting the "head" or "arrow" of v to the line segment PQ, so that it is perpendicular. Let the angle that v makes with PQ (the right next to P) be K.

By trigonometry, we know that the cos function of K is as follows:

cosK = adj / hyp

Now, waving my hands a bit, let u be the vector that PQ makes: u = [-3 - 1, 5 - 3] = [-4, 2]. Consequently, where ||v|| denotes the magnitude of v (we define ||p|| = ||[p₁, p₂]|| = \sqrt{p₁²+p₂²}), and w denotes the component of v the lies on PQ,

cosK = ||w|| / ||v||

so,

||w|| = ||v|| cosK

Now, we know that w has to be in the direction of u, because we want work. Because (1 / ||u||)u is a unit vector (has magnitude 1, just like how we used i, j), and that it is clear that w is parallel to u (a numerical multiple of it), we see that (with some more hand waving)

w = ((||v|| cosK) / ||u||) u

We proceed as follows. The expression above is the same as multiplying 1 by it, which happens to be equal to ||u|| / ||u||, so we get

w = ((||v|| ||u|| cosK) / ||u||²) u

A theorem of linear algebra is that (full on hand waving now)

||v|| ||u|| cosK = v * u

which imples that

w = ((v * u) / ||u||²) u

This is often call the projection of v on u. Consequently,

w = (([2, 3] * [-4, 2]) / ((-4)²+2²))[-4, 2] = -1/10[-4, 2] = [2/5, -1/5]

By defintion, then

work = W = [2/5, -1/5] * [-4, 2] = -2

A few remarks. First, not sure about the negative, haha (it seems from the comments that the negative is correct, actually). Also, this solution is actually a derivation of projections, so you wouldn't need to write this down. However, its handy in the way that if I made this understandable enough, you have no need to memorize the formula for work, but can easily derive it when need be. (Memorizing formulas can be a problem, because you can easily misapply them.) Oh, and I forgot the most important remark: you can use what we developed as a formula. That is, physically, the work done by a vector v moving a particle from a point P to a point Q is calculated by the equation

W = ((v * u) / ||u||²) u

where u represents the vector generated by PQ.

Hopefully this helped, or at least gave you some insight into why work is what it is! (Assuming I didn't speak complete garbage, haha.)

Thanks! That really helped me grasp it better.

Maybe I wont fail my test tomorrow haha

It can be negative, and that's seen by the definition you gave Work is the mechanical energy transferred by a given force from one system to another one (from your hand to an object for example), in a given displacement. When it's positive, it means energy entered the system you're exerting the force in, and when it's negative it means energy left the system you're exerting the force in (for example a friction force - it makes the body lose velocity, hence it loses kinetic energy)

Here's some advice, don't take Calculus until college unless you are REALLY good at math. You better be making it through pre-calc with no effort to be able to do good in AP Calc. Take something else, I really wish I did. Hell I'm gonna take AP Stat next year instead of Linear Algebra (no idea why my teacher would reccomend me for that).

In my opinion and experience, if the calculus course is more computational based and doesn't emphasize on proofs and theory, then I don't think you need too much background experience to do that kind of calculus. Of course, that's not to say it will be easy (it would be a nasty amount of memorization, probably, but that really happens with any computational-based math course, IMO)

Also, if its a decent course on linear algebra, then it generally acts as a decent transitional course in mathematics. That is, if its a linear algebra course taught so to be very oriented on theory, its a good transition to "actual mathematics" if your experience was mostly computational math. However, a computational one will just svck, and be not nearly as beneficial, IMO.

What I would recommend as a great way to develop a solid foundation in math would be to read the first to parts of Calculus by Michael Spivak. Its generally a very difficult book, but its very rewarding, IMO. (The first two parts have nothing to do with calculus, where some topics include the axioms of the real number system, graphs, limits, functions...)

No problem! I'm glad it was of some use... Hopefully I didn't make too many mistakes, haha

My test is tomorrow, wish me luck.

I'll probably pop in and out of this thread occasionally to get some more help.

