HOMEWORK 4 SOLUTIONS. All questions are from Vector Calculus, by Marsden and Tromba

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1 HOMEWORK SOLUTIONS All questions are from Vector Calculus, by Marsden and Tromba Question :..6 Let w = f(x, y) be a function of two variables, and let x = u + v, y = u v. Show that Solution. By the chain rule, w u v = w w y. Thus, i.e., w u v = u w v = w v + w v = w x w y. ( ) w = v u (w x w y ) = u w x u w y ( wy u + w y u = w x u + w x u = w xx + w xy (w yx + w yy ) = w xx w yy w u v = w w y. ) Question :.. (a) : Show that the function g(x, t) = + e t sin x satisfies the heat equation: g t = g xx. [Here g(x, t) represents the temperature in a metal rod at position x and time t.] (b) : Sketch the graph of g for t. (Hint: Look at sections by the planes t =, t =, and t =.) (c) : What happens to g(x, t) as t? Interpret this limit in terms of the behavior of heat in the rod. Solution. (a) : Since g(x, y) = + e t sin x, then g t = e t sin x, g x = e t cos x, and g xx = e t sin x. Therefore, g t = g xx. (b) : The graph of g is shown in Figure. Date: Math c Practical, 8.

2 HOMEWORK SOLUTIONS t = t = t = z x t Figure. The graph of g at t =,, and. (c) : Note that lim g(x, t) = lim ( + t t e t sin x) = This means that the temperature in the rod at position x tends to be a constant (= ) as the time t is large enough. Question :.. Determine the second-order Taylor formula for f(x, y) = x + y + about x =, y =. Solution. We first compute the partial derivatives up through second order: f x = f xy = f xx = f yy = x ( + x + y ), f y y = ( + x + y ) 8xy ( + x + y ), f 8xy yx = ( + x + y ) ( + x + y ) + 8x ( + x + y ) ( + x + y ) + 8y ( + x + y ).

3 HOMEWORK SOLUTIONS Next, we evaluate these derivatives at (, ), obtaining and f x (, ) = f y (, ) =, f xy (, ) = f yx (, ) = f xx (, ) = f yy (, ) =. Therefore, the second order Taylor formula is where h = (h, h ) and where f(h) = h h + R (, h), R (, h) h as h. Question :..6 Determine the second-order Taylor formula for the function expanded about the point x =, y =. f(x, y) = e (x ) cos y Solution. The ingredients needed in the second-order Taylor formula are computed as follows: f x = (x )e (x ) cos y f y = e (x ) sin y f xx = e (x ) cos y + (x ) e (x ) cos y f xy = (x )e (x ) sin y = f yx f yy = e (x ) cos y. Evaluating the function and these derivatives at the point (, ) gives f(, ) = f x (, ) = f y (, ) = f xx (, ) = f xy (, ) = f yx (, ) = and f yy (, ) =. Consequently, the second order Taylor formula is where h = (h, h ) and where f(h) = + h h + R ((, ), h), R ((, ), h) h as h. Question 5:..7 Find the critical points for the function f(x, y) = x + xy + x + y + y +. and then determine whether they are local maxima, local minima, or saddle points.

4 HOMEWORK SOLUTIONS Solution. Here, We have = 6x + y +, =, y = x + y +. y = when x = y = /. Therefore, the only critical point is ( /, /). Now, ( /, /) = 6, f y ( /, /) =, and f y ( /, /) =, which yields D = 6. = >. Therefore ( /, /) is a local minimum. Question 6:..7 Find the local maxima and minima for z = (x +y )e x y. Solution. We first locate the critical points of f(x, y) = (x + y )e x y. f(x, y) = e x y (x( y x )i + y( x y )j) Thus, f(x, y) = if and only if (x, y) = (, ), (, ±), or (±, ). To determine whether they are maxima or minima, we need to calculate the second partial derivatives. = ( + x y + x (6y 5))e x y = ( 5y + 6y + x (y ))e x y, and y y = (y + x )e x y. Therefore, f (, ) = e, f y (, ) = 6e, and f y (, ) =, which yields D = (e)(6e) = e >, and (, ) is a local minimum. (, ±) =, f y (, ±) =, and f y (, ±) =, which yields D = ( )( ) = >, and (, ±) are local maxima. (±, ) =, f y (±, ) =, and f y (, ±) =, which yields D = ( )() = 6 <, and (±, ) are saddle points. Question 7:..5 Write the number as a sum of three numbers so that the sum of the products taken two at a time is a maximum. Solution. Let the three numbers be x, y, z. Thus, We want to find the maximum value for We differentiate to get x + y + z =, z = x y. S(x, y) = xy + yz + xz = xy + (x + y)( x y) = x xy y + x + y. S S = x y +, y = x y +. These vanish when x = y =, then z = (x + y) =. Therefore, when x = y = z = is the only critical point. The condition x, y, z describes a cube in R and on the boundary of the cube (either x =, x =, y =, y =, z =, z = ), S is zero. Therefore the maximum of S occurs on the interior of this cube, i.e., at a local maximum. Since x =, y =, z = is the only critical point, it must be a maximum.

5 HOMEWORK SOLUTIONS 5 Question 8:.. Find the extrema of f(x, y) = x y subject to the constraint x y =. Solution. By the method of Lagrange multipliers, we write the constraint as g =, where g(x, y) = x y and then write the Lagrange multiplier equations as f = λ g. Thus, we get = λ x = λ y x y =. First of all, the first two equations imply that x and y. Hence we can eliminate λ, giving x = y. From the last equation this would imply that =. Hence there are no extrema. Question 9:.. Let P be a point on a surface S in R defined by the equation f(x, y, z) =, where f is of class C. Suppose that P is a point where the distance from the origin to S is maximized. Show that the vector emanating from the origin and ending at P is perpendicular to S. Solution. We want to maximize the function g(x, y, z) = x + y + z subject to the constraint f(x, y, z) =. Suppose this maximum occurs at P = (x, y, z ), then by the method of Lagrange multipliers we have the equations x = λ { f(x, y, z )} y = λ { f(x, y, z )} z = λ { f(x, y, z )} where { f(x, y, z )} i denotes the ith component of f(x, y, z ), i. If v = (x, y, z ) is the vector from the origin ending at P, then these equations say that v = ( ) λ f(x, y, z ). But f(x, y, z ) is perpendicular to S at P, and since v is a scalar multiple of f(x, y, z ) it is also perpendicular to S at P. Question :..8 A company s production function is Q(x, y) = xy. The cost of production is C(x, y) = x + y. If this company can spend C(x, y) =, what is the maximum quantity that can be produced? Solution. We want to maximize Q subject to the constraint C(x, y) =. Since both x, y, this imposes the condition that x 5, y /. Thus, we wish to maximize Q on the line segment x+y =, x, y. If the maximum occurs at an interior point (x, y ) of this segment, then Q(x, y ) = λ C(x, y ); that is, y = λ x = λ x + y =. Thus 6λ + 6λ =, λ = 5/6, y = 5/, x = 5/, Q(x, y ) = 5/6. The value of Q at the endpoints of this segment are Q(, ) = = Q(5, ). Consequently the maximum occurs at (5/, 5/) and the maximum value of Q is 5/6.

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