Solving Euler Theorem Doubts with Partial Derivatives

Click For Summary
The discussion focuses on applying Euler's theorem to solve two problems involving partial derivatives. The first problem requires demonstrating that a specific equation involving second derivatives equals zero, while the second problem involves proving a different equation equals a specific value. Participants emphasize the importance of differentiation and suggest using LaTeX for clarity in mathematical expressions. There is also a note that the term "Euler Theorem" can refer to multiple concepts, indicating a need for clarity on which theorem is being referenced. Overall, the thread highlights the application of partial derivatives in solving complex equations.
Nina2905
Messages
1
Reaction score
0
firstly, all d's i use will mean partial derivative 'do' because i don't have the font installed. sorry :(

please help me with these.. u got to use euler theorem
1. if z= xf(y/x) + g(y/x), show that x2(d2z/dx2) + 2xy(d2z/dxdy) + y2(d2z/dy2) =0
2. if z= (xy)/(x-y), PT (d2z/dx2) + 2(d2z/dxdy) + (d2z/dy2) = 2/(x-y)

thanks...
 
Physics news on Phys.org
First, it's not a matrer of having "fonts" installed, just use LaTex with [ tex ] and [ /tex ] (without the spaces) beginning and ending. To see LaTex commands, click on any formula on this board.

I'm not sure which "Euler Theorem" you mean (there are many). It looks to me like like you only need to differentiate.

If z= xf(y/x)+ g(y/x), then
\frac{\partial z}{\partial x}= f(y/x)+ x f'(y/x)(-y/x^2)+ g'(y/x)(-y/x^2)
by the chain rule. Doing that again,
\frac{\partial^2 z}{\partial x^2}= [f'(y/x)(-y/x^2)]+ [f'(y/x)(-y/x^2)+ xf"(y/x)(-y/x^2)^2+ xf'(y/x)(2y/x^3)]+ g"(y/x)(-y/x^2)^2+ g'(y/x)(-2y/x^3)]
Of course, that can be simplified a lot.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 42 ·
2
Replies
42
Views
4K
Replies
6
Views
2K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K