Recent content by math_maj0r

A rocket ship of length L (in the rocket's frame) leaves the Earth at speed v. A light signal is sent after it which arrives at the rocket's tail at t'=0 and t=0 according to Earth clocks. My question is: Why is it that the total time (according to earth) for the signal to get to the head and...

Homework Statement Two bulbs rated 25 W - 110 V and 100 W - 110 V are connected in series to a 230 V supply. What will happen? > > Homework Equations p = vi. p = v2/r The Attempt at a Solution I am not sure I know how to interpret it correctly. To find the current through...

What does this symbol mean? a^T

If n*n matrix, can row space ever be equal to null space? P.S.: this is NOT a homework question. It's a general question to get the concepts straight in my head.

thanks a lot!

How do you know if the solution set of a matrix is a point, line, plane or cube? How do you know the dimension of the solution set? P.S.: This is NOT a homework question. It's a general question about something I'm not 100% clear about

how do you know this: V(obj)= 0.127 ? and is this the volume of the logs or of the children?

thank you so much :)

thank you so much :)

Problem: evaluate the double integral of yz over that part of the plane z=y+3 which is inside the cylinder x2+y2=1 I evaluated with respect to z, from z=0 to z=y+3 I got (y3+9y+6y2)/2. Then I integrated this over x2+y2=1. To do that, I switched to polar coordinates, letting x=rcos(theta)...

the problem: evaluate the following integral by making appropriate change of variables. double integral, over region R, of xy dA R is bounded by lines: 2x - y = 1 2x - y = -3 3x + y = 1 3x + y = -2 my attempt: let 2x - y = u, and let 3x + y = v then the new region in (u,v) coordinates...

i can't figure out how to take into account the line segments from (0,0) to (2,0), and from (square root of 2, square root of 2) to (0, 0) what's the most efficient way to set up the path integral using green's thm?

The question did say that the curve is bounded by x = 0 and y = 0. I didn't mention it because I mistakenly thought it was obvious. Anyway, so to convert [y2 - 2y + x2] dx dy into polar coordinates I did: x = 2cos(theta), y = 2sin(theta), then got: [r2-4sin(theta)]r dr d(theta) Is that...

the curve doesn't consist of the whole circle. it is only part of the circle where theta is from 0 to pi/4

Curve consists of part of a circle x2 + y2 = 4 theta from 0 to pi/4 Vector field F = <y2-yx2 , yx2> (let the i component be P, and the j component be Q) I used the following formula: double integral of [(partial derivative of Q with respect to x) - (partial derivative of P with respect...