Somehow, everyone in my class gets this, but me. I don't know how I missed it, but I've never learned this stuff... I'm past the proof, but now I have another problem and just don't understand what it means.
The book we're using is the third edition of Abstract Algebra, I.N. Herstein. The...
I'm trying to study a proof for a quiz right now, and even though I have the answer, I don't get it because I don't understand what an orbit is. I got no definition in class, and can't find anything in the book. I looked on wikipedia, but that was completely over my head.
Can someone please...
I have M= CINT d(x,y,z) ds
v(t) = 2tj + 2k ; |v| = 2root(t2 + 1)
M = 0S1 (3/2)t 2root(t2 + 1) dt
Use u substitution- u=(t2 + 1) du= 2t dt
(3/2) INT U1/2 du
(3/2) (1/{2root(t2 + 1)}) from 1 to 0
and i got (3root2 - 6)/8 but I should have gotten 2root2 - 1
I am skipping that problem and one very similar to it. Hopefully he will not put them on the exam, but I cannot afford to look at this problem any longer. It's been over an hour, and I'm down to 5 hours before my exam.
Now I am working on:
Find the mass of a wire that lies along the...
For the most recent problem,
Integrate f over the given curve:
f (x,y) = x3/y
C: y = x2/2
0<= x <= 2
is r(t) = 2ti + 2tj correct? Then |v| = 2root2. I believe t is in the interval [0,2]. My line segment goes from (0,0) to (2,2) if I plug x=0,2 into the y= (x^2)/2.
The answer...
I just realized that completely changes my integral, but I still got the right answer!
INT 1-t+2-3t+3-2t root14 dt
root14 INT 6(1-t) dt
6root14 INT 1-t dt
6root14 [t-.5t2] from 1 to 0
6root14 * .5 = 3root14
Okay. I think I have r(t), but it isn't pretty.
r(t)= (1-t)i + (2-3t)j + (3-2t)k
v(t)= -1i -3j -2k
|v| = root(1+9+4) = root14 (AHA!)
Is that right? And t goes from 0 to 1 because if you plug in 0 you get one endpoint of the segment and if you plug in 1 you get the other endpoint.
Next problem. Three more and then next section.
Integrate f over the given curve:
f (x,y) = x3/y
C: y = x2/2
0<= x <= 2
I don't even know where to start! There's no example like this in the text. I have nothing to follow.
Okay, that last one wasn't so bad once I got into it.
r(t) is given. From that |v(t)| = root3
Interval is given: [a,b]
Sub t into each x, y, and z. Reduce the equation and you get:
root3 aSb (t-1) dt
root3 [ln b - ln a] = root3 ln(b/a) which is the answer.
Is my technique...
The next one is even worse. I'm going to post it before I work on it to give you some time to look at it also. You are helping me so much! I wish there was some way I could repay you!
Integrate f(x,y,z) = (x + y + z)/(x2 + y2 + z2) over the path r(t) = ti + tj + tk, 0< a<= t <= b.
Logically, I want to say r(t) = ti + 2tj + 3tk because one of the points is (1,2,3), but there's no way it's that simple... It has to have 1,2, and 3 for coefficients, because that gives root14.
Is it as simple as (x1-x0)ti + (y1-y0)ti + (z1-z0)ti? So, (0-1) for x, (-1-2) for y, and (1-3) for z? Or am I backwards? I've been awake for nearly 24 hours already. My brain isn't functioning fully
All is says that makes any sense whatsoever is the snippet I put in the first post.
1. Find a smooth parametrization of C
r(t) = g(t)i + h(t)j + k(t)k, t in the interval [a,b]
2. Evaluate the integral as
Sc f(x,y,z) ds = bSa f (g(t),h(t),k(t))|v(t)| dt
Okay. I have a theory on that last one. At least, I get the right answer. Again, is my technique correct?
f(x,y,z)= x+y+z
r(t) = ti + 3tj + 2tk (the distance between corresponding coordinates in the points given.)
v(t) = i + 3j + 2k |v(t)| = root14 (YAY)
I'm assuming t...