SOLUTIONS FOR HOMEWORK SET #10.3 FOR MATH 550

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1 SOLUTIONS FOR HOMEWORK SET #10.3 FOR MATH 550 p159,#4: Verify the chain rule for, where hx,y fux,y,vx,y and fu,v u2 + v 2 u 2 v 2, ux,y e x y, vx,y e xy. Solution: First we will work out what the chain rule looks like in this context. The mapping h: x,y hx,y is defined to be the composition of two mappings, h f g: ux,y g: x,y, and f : u,v fu,v. vx, y We agree to write fu,v or f u,v interchangeably. We only use the second notation to match with describing the first mapping as g: R 2 R 2 and the second mapping as f : R 2 R, since we have stated that R n is a set of column vectors not row vectors. The chain rule states that Df g x,y Dfg x,y Dg x,y. Of course Dh,, whereas the problem only asks for the first component of this. We have Dg Df. Dg gets evaluated at x,y whereas Df gets evaluated at u,v g x,y. The chain rule says then that Dh Df Dg + + Comparing the first components of these row vectors we see that the chain rule claims that +. evaluated at the appropriate arguments. So to verify the chain rule in this case we should compute both sides of this equation and show that they are equal. We will start with the left side of the equation. hx,y fux,y,vx,y ux,y2 + vx,y 2 ux,y 2 vx,y 2 e x y 2 + e xy 2 e x y 2 e xy 2 1.

2 2 hx,y e 2x 2y + e 2xy e 2x 2y e 2xy e 2x 2y + e 2xy e 2x 2y e 2xy 1 e 2x 2y e 2xy 2 e 2x 2y + e 2xy e 2x 2y e 2xy e 2x 2y e 2xy 2 [ 2e 2x 2y + 2ye 2xy e 2x 2y e 2xy ] e 2x 2y + e 2xy 2e 2x 2y 2ye 2xy Now we will work on the right side of the equation. fu,v u2 + v 2 u 2 v 2 2uu2 v 2 u 2 + v 2 2u u 2 v 2 2 e x y,e xy 2e x y e 2x 2y e 2xy e 2x 2y + e 2xy 2e x y e 2x 2y e 2xy 2 2vu2 v 2 u 2 + v 2 2v u 2 v 2 2 e x y,e xy 2exy e 2x 2y e 2xy e 2x 2y + e 2xy 2e xy e 2x 2y e 2xy 2 ux,y e x y e x y vx,y e xy yexy So combining everything on the right side we get e x y,e xy + e x y,e xy 2e x y e 2x 2y e 2xy e 2x 2y + e 2xy 2e x y [ e x y ] e 2x 2y e 2xy 2 + 2exy e 2x 2y e 2xy e 2x 2y + e 2xy 2e xy e 2x 2y e 2xy 2 [ye xy ] If we apply the distributive law to this, it can be seen to be equal to the left side, as desired. In the calculus of one variable the chain rule is introduced as a method of computing derivatives of complex functions in terms of

3 derivatives of simpler functions. In our context of several variables it is less about computing the derivative in the easiest or shortest manner, and more about an important structural relationship satisfied by derivatives of compositions of functions. p160,#8: Let f : R 3 R be differentiable. Making the substitution x ρ sin φ cos θ, y ρ sin φ sin θ, z ρ cos φ spherical coordinates into fx,y,z, compute,, in terms of,,. Solution: The statement of this problem straight out of the text contains a common, but unfortunate, abuse of notation, one which can cause fairly severe confusion in applications. f is initially defined as a function of x,y,z R 3, so the partial derivatives,, are unambiguous. This abuse of notation attempts to also think of f as a function of ρ,φ,θ 0, 0,π π,π R 3, but this is overloading the symbol f and introducing an ambiguity in the notation for partial derivatives,,. Warning about partial derivative notation: Be very careful when interpreting to identify exactly what function f stands for, what all its independent variables are, and to check that u is one of them. When changing variables in applications sometimes out of laziness or to avoid introducing yet another symbol function symbols are given multiple conflicting meanings. So we introduce the mapping g: 0, 0,π π,π R 3 by the rule x y g ρ,φ,θ g1 ρ,φ,θ ρ sin φ cos θ g 2 ρ,φ,θ ρ sin φ sin θ. z g 3 ρ,φ,θ ρ cos φ Here we are using x,y,z as dependant variable names for the three component functions g 1,g 2,g 3 of the vector-valued function g. Define the mapping h: 0, 0,π π,π R to be h f g. h ρ,φ,θ f ρ sin φ cos θ,ρsin φ sin θ,ρcos φ, so when the problem asks for,, it really means,,, which are the components of the row vector Dh. We use the chain rule to compute it: Dh ρ,φ,θ Dfg ρ,φ,θ Dg ρ,φ,θ, i.e. 3

