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### math solution 3

Course: MATH 2220, Spring 2007
School: Cornell
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Word Count: 1318

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due Homework 7th February Solutions to compulsory questions Problem (2.2 Exercise 8b). Compute the limit sin(xy) y (x,y)(0,0) lim Solution. Using the substitution u = xy we have lim (x,y)(0,0) if it exists. sin(xy) sin u = lim = 1. u0 u (xy) Therefore, by the properties of limits, lim (x,y)(0,0) sin(xy) sin(xy) = lim x y (xy) (x,y)(0,0) = lim (x,y)(0,0) x lim (x,y)(0,0) sin(xy) (xy) = 0. Alternatively,...

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due Homework 7th February Solutions to compulsory questions Problem (2.2 Exercise 8b). Compute the limit sin(xy) y (x,y)(0,0) lim Solution. Using the substitution u = xy we have lim (x,y)(0,0) if it exists. sin(xy) sin u = lim = 1. u0 u (xy) Therefore, by the properties of limits, lim (x,y)(0,0) sin(xy) sin(xy) = lim x y (xy) (x,y)(0,0) = lim (x,y)(0,0) x lim (x,y)(0,0) sin(xy) (xy) = 0. Alternatively, by expanding sin(xy) into a Taylor series we have (xy) - sin(xy) = lim y (x,y)(0,0) (x,y)(0,0) lim 1 3 3! (xy) + y = lim (x,y)(0,0) x 1- (xy)2 + 3! = 0. Problem (2.2 Exercise 10c). Compute the limit (x - y)2 (x,y)(0,0) x2 + y 2 lim Solution. Along the line y = x we have f (x, x) = and along the line y = -x we have f (x, -x) = Therefore the limit does not exist. Problem (2.2 Exercise 15). If f : Rn R and g : Rn R are continuous, show that the function Rn - R f2 + g : x - [f (x)]2 + g(x) is continuous. Solution. If a Rn then xa if it exists. (x - x)2 x0 = 0 - 0, x2 + x2 (x + x)2 x0 = 2 - 2. x2 + x2 lim (f 2 + g)(x) = lim [f (x)]2 + g(x) xa 2 (definition of f 2 + g) (using limit properties) (since f, g continuous) (definition of f 2 + g). = xa lim f (x) 2 + lim g(x) xa = [f (a)]2 + g(a) = (f + g)(a) Therefore f 2 + g is continuous at each a Rn , and so f 2 + g is continuous on Rn . 1 Problem (2.3 Exercise 2ac). Evaluate the partial derivatives z/x and z/y for the functions: (a) z = a2 - x2 - y 2 at (0, 0) and at (a/2, a/2). at (2/b, 0). (c) z = eax cos(bx + y) Solution. (a) We compute z =- x Therefore z z (0, 0) = (0, 0) = 0 x y (c) We compute z = eax (a cos(bx + y) - b sin(bx + y)) x Therefore z (2/b, 0) = ae2a/b x and and z = -eax sin(bx + y). y and z z 1 (a/2, a/2) = (a/2, a/2) = - . x y 2 x a2 - x2 - y2 and z =- y y a2 - x2 - y 2 z (2/b, 0) = 0. y Problem (2.3 Exercise 6a). Compute the plane tangent to the graph of f (x, y) = xy at the point (x, y) = (0, 0). Solution. The tangent plane to f (x, y) at (x, y) = (a, b) is z = f (a, b) + fx (a, b)(x - a) + fy (a, b)(y - b). We have f (0, 0) = 0, fx (0, 0) = f (0, 0) = 0, x and similarly fy (0, 0) = 0. So the tangent plane is the plane z = 0. Problem (2.3 Exercise 8ad). Compute the matrix of partial derivatives of (a) f (x, y) = (ex , sin(xy)), and (d) f (x, y, z) = (x + z, y - 5z, x - y). Solution. (a) The matrix is (ex ) x (sin(xy)) x (ex ) y (sin(xy)) y = ex 0 . y cos(xy) x cos(xy) (b) The matrix is (x+z) x (y-5z) x (x-y) x (x+z) y (y-5z) y (x-y) y (x+z) z (y-5z) z (x-y) z 1 0 1 = 0 1 -5 . 1 -1 0 2 Problem (2.3 Exercise 12c). Use a linear approximation of an appropriate function to estimate (4.01)2 + (3.98)2 + (2.02)2 . Solution. The natural choice of function is f (x, y, z) = The linear approximation of f at (4, 4, 2) is f (x, y, z) f (4, 4, 2) + fx (4, 4, 2)(x - 4) + fy (4, 4, 2)(y - 4) + fz (4, 4, 2)(z - 2) for (x, y, z) near (4, 4, 2). We calculate f (4, 4, 2) = 6, and so 2 fx (4, 4, 2) = , 3 2 fy (4, 4, 2) = , 3 and 1 fz (4, 4, 2) = , 3 x2 + y 2 + z 2 . 2 2 1 w = 6 + (x - 4) + (y - 4) + (z - 2). 