I found AP calc easier than pre-calc. Though my pre-calc teacher just sat there and read the book from her desk so it was pretty much just a self taught course. The concepts in Calc 1 are fairly easy to grasp. I didn't take BC (not offered by my school) but I guess that would be to much for most people for a HS year.

Side note: I wish I did vectors in Pre-calc, that would have made college physics easier.

Ok, can't figure out how to do this one. Once again, she hasn't taught us how, but expects us to know how to.

You can develop another formula for this one by basic trigonmetry, but I think there might be one or two (not complex) theorems of linear algebra that might be handy in the derivation. Actually, the formula ends up being what is known as a third degree determinant (the magnitude of the cross product of the two given vectors), but that's not necessary to know.

I'll follow what a book of mine does. First, though, I'll be quick and give the quick answer:

Given two vectors u and v that are in 3-space (i.e. in R³) that determine a parallogram (in R³) (i.e. are the two different sides

that make the parallelogram), we find the "area" A of the paralleogram

from the magnitude of their cross product: A = ||u x v||.

Just in case you don't know what a cross product is:

Given two vectors u = [u₁, u₂, u₃] and v = [v₁, v₂, v₃] that are in 3-space (i.e. in R³),

we might define their cross product, denoted u x v, as u x v = [u₂v₃-v₂u₃, v₁u₃ - u₁v₃, u₁v₂-v₁u₂].

The above is really a horrendous definition. However, I would have to go into determinants (which aren't actually hard) to give a better one.

So, the solution would go as follows.

A = ||[1, 2, 2] x [1, 0, 1]|| = ||[(2)(1) - (0)(2), (1)(2) - (1)(1), (1)(0) - (1)(2)]|| =

=||[2, 1, -2]|| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

The niceness of the result seems to suggest I did it right, haha.

Also, I unfortunately don't have time to go into the theory behind it, but I might be able to later. But, a few remarks. One, you might have noticed that I said "...might define." That was just to emphasize that the definition I gave wasn't a good one, nor common (but it seems to be for high schools, as they don't generally seem to deal with determinants). Also, I might have made a mistake writing down the definition of a cross product, but I think I got it right. Also, just to make sure I'm not causing any confusion: a vector might be denoted in bold type, a variety of different arrows place on top of it, or just written normally, and for a general vector v, we might denote its components as v₁ through whatever space it is in, so, for example, in 3-space,

v = [v₁, v₂, v₃] = v₁i + v₂j + v₃k = v₁i + v₂j + v₃k

Once again, thank you so much.

You saved me on that test today. She does 10 question tests, so if you miss more than 1 you're screwed. That was one of the ones I didn't know before.

Math. I hate it. And svck at it.

But those two are looped. I only svck because I hate it and don't want to learn, and I only hate it because I svck at it and it's hard.

Endless spiral!

No problem,

Oh dear... Yeah, I don't like it when tests are graded in a way that if you lose just one mark, it amounts to sometimes 4% the total, D=

Too bad your teacher doesn't go through this stuff with you and your class, . I find it horribly annoying to learn math (or anything else, I guess) if you're just having facts thrown at you without explanation (and even worse when they don't have the kindness to give you the facts, haha)

Ahhhh, thanks!

Yeah, Physics... I get lost in there, haha. Especially when they talk about different systems, (me being angry at the systems). That often confuses me, D=

Yeah I'm doing AP Calc AB/BC so it's every day and I just couldn't handle it, I've pretty much given up, when I gave up on it though I just focused on my other classes and my overall GPA came out pretty good considering it counts as failing two classes. Also, I know how you feel, my AP Bio teacher does that, but even less, we just get packets. There's already been a class when we had no teach (not even a sub) and the class after that we just watched Fantastic Four, pretty fun class though.


Nah it's okay, I don't plan on doing anything based in math anyway, I'll just do my required courses in college, try to slide by with minimal effort and focus on my other courses. I mean I'm good at math (certainly better than the average american) but I'm not good/motivated enough to be good in AP Calc AB/BC, the homework is long and everyday, so many proofs and theories to memorize, I gave up. At least we're finished with everything now though, series and sequences was way too long.