2220hw2sol

2220hw2sol - Math 2220 Problem Set 2 Solutions Spring 2010...

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Unformatted text preview: Math 2220 Problem Set 2 Solutions Spring 2010 Section 2.4 : 10. Determine all the second-order partial derivatives, including the mixed partials, for the function f ( x, y ) = cos( xy ). Solution. We have f x = − y sin( xy ) and f y = − x sin( xy ). The second order derivatives are f xx = − y 2 cos( xy ) f xy = − sin( xy ) − xy cos( xy ) f yx = − sin( xy ) − xy cos( xy ) f yy = − x 2 cos( xy ) . The two mixed partial derivatives f xy = f yx are equal because all these partial derivatives are continuous. 14. Do the same thing for f ( x, y ) = e x 2 + y 2 . Solution. We have f x = 2 xe x 2 + y 2 and f y = 2 ye x 2 + y 2 . The second order derivatives are f xx = 2(1 + 2 x 2 ) e x 2 + y 2 f xy = f yx = 4 xye x 2 + y 2 f xx = 2(1 + 2 y 2 ) e x 2 + y 2 . 18. Consider the function F ( x, y, z ) = 2 x 3 y + xz 2 + y 3 z 5 − 7 xyz . (a) Find F xx , F yy , and F zz . Solution. The first partial derivatives are F x = 6 x 2 y + z 2 − 7 yz F y = 2 x 3 + 3 y 2 z 5 − 7 xz F z = 2 xz + 5 y 3 z 4 − 7 xy and the second order derivatives are F xx = 12 xy F yy = 6 yz 5 F zz = 2 x + 20 y 3 z 3 . (b) Calculate the mixed partials F xy , F yx , F xz , F zx , F yz , and F zy and verify Theorem 4.3 (equality of mixed partials). Solution. We have F xy = 6 x 2 − 7 z F yx = 6 x 2 − 7 z F xz = 2 z − 7 y F zx = 2 z − 7 y F yz = 15 y 2 z 4 − 7 x F zy = 15 y 2 z 4 − 7 x (c) Is F xyx = F xxy ? Could you have known this without resorting to calculation? Solution. F xyx = 12 x and F xxy = 12 x . They are equal since they are mixed partial derivatives of the function F x , namely ( F x ) yx and ( F x ) xy . 1 Math 2220 Problem Set 2 Solutions Spring 2010 (d) Is F xyz = F yzx ? Solution. F xyz = − 7 and F yzx = − 7. They are the same because F xyz = ( F xy ) z = ( F yx ) z = F yxz = ( F y ) xz = ( F y ) zx = F yzx . 20. The partial differential equation ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 = 0 is known as Laplace’s equa- tion (in three variables). Any function f of class C 2 satisfying this equation is called a harmonic function . (a) Is f ( x, y, z ) = x 2 + y 2 − 2 z 2 harmonic? What about f ( x, y, z ) = x 2 − y 2 + z 2 ? Solution. We have f xx = 2, f yy = 2 and f zz = − 4. Since f xx + f yy + f zz = 2+2+( − 4) = 0 the function f ( x, y, z ) = x 2 + y 2 − 2 z 2 is harmonic. The function g ( x, y, z ) = x 2 − y 2 + z 2 is not harmonic since g xx + g yy + g zz = 2 negationslash = 0....
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This note was uploaded on 04/19/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell.

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2220hw2sol - Math 2220 Problem Set 2 Solutions Spring 2010...

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