Math F412: Homework 5 Solutions March 8, a) Find a smooth 2π-periodic function h(θ) that has the following properties.

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1 All parts of tis omework to be completed in Maple sould be done in a single workseet. You can submit eiter te workseet by or a printout of it wit your omework. 1. Te point of tis problem is to find a continuous function f R R tat as directional derivatives defined at (0, 0) but is not differentiable tere. a) Find a smoot π-periodic function (θ) tat as te following properties. (θ) 0 for 0 θ π. (θ) 0 for π θ π. (θ) vanises exactly at te multiples of π/ (0) > 0. Hint: First find a function g(θ) suc tat g(θ) 0 and g(θ) vanises at multiples of π/. b) Find a function f (x, y) satisfying f (cos(θ), sin(θ)) = (θ) f (cx, cy) = c f (x, y) for any c R. c) w tat your function is continuous at (0, 0). d) Compute te directional derivative of f at (0, 0) in te direction (a, b). e) Wat are te partial derivatives f / x and f / y at (0, 0)? Don t work ard! f) Explain wy f cannot be differentiable at (0, 0). g) Make a elpful plot in Maple of te function f (x, y). ) Using Maple or oterwise, compute f / y along te x-axis and along te y-axis. Explain wy tis computation sows tat f / y is not continuous at (0, 0).

2 lution, part a: Let (θ) = cos (θ) sin(θ). Tis as te desired properties. lution, part b: Let Ten f ((θ)) = Note also tat if c 0 ten Moreover, f (0x, 0y) = f (0, 0) = 0 = 0 f (x, y). x y x f (x, y) = +y (x, y) (0, 0) 0 (x, y) = (0, 0) cos (θ) sin(θ) cos (θ) + sin (θ) = cos (θ) sin(θ) = (θ). f (cx, cy) = c3 x y = c f (x, y). c x + y lution, part c: Let є > 0. Let δ = є. Suppose 0 < (x, y) (0, 0) < δ. Ten x y f (x, y) f (0, 0) = f (x, y) = x x + y x + y x x + y < δ = є. f is continuous at (0, 0). lution, part d: Let v = (a, b). Ten f ((0, 0) + v) f (0, 0) f (v) v f (0, 0) = = = f (v) = f (v) = a b 0 0 a + b. lution, part e: Te partial derivatives f / x and f / y are just te directional derivatives in te directions (1, 0) and (0, 1) respectively. By part (d), tese are bot 0. lution, part f: If f were differentiable at (0, 0), te directional derivatives at (0, 0) in te direction v could by computed by v f (0, 0) = D f (0, 0)v = [0, 0]v = 0. But part (c) indicates tere are non-vanising directional derivatives. lution, part g: See Maple workseet. lution, part : See Maple workseet.

3 . Suppose U is an open subset of R n, α is a differentiable curve in U, t 0 is in te domain of α, and f U R is differentiable at α(t 0 ). w tat f α is differentiable at t 0 and ( f α) (t 0 ) = D f (α(t)) α (t 0 ). Hint:You know tat f (α(t 0 ) + v) = f (α(t 0 )) + D f (t 0 ) v + R(v). w tat R(v) can be written in te form R(v) = S(v) v were S(0) = 0 and S is continuous at 0; to do tis define S(v) = R(v) / v for v 0 and S(0) = 0. Since α is differentiable at t 0, form some remainder function r() were Since f is differentiable at α(t 0 ), for some remainder term R(v) suc tat α(t 0 + ) = α(t 0 ) + α (t 0 ) + r() r() = 0. 0 f (α(t 0 ) + v) = f (α(t 0 )) + D f (t 0 ) v + R(v) We define R(v) / v v 0 S(v) = 0 v = 0. Ten S is continuous at 0, by te it noted earlier in (1). Now But R(v) = 0. (1) v 0 v ( f α) f (α(t 0 + )) f (α(t 0 )) (t 0 ) =. 0 f (α(t 0 + )) f (α(t 0 )) = f (α(t 0 ) + α (t 0 ) + r()) f (α(t 0 )) f (α(t 0 + )) f (α(t 0 )) Since 0 r()/ = 0, to sow tat = f (α(t 0 )) + D f (α(t 0 )) [α (t 0 ) + r()] + R(α (t 0 ) + r()) f (α(t 0 )) = D f (α(t 0 )) α (t 0 ) + D f (α(t 0 )) r() + R(α (t 0 ) + r()). = D f (α(t 0 )) α (t 0 ) + D f (α(t 0 )) r() ( f α) (t 0 ) = D f (α(t)) α (t 0 ) 3 + R(α (t 0 ) + r()).

4 it is enoug to sow tat or equivalently tat But R(α (t 0 ) + r()) R(α (t 0 ) + r()) = 0 0 R(α (t 0 ) + r()) = 0. 0 = S(α (t 0 ) + r()) α (t 0 ) + r() = S(α (t 0 ) + r()) α (t 0 ) + r(). Since 0 α (t 0 ) + r() = 0, and since S is continuous at 0, 0 S(α (t 0 ) + r()) = 0. Moreover, [α (t 0 ) + r() 0 ] = α (t 0 ). R(α (t 0 ) + r()) = 0 α (t 0 ) = Let S = {(x, y, z) x + y = 1}. Find a single surface patc (x, U) suc tat x(u) = S. You must verify tat tis map satisfies all properties of being a surface patc, including smootness of te inverse. Let U = R {(0, 0)}, so U is open. Define x U R 3 by u x(u, v) = ( u + v, v u + v, ln(u + v )). Te component functions of x are evidently infinitely differentiable, so x is smoot. It is also clear tat x(u, v) S for all (u, v) U. We will sow tat it is bijective onto S by exibiting an inverse function. Define F R 3 R by F(x, y, z) = (x exp(z/), y exp(z/)). Ten if (u, v) U, u v F(x(u, v)) = ( u + v, u + v ) = (u, v). u + v u + v Also, if (x, y, z) = S, x(f(x, y, z)) = x exp(z/) exp(z/) x + y, y exp(z/) exp(z/) x + y, ln(x exp(z) + y exp(z)) = (x, y, ln(exp(z))) = (x, y, z), 4

5 since x + y = 1. Tis sows tat x is bijective onto S. Since F is defined on all of R 3 and is smoot, w as a smoot inverse. 4. Oprea.1.11 Tis exercise as minor mistakes in it. Part of your job is to find and correct tem. We assume additionally tat al pa(u) = (g(u), (u)) is regular and (u) 0 for all u. Let x(u, v) = (g(u), (u) cos(v), (u) sin(v)). Ten x u (u, v) = (g (u), (u) cos(v), (u) sin(v)) x v (u, v) = (0, (u) sin(v), (u) cos(v)). x u (u, v) x v (u, v) = ((u) (u) [sin(v) + cos(v) ], (u)g (u) cos(v), (u)g (u) sin(v)) = (u) ( (u), g (u) cos(v), g (u) sin(v)). Now if x u (u, v) x v (u, v) = 0 ten eiter (u) = 0, or ( (u), g (u) cos(v), g (u) sin(v)). We know tat (u) 0, so (u) = 0 and (g (u) cos(v)) + (g (u) sin(v)) = g (u) = 0. But α(t) = (g(t), (t)) is a regular curve, so α (t) (0, 0) so at any t, at least one of g(t) or (t) is non-zero. 5. Oprea.1.0 See workseet. 6. Oprea.1.. Make a plot in Maple tat exibits te surface as a (singly) ruled surface. Hint: look closely at te documentation for plot3d for instructions on ow to plot a surface patc. See workseet 5