Helloo

UNIVERSIDAD CARLOS III DE MADRID MATHEMATICS II EXERCISES (SOLUTIONS ) CHAPTER 4: Higher order derivatives 4-1. Let u : R2 → R be defined by u(x, y) = ex sin y. Find all the second partial derivatives D2 u, and verify Schwarz’s Theorem. Solution: The partial derivatives of the function u(x, y) = ex sin y are ∂u = ex sin y, ∂x Therefore the Hessian is ex sin y ex cos y ex cos y −ex sin y ∂u = excos y ∂y SPRING 2010

4-2. Consider the quadratic function Q : R3 → R defined by Q(x, y, z) = x2 + 5y 2 + 4xy − 2yz. Compute the Hessian matrix D2 Q. Solution: The gradient of Q is (x2 + 5y 2 + 4xy − 2yz) = (2x + 4y, 10y + 4x − 2z, −2y) The Hessian matrix of Q is  2  4 0 4-3. Let f (x, y, z) = ez +
1 x

4 10 −2

 0 −2  0

+ xe−y , for x = 0. Compute ∂2f , ∂x2 ∂2f , ∂x∂y ∂2f , ∂y∂x ∂2f ∂y2
1 x

Solution: The partial derivatives of the function f (x, y, z) = ez + ∂ 2 f (x, y, z) ∂x2 2 ∂ f (x, y, z) ∂x∂y ∂ 2 f (x, y, z) ∂y∂x ∂ 2 f (x, y, z) ∂y 2 = 2 x3

+ xe−y are

= − e−y = − e−y =xe−y

4-4. Let z = f (x, y), x = at, y = bt where a and b are constant. Consider z as a function of t. Compute terms of a, b and the second partial derivatives of f: fxx , fyy and fxy .Solution: Since the function is of class C 2 , we may apply Schwarz’s Theorem. d (f (at, bt)) =afx (at, bt) + bfy (at, bt) dt d2 (f (at, bt)) =a2 fxx (at, bt) + 2abfxy (at, bt) + b2 fyy (at, bt) dt2

d2 z dt2

in

1

2

4-5. Let f (x, y) = 3x2 y + 4x3 y 4 − 7x9 y 4 . Compute the Hessian matrix D2 Q.. Solution: The gradient of f is f (x, y) = 6xy + 12x2 y 4 − 63x8 y 4 , 3x2 + 16x3 y 3 − 28x9 y 3The Hessian matrix of f is H(x, y) = 6y + 24xy 4 − 504x7 y 4 6x + 48x2 y 3 − 252x8 y 3 6x + 48x2 y 3 − 252x8 y 3 48x3 y 2 − 74x9 y 2

4-6. Let f, g : R2 → R be two functions whose partial derivatives are continuous on all of R2 and such that there is a function n h : R2 → R such that (f, g) = h, that is, f (x, y) = ∂h (x, y) ∂x g(x, y) = ∂h (x, y) ∂y

at every point (x, y) ∈ R2 . Whatequation do ∂f ∂y satisfy? Solution: On the one hand, we have that ∂ ∂h ∂2h ∂f ∂x = = ∂y ∂y ∂x∂y On the other hand, we see that ∂ ∂h ∂y ∂g = ∂x ∂x = ∂2h ∂y∂x and ∂g ∂x

Since the functions f and g have continuous partial derivatives on all of R2 , the function h is of class C 2 . By Schwartz’s Theorem, we conclude that ∂2h ∂2h = ∂x∂y ∂y∂x That is, ∂f ∂g = ∂y ∂x

4-7. The demand function of a consumerby a system of equations of the form ∂u ∂x ∂u ∂y p1 x + p2 y = λp1 = λp2 = m

where u(x, y) is the utility function of the agent, p1 and p2 are th prices of the consumption bundles, m is income and λ ∈ R. Assuming that this system determines x, y and λ as functions of the other parameters, determine ∂x ∂p1 Solution: First we write the system as f1 f2 f3 ≡ ≡ ≡ ∂u − λp1 = 0 ∂x ∂u − λp2 = 0 ∂y p1x + p2 y − m = 0

3

and compute ∂ (f1 , f2 , f3 ) = ∂ (x, y, λ)
∂2u ∂x2 ∂2u ∂x∂y ∂2u ∂x∂y ∂2u ∂y 2

p1

p2

−p1 −p2 0

=

∂2u 2 ∂2u ∂2u p − p1 p2 + 2 p2 2 2 ∂x ∂x∂y ∂y 1

We suppose that this determinant does not vanish and that we may apply the mean value Theorem. Differentiating with respect to p1 (but assuming now that x, y, λ depend on the other parameters) we obtain ∂ 2 u ∂y∂λ ∂ 2 u ∂x + − p1 − λ = 0 2 ∂p ∂x ∂x∂y ∂p1 ∂p1 1 ∂ 2 u ∂x ∂ 2 u ∂y ∂λ + 2 − p2 = 0 ∂x∂y ∂p1 ∂y ∂p1 ∂p1 ∂x ∂y x + p1 + p2 =0 ∂p1 ∂p1 which may be written as ∂ 2 u ∂x ∂x2 ∂p1 ∂ 2 u ∂x ∂x∂y ∂p1 ∂x p1 ∂p1 The unknowns of the above system are ∂x , ∂p1 We see that the determinant of the system is ∂ (f1 , f2 , f3 ) ∂ (x, y, λ) Using Cramer’s rule we see that, λ 0 −x
∂2u 2 ∂x2 p2 ∂2u ∂x∂y ∂2u ∂y 2

∂2 u ∂y ∂λ − p1 = λ ∂x∂y ∂p1 ∂p1 ∂ 2 u ∂y ∂λ + − p2 = 0 ∂y 2 ∂p1 ∂p1 ∂y + p2 = −x ∂p1 + ∂y , ∂p1 ∂λ ∂p1

∂x = ∂p1

p2

−p1 −p2 0 +
∂2u 2 ∂y 2 p1

−

∂2u ∂x∂y p1 p2

=

∂ u λp2 + m ∂x∂y p2 − m ∂ u p1 2 ∂y 2 ∂2u 2 ∂x2 p2

2

2

−

∂2u ∂x∂y p1 p2

+

∂2u 2 ∂y 2 p1

4-8. Consider the system of equations z 2 + t − xy zt + x2 (a) (b) (c) (d) = = 0 y2

Prove that it...