Recent content by beer

Are you given three points per each triangle? ie six points total? Also are you finding IF they intersect or WHERE they intersect? You know that the triangles exist on specific planes, and that planes stretch to infinity. One way to determine if the triangles DO NOT intersect is to determine...

I'm at it again today. Helping the same friend (plus another) study for their final calculus exams; it's a good refresher for me as well. (I'm a senior industrial engineering major so I'm "done" with calculus, and it isn't used much in our upper level classes nor professionally to the best of...

I drew it out by hand and by golly I think I've got it. This is the "projection" of the triangle on to the parabolic cylinder, yes? So the "stretched" area of the triangle as it would appear on the parabola z = y2Part of my confusion was probably incorrectly calculating the area of the region...

Yes I'm familiar with those concepts. Interestingly some more complex integrals are easier for me to visualize. I'm likely over thinking this and confusing myself. y2 represents a parabola (or more technically a parabolic cylinder) that is orthogonally oriented to the triangle formed by the...

That's a fantastic question. I multiplied the "divided region" together, rather than added it. The area of R is 24. :p We didn't catch on to that earlier, but it is kind of irrelevant to my primary question. Where does y2 fit in - graphically speaking - with the region R?

Hello, I was helping my friend prepare for a calculus exam today - more or less acting as a tutor. He had the following question on his exam review: ∫∫R y2 dA Where R is bounded by the lines x = 2, y = 2x + 4, y = -x - 2I explained to him that R is a triangle formed by all three of those...

It's always a pleasure to check old replies and see what I thought was confusing a few weeks ago now makes great sense. :) Thanks for the help guys! I have my electrostatics exam tomorrow night. I feel mostly prepared for it.

Essentially, it would seem, if the two are connected then lambda is equal to some arrangement of q and length. Charge per unit length. q/L perhaps. Is it really that simple? ?:)

The only time I've repeatedly used lambda was in the study of mechanics - primarily statics and the like, where lambda was, essentially, a vector divided by it's magnitude. In the case of my post above, what is lambda equal too? λ = E 2π ε0r That's just algebraic rearrangement, though. In...

This was moved but it isn't a homework question. I answered the question already... I'm looking for discussion on ways to deduce one formula from the other. This has little to do with the question. Can this be moved back to the correct section?

Given the following information: An infinite line of charge lies long the z-axis. The electric field a perpendicular distance 0.490m from the charge is 770N/C . We're asked to do the following: How much charge is contained in a section of the line of length 1.20cm ? The answer is 2.52*10^-10...

I'm studying electrostatics and magnetism right now in Physics II for Engineers. I'm doing ok with most of the concepts, doing well in class, etc. I'm missing (or at least not 100% sure) of a basic part of the concept here. I feel like I'm correct in assuming that an object with no charge is...

So, given my example, with the resultant scalar product of 4.5, this would be vector A projected on the vector B. I can say that, I can calculate it... but what property is it projecting? There is nothing coming from this calculation except a number (a scalar) as far as I can tell. It doesn't...

I've read that projection part before. Maybe it will clear up later; I was hoping for some alternate - perhaps watered down - explanation of it in non-technical terms. "The projection of vector A on to vector B" doesn't mean anything to me at this point... what is it projecting? It's...

I'm taking "physics for engineers" right now - the condensed 4 hour summer course over 7 weeks. I'm doing fine in the class. I feel confident about the ideas and concepts we've covered so far, sure enough, but I'm having a hard time grasping the concept (geometrically at least) of the...