4 4 It is well-known that and g2 are synonymous, since y g2 ρ,φ,θ. Traditional notation is loath to introduce the symbol g i for the ith component function of the vector-valued function g. So now we compute: Hence ρ,φ,θ sin φ cos θ ρ cos φ cos θ ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ ρ sin φ 0 g ρ,φ,θ sin φ cos θ ρ cos φ cos θ ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cosφ ρ sin φ 0 I will assume you can do this row vector matrix multiplication, and will leave the answer in this shorter form. p160,#10: Let f u,v,w e u w, cosv + u + sinu + v + w and g x,y e x, cosy x,e y. Calculate f g and Df g 0, 0. Solution: f g x,y fg x,y f e x, cosy x,e y e ex e y coscosy x + e x + sine x + cosy x + e y By the chain rule Df g 0, 0 Dfg 0, 0 Dg 0, 0. Computing: ex g x,y cosy x e y e0 g 0, 0 cos e 0 1 Dg x,y ex 0 siny x siny x 0 e y Dg 0, 0 e0 0 sin0 0 sin e e f u,v,w u w cosv + u + sinu + v + w

5 e u w 0 e u w Df u,v,w sin v + u sin v + u cos u + v + w + cos u + v + w + cos u + v + w e e 1 1 Df 1, 1, 1 sin sin cos cos cos sin 2 + cos 3 sin 2 + cos 3 cos Df g 0, sin 2 + cos 3 sin 2 + cos 3 cos the answer! sin 2 + cos 3 cos 3 Df u,v,w is a 2 3 matrix; what appears above to be a 3rd row is a vertical continuation of the 2nd row. 5 p160,#14: Let f : R n R m be a linear mapping so that by a previous homework problem Dfy f for all y R n. We can think of Dfy as the m n matrix of partial derivatives or as the linear mapping R n R m it induces. Check the validity of the chain rule directly for linear mappings. Solution: Let g: R p R n be another linear mapping. We claim that f g: R p R m is linear. To check this suppose x,y R p and a R. Then f gxa + y fgxa + y fgxa + gy fgxa + fgy f gxa + f gy. These equalities are because of the definition of f g, the linearity of g, and the linearity of f. Thus f g is linear. By the earlier homework problem Dgx g and Df gx f g. Hence for any y R n Df gx f g Dfy Dgx DfgxDgx for all x R p, as the chain rule claims. Note that as linear mappings we could write Dfgx Dgx whereas as matrices we usually write DfgxDgx. The product of the matrices induces the composition of the linear mappings. Because of this people often denote the composition of linear mappings without writing.

6 6 p161,#20: Consider the function f x,y Show that a and { xy 2 x 2 +y 2 x,y 0, 0 0 x,y 0, 0. exist at 0, 0. b If gt at,bt for all t R, and for constants a,b R, ab 0, then f g is differentiable and Df g0 ab 2 /a 2 + b 2 but Dfg0Dg0 0. Hence the conclusion of the chain rule fails to hold, so f must not be differentiable at 0, 0. Solution: This was done in class. a 0, 0 lim h 0[f0 + h, 0 f0, 0]/h lim h 0 [h0 2 /h ]/h lim h 0 0/h 0. So it exists. Similarly, 0, 0 lim h 0[f0, 0+h f0, 0]/h lim h 0 [0h 2 /0 2 + h 2 0]/h lim h 0 0/h 0. So it exists. { atbt 2 t 0 at b f gt f at,bt 2 +bt 2 0 t 0 ab2 t/a 2 + b 2. This function is clearly differentiable and Df gt ab 2 /a 2 + b 2 for all t, including t 0. Dgt a,b for all t, including t 0. Dfg0 Df 0, 0 0, 0, 0, 0 0, 0 from part a. Performing the matrix product: Dfg0Dg0 0, 0 a,b 0 as desired. So to summarize: Df g0 ab 2 /a 2 + b 2 0 Dfg0Dg0, violating the conclusion of the chain rule.