I have this statement:
Find the most general form of the fourth rank isotropic tensor. In order to do so:
- Perform rotations in ## \pi ## around any of the axes. Note that to maintain isotropy conditions some elements must necessarily be null.
- Using rotations in ## \pi / 2 ## analyze the...
The issue here is that I don't know how to operate the final equations in order to get the phase diagram. I suppose some things are held constant so I can get a known curve such as an ellipse.
I attach the solved part, I don't know how to go on.
I think it's fine, but I agree with @haruspex that you have to check the factorisation of ##y''##.
Just a friendly reminder: there are critical points as well where the derivative doesn't exist. In this case, the first derivative exists for every ##x##, so you don't have to matter.
As @Mark44 says, its the same expression. An integer ##k## could be positive or negative, so if you take ##2k## or ##-2k##, both determines the same set of integers, which is positive and negative even integers.
##k=\left\lbrace...,-3,-2,-1,0,1,2,3,...\right\rbrace ##...
Be careful!
There are 2 types of critical values. First type are the ones that makes ##f'(x)=0## or where the first derivative doesn't exist. Second type are the ones that makes ##f''(x)=0## or where the second derivative doesn't exist.
First type critical values could be extremes (maximum or...
I think is correct. The second derivative in ##x>0## equals ##0## and in ##-6<x<-2## the function is concave up.
## x=-2 ## would be the point of inflection.
I copy again the statement here:
So, I think I solved parts a to c but I don't get part d. I couldn't even start it because I don't understand how to set the problem.
I think it refers to some kind of motion like this one in the picture, so I'll have a maximum and a minimum r, and I can get...