Recent content by KUphysstudent

Oh it was this simple. I was afraid to get guess but thanks really helped me :)

Yes it is, how did you know? :P

Limh→0+ (f(rh,h))/h Is the f(rh,h) part the same as f(r+h)-f(h)? I have never seen this before and googling for a long time didn't help, there are no videos with this notation and it's not in my book so, am I just to assume it is? because it doesn't look like it should be the same. Anyone know...

That makes sense. so you can basically check every time you have manipulated the equation to see if the equality still is correct, if not you did it wrong.

Yea I get the algebra part, I just kept making the same mistake. I think I do, the bold line is the one you get from the inequality and the green area is to depict the positive x- and y-values. so you put in an x-value and get a y-value and in the first quadrant it is everything below the...

Yea okay that makes sense. Okay I'm still very confused, there must be something fundamentally wrong with my results. So if I start out with 5xy-2y2 ≥ 0 → 5xy ≤ -2y2 → 5x ≤ -2y → -5x/2 ≥ y , so this is wrong but what is the problem? I cannot remember ever learning about inequalities in school...

I mean, don't we already assume that the number is positive since it under the radical? meaning you can basically throw the negative symbol away to begin with so we have 5xy≥2y2

No matter how I solve the inequality there will be a negative sign, so I get the complex part of the function if I try would sketch it. So wouldn't sketching it and taking everything the outside the complex part be a thing you could do? Just a thought I'm going to look at the factoring part :)

Yea signs in inequalities shift direction if you multiply or divide with a negative number, I knew there was something so I googled it before hand. If x ≥ -2y/5 or -y ≥ 5x/2 tells me where the function is defined, well then that is its domain. What format is commonly used to denote it? I think...

Homework Statement Find the domain of f(x,y)=sqrt(5xy-2y^2) Homework EquationsThe Attempt at a Solution As far as I understand it's all about solving inequalities when you have this kind of problem. The problem doesn't state it is in ℝ2 but I'm pretty sure it is. So no negative numbers under...

oh, that explains a lot. Thanks you two I was getting frustrated :)

Homework Statement So it is pretty straight forward, solve this. z2+2(1-i)z+7i=0 Homework Equations z2+2(1-i)z+7i=0 (-b±√(b2-4ac))/2a The Attempt at a Solution So what I would do first is solve 2(2-1)z, I get (2-2i)z=2z-2iz we now have z2-2iz+7i+2z=0 Now I don't really know what to do because...

if you calculate the integral with this step, (integralsign)∫ t3 dt + 3t2 t2 dt from -1 to 2 you get 25.25 I forgot to add that if you graph y=x^3 and look at the area between x = 1 and x = 2 it looks like it adds to about 4, which is quite a bit away from 25.25

Homework Statement Calculate the line integral ° v ⋅ dr along the curve y = x3 in the xy-plane when -1 ≤ x ≤ 2 and v = xy i + x2 j. Note: Sorry the integral sign doesn't seem to work it just makes a weird dot, looks like a degree sign, ∫.2. The attempt at a solution I have to write something...

<Mentor note: moved from a technical forum and therefore without template>So I´m trying to understand how to use the equation for finding the gradient in spherical coordinates, just going from cartesian to spherical seemed crazy. Now I´m at a point where I want to try out what I have read and I...