Recent content by Alex Bard

Thanks for answering mfb, First I noticed I made a mistake, in spherical coordinates x= ρ sin\phicosθ y= ρ sin\phisinθ so now when I substitute that into the formula, I will get √[(ρ sin \phi cosθ)^2 + (ρ sin \phi sinθ)^2] What I get stuck with is I can't get rid of the square...

Homework Statement Evaluate the iterated integral ∫ (from 0 to 1) ∫ [from -sqrt(1-x^2) to sqrt(1-x^2) ] ∫ (from 0 to 2-x^2-y^2) the function given as √(x^2 + y^2) dz dy dx The Attempt at a Solution I changed the coordinates and I got the new limits as ∫(from 0 to pi) ∫(from...

Thank you for your help and I believe I got the problem already solved on paper, just do not know how to mark the post as SOLVED as I can't edit my original Post

Edited the above post for correction. Thank you for point out my mistake and I will be more careful going forward.

Okay, so in that respect, this is what the problem should look like. Let f (x, y) = cos(xy) + y cos(x); = [-y sin(xy) -y sin (x)](1) + [-x sin(xy) + cos(x)](-2v) substituting u & v for their respective values of x & y: = [ (-u-v^2)(sin((u^2+v)(u-v^2)) - (u - v^2) sin((u^2 + v)(u - v^2))] +...

Thank you! Finally I can put this problem to rest, I believe. Guess I'll have to look at it with fresh eyes in the morning.

∂u/∂v means the change in u in respect to change in v. But isn't it ∂x/∂v? As in the change in x in respect to v? so it would be just 1? considering the u^2 is a constant it would be 0 and v would become 1, correct?

Okay, I see that it should be (2u + 1) in ∂x/∂v and I see, it was an ill placed bracket, the y should NOT be distributed across both terms. It should be: = [(y)-sin(xy) - y sin(x)] (2u + v) Am I correct in both of those assessments?

Sir, I believe I've cleaned it up right before your comment. Per haps it looks better now? Let me know if its still messy and I will attempt to break it up. Also, you are correct, i did mess that part up and re-did my math as a result. Thank you for bringing it to my attention.

So then it's simply square both and add under a square root?

Apologize for the messiness, I rewrote and did my best to follow the brackets and (), please let me know if it doesn't make sense for me to go back and clean it up even further.

Okay sir, I'm stumped. Find the Directional derivative in the direction of fastest increase at the point (√5, -2) The Direction of the FASTEST increase is Gradient, which I've found is ∇f(√5, -2) = <\frac{-4}{21}, \frac{-4√5}{21}> The magnitude of the gradient is the maximal...

Yes sir, for some reason seeing the integral sign at first looked like a f to me, as in function. Fixed the notation. and

I fixed the 25 to 21. This is the theorem from the book. I actually didn't feel too comfortable with the answer myself and have been searching the internet for a better explanation. I understand that the gradient is the direction of the maximum increase, and is orthogonal to the level...

Where should I send the cake? LCKurtz, I believe I owe you a cup of coffee or something. Thank you for your input and gentle guidance. Let f(x, y) = tan^{-1} \frac{y^2}{x} A. fx(√5, -2) = \frac{1}{1 + \frac{y^4}{x^2}} * \frac{y^2}{-x^2} → \frac{-y^2}{1 + y^4x^{-2}} * \frac{1}{x^2} →...