# TD_6 - ondes u 00 tt = a 2 u 00 xx a u = sin kx sin akt b u...

This preview shows pages 1–2. Sign up to view the full content.

TD 6 Mohammed Saddoune AUTOMNE 2009 Exercice 1 Soit la fonction f ( x, y ) = e x + y a) Donner une fonction linéaire L ( x, y ) qui approxime f ( x, y ) au point P o = (0 , 0) . b) Trouver l’équation du plan tangent à la surface d’équation z = e x + y au point (0,0,1). Exercice 2 Soit f ( x, y ) une fonction satisfaisant 2 f ∂x 2 = 12. (*) En utilisant le changement de variables x = s + t et y = s - t , Transformer l’équation (*) dans le systeme de coordonnées { s, t } . Exercice 3 Donner l’expression de ∂z ∂x et ∂z ∂y pour: a) z = f ( x ) g ( y ) . b) z = f ( x/y ) . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Exercice 4 Montrer que chacune des fonctions suivantes est une solution de l’équation des
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ondes u 00 tt = a 2 u 00 xx a) u = sin ( kx ) sin ( akt ) . b) u = ( x-at ) 6 + ( x + at ) 6 . Exercice 5 Expliquer pourquoi la fonction est diﬀérentiable au point donné. Chercher ensuite la linéarisation L ( x,y ) de la fonction en ce point a) f ( x,y ) = yln ( x ) au point (2 , 1) . b) f ( x,y ) = e x cos ( xy ) au point (0 , 0) . 2...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern