Proving Characteristics of Partial Derivatives

[TranscendArcu]

Homework Statement

Let F(x,y) be a twice differentiable function such that

4 * F_x² + F_y² = 0.

Set x = u² - v² and y = u*v. Show that

F_u² + F_v² = 0

The Attempt at a Solution

F_u = F_x*2u + F_y*v
F_v = F_x*-2v + F_y*u

F_u² = 4v²F_x²-4F_xF_y*u*v+F_y²u²

F_v² = 4u²F_x²+4F_xF_y*u*v+F_y²v²

Adding these two together gives

4F_x²*(u²v²) + 4F_y²*(u²v²),

Which from our first equation, can clearly equal zero if we factor out the (u²v²) and divide. Then, we see that, indeed, F_u² + F_v² = 0.

Sound about right?

 

[SammyS]

Homework Statement

Let F(x,y) be a twice differentiable function such that

4 * F_x² + F_y² = 0.

Set x = u² - v² and y = u*v. Show that

F_u² + F_v² = 0

The Attempt at a Solution

F_u = F_x*2u + F_y*v
F_v = F_x*-2v + F_y*u

[STRIKE]F_u² [/STRIKE]= 4v²F_x²-4F_xF_y*u*v+F_y²u² = F_v²

F_v² = 4u²F_x²+4F_xF_y*u*v+F_y²v² = F_u²

Adding these two together gives

4F_x²*(u²+v²) +[STRIKE] 4 [/STRIKE] F_y²*(u²+v²),

Which from our first equation, can clearly equal zero if we factor out the (u²+v²) and divide. Then, we see that, indeed, F_u² + F_v² = 0.

Sound about right?

You have some typos in your post.