3 3 3 Substituting in (x, y, z) = (4.01, 3.98, 2.02) gives f (4.01, 3.98, 2.02) 6 + (In fact, f (4.01, 3.98, 2.02) = 6.0000749995 . . .). Problem (Tangent plane). Compute the tangent plane to f (x, y) = at the point (1, 0, 0). Solution. The tangent plane is z = f (1, 0) + fx (1, 0)(x - 1) + fy (1, 0)(y - 0). We calculate y(y 2 - x2 ) f = 2 x (x + y 2 )2 fx (1, 0) = 0 So the tangent plane is the plane z = y. Problem (2.4 Exercise 10). Compute the tangent vector to the path c(t) = (3t2 , t3 ). Solution. tangent The vector is c (t) = d(3t2 ) d(t3 ) , dt dt = (6t, 3t2 ) = (6t)i + (3t2 )j. f x(x2 - y 2 ) = 2 . y (x + y 2 )2 fy (1, 0) = 1. x2 xy + y2 4 2 2 - + = 6. 300 300 300 and Therefore and 3 Problem (EP2, Limits existing versus limits existing along lines). Let f (x, y) = (x2 -y)2 x4 +y 2 if (x, y) = (0, 0), if (x, y) = (0, 0). 1 (a) Let k be a constant. Compute limx0 f (x, kx) and limy0 f (0, y). (b) Compute limx0 f (x, x2 ). (c) Does lim(x,y)(0,0) f (x, y) exist? Explain your answer. Solution. (a) We calculate x0 lim f (x, kx) = lim (x - k)2 (0 - k)2 (x2 - kx)2 = lim 2 = = 1, x0 x + k 2 x0 x4 + k 2 x2 0 + k2 lim f (0, y) = lim (0 - y)2 = 0. y0 0 + y 2 and y0 (b) We calculate (x2 - x2 )2 = 0. y0 x0 x4 + x4 (c) No, the limit does not exist, because there are two paths of approach to (0, 0) along which f (x, y) approaches different values. In particular, note that f (x, y) is not continuous at (0, 0), because lim f (x, y) does not exist. lim f (x, x2 ) = lim (x,y)(0,0) Solutions to other selected problems Problem (EP1, Extending a function to make it continuous). How can f (x, y) = x2 + y 2 - x3 y 3 x2 + y 2 be defined at (0, 0) so that the resulting function is continuous on R2 ? Solution. The function is continuous for all (x, y) = (0, 0), since it is quotient of polynomials with nonzero denominator whenever (x, y) = (0, 0). Our only chance to define f (0, 0) so that f is continuous at (0, 0) is to set f (0, 0) = lim (x,y)(0,0) f (x, y) if the limit exists. If this limit exists then f will be continuous by construction, and if this limit does not exist then it is impossible to define f (0, 0) to make f continuous. In fact the limit does exist, and it equals 1, since x2 + y 2 - x3 y 3 x3 y 3 = lim 1- 2 2 + y2 x x + y2 (x,y)(0,0) (x,y)(0,0) lim =1- (x,y)(0,0) x2 lim x3 y 3 = 1 - 0 = 1. + y2 This final limit can be calculated a few ways. For example, making the substitution x = r cos and y = r sin and observing that (x, y) (0, 0) if and only if r 0 gives x3 y 3 r6 cos3 sin3 = = r4 cos3 sin3 x2 + y 2 r2 So we define f (0, 0) = 1. 4 - r0 0. Problem (2.2 Exercises 9b and 10b). Compute the limits cos(xy) - 1 lim (xy)2 (x,y)(0,0) if they exists. Solution. For the first limit, making the substitution u = xy and using L'H^pital's rule gives o cos(xy) - 1 cos u - 1 - sin u 1 = lim = lim =- . 2 2 u0 u0 (xy) u 2u 2 (x,y)(0,0) lim For the second limit, along the line x = 0 we have f (0, y) = 1-1-0 y0 = 0 - 0, 2 0+y x2 2 and cos x - 1 - lim x4 + y 4 (x,y)(0,0) x2 2 while along the path y = 0 repeated uses of L'H^pital's rule gives o cos x - 1 - lim f (x, 0) = lim x0 x0 x4 - sin x - x = lim x0 4x3 - cos x - 1 = lim x0 12x2 sin x = lim x0 24x 1 = . 24 So the limit does not exist. Problem (2.3 Exercise 4d). Show that f (x, y) = is differentiable at each point of its domain. Solution. The domain of f is D = {(x, y) R2 | (x, y) = (0, 0)}. For (x, y) D the partial derivatives are computed as f y3 = 2 x (x + y 2 )3/2 and f x3 = 2 . y (x + y 2 )3/2 xy x2 + y 2 Therefore both of the partials are continuous on D (since they are quotients of continuous functions with nonzero denominator), and so by a theorem, f is differentiable on D. Notice how useful the theorem continuous partials = differentiable is. If you wanted to show that f (x, y) is differentiable directly, you would need to show that xy lim (x,y)(a,b) x2 +y 2 - ab a2 +b2 + b3 (x (a2 +b2 )3/2 - a) + a3 (y (a2 +b2 )3/2 - b) =0 (x - a)2 + (y - b)2 whenever (a, b) D, which is an unpleasant task to say the least! 5
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