Recent content by Anakin_k

Well thanks for clarifying that up you guys. I have one more similar question with which I need some help. I've gotten a bit further with this one but not enough: I concluded that the limit equals 0 by trying a few paths (hopefully this time it was right): \lim_{\{x,y\}\to \{1,0\}} \...

I updated my post above, to show that I did try some other paths too. So if both limits are different, then it would mean that the limit does not exist. I guess that WA isn't perfect after all. :)

Hi, If we are approaching from the path x = y^3, then lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3 = lim (x,y) -> (0,0) of (y)*y^2 / y^3+y^3 = = lim(x,y) -> (0,0) of y^3 / 2y^3. = 1/2 That seems to make sense. However, if we use the path y=x: lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3 = lim...

Homework Statement Evaluate or show that the limit DNE. Limit as (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3. The Attempt at a Solution I tried approaching from multiple paths, and it seems that the limit is equal to 0. I used the delta-epsilon method to prove the limit but I've been stuck so far...

That was extremely thorough Mathstatnoob, thank you for taking the time to post the answer. I appreciate it very much. Through your steps, I've learned how to manipulate little pieces here and there to arrive at the needed answer. :) If anyone could confirm my Q1 and Q5 solutions, I'd be...

Hello everyone, I am in need of a little assistance. I have a homework assignment due soon that consists of 5 questions. Of which, I have done all but the 4th one. I did start on it but I'm not sure where to go from there. I also would like for someone to confirm my solutions for question 1 and...

.1^(x-1) < .1 log (.1)^(x-1) < log .1 (x-1)(log .1) < log .1 Expand and solve.

I think it should be dV/dT = dH/dT * dV/dH rather than what you had, Mandeep.

What was the answer you got? I got 0.071 m/s.

sin^3(x)-cos^3(x) ----------------- sin(x) - cos(x) Try factoring the numerator, it may help you.

Oh I got it! Gee, thanks for all the kind help Rock, I appreciate it. :)

In that case: Ax+By+Cz+D=0 (5)(2)+(-4)(-1)+7(8)+D=0 10+4+56+D=0 D=-70 Therefore, the Cart. Eq is 5x-4y+7z-70=0. Right? And any luck on the 2nd question?

That makes sense. Perhaps, they are the same? Just to be sure in the equation [x,y,z] = (-1,-2,-3) + s[5,-4,7], [5,-4,7] is the direction vector right? If that is the case, could we use Ax+By+Cz+D=0 and plug in 5 for A; -4 for B; 7 for C? And x,y,z would be 2, -1 and 8 respectively to solve...

Um I THINK they are parallel? Something in my head pops up about the normal being perpendicular to the line. So if the line is perpendicular to the plane, they are parallel? I'm likely wrong.

Homework Statement Write the Cartesian equation for the plane containing the point (2,-1,8) and perpendicular to the line [x,y,z] = [1,-2,-3] + s[5,-4,7]. The Attempt at a Solution The situation is that I have my Calc. + Vectors exam tomorrow morning and I'm just going through some...