Answer sheet: Second Midterm for Math 2339

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1 Answer sheet: Second Midterm for Math 2339 October 26, 2010 Problem 1. True or false: (check one of the box, and briefly explain why) (1) If a twice differentiable f(x,y) satisfies f x (a,b) = f y (a,b) = 0, and f xx (a,b)f yy (a,b) > f 2 xx(a,b). Then f must have a local minimum at (a,b). false Even with the correct formula f xx (a,b)f yy (a,b) > fxy(a,b), 2 it is still not sufficient to guarantee a local minimum without f xx (a,b) > 0. (2) If f(x,y,z) = ln(x 2 y 2 z 2 ), then f(x,y,z) = 2 x + 2 y + 2 z. false, f(x,y,z) = 2 x, 2 y, 2 z (3) There exists a function f(x,y) such that f x = sin(xy) and f y = cos(xy). false Note that f xy = x cos(xy) but f yx = y sin(xy), this contradicts the Clairaut s Theorem. (4) If a 3 1 a 3 2 = 2, then u = e a 1x a 2 y 3 u is a solution of x + 3 u 3 y = 2u. 3 true, Note 3 u x 3 = a3 1u, 3 u y 3 = a3 2u and a 3 1 a 3 2 = 2. (5) If f(x,y,z) = sin(x) + cos(y) + sin(z), and u is a unit vector, then 3 D u f(x,y,z) 3. true Note that D u f(x,y,z) = f u = cos(x), sin(y), cos(z) u = cos(x), sin(y), cos(z) cos 2 (x) + sin 2 (y) + cos 2 (z) 3 1

2 Problem 2. Let f(x,y,z) = z x 2 y 2 + ln(x 2 + y 2 + z 2 1). (a) Find the domain of f. Need z x 2 y 2 0 and x 2 + y 2 + z 2 1 > 0. Therefore the domain is dom(f) = {(x,y,z) x 2 + y 2 + z 2 > 1,z x 2 + y 2 }. (b) Evaluate f(1, 0, e). f(1, 0, e) = e 1 + ln(e 2 ) = e (c) Find the range of f. Note that f(0, 0, ), and when z = x 2 + y 2 and let z + z ( ), i.e., z is close to the positive solution of z 2 + z 1 = 0, or +, z then f Therefore the range of f is (, ).

3 Problem 3. Let f(x,y) = x y. (a) (b) Find x Find 2 f x 2 and y and 2 f y 2 x = yxy 1 = x y ln(x) y 2 f x 2 = y(y 1)x y 2 2 f y 2 = x y (ln(x)) 2 (c) Find 2 f x y and 2 f y x 2 f = x y 1 + yx y 1 ln(x) = x y 1 (1 + y ln(x)) x y 2 f y x = yxy 1 ln(x) + x y 1 x = xy 1 (y ln(x) + 1) They should be the same by the Clairaut s theorem.

4 Problem 4. Let f(x,y,z) = ln (x 2 + y 2 + z 3 ), x = r cos(θ), y = r sin(θ), z = rθ. Use the chain rule to find r and θ. r = x x r + y y r + z z r 1 = x 2 + y 2 + z 3(2x cos(θ) + 2y sin(θ) + 3z2 θ) θ = x x 1 = = θ + y y θ + z z θ x 2 + y 2 + z 3( 2xr sin(θ) + 2yr cos(θ) + 3z2 r) r x 2 + y 2 + z 3( 2x sin(θ) + 2y cos(θ) + 3z2 )

5 Problem 5. Let F be a differentiable function. Define the function G as G(s,t) = F( cos(s 2 t 2 ), sin(s 2 t 2 ) ). Does the function G satisfy the following differential equation t G s + s G t = 0? Let x = cos(s 2 t 2 ), y = sin(s 2 t 2 ), then By the chain rule, G(s,t) = F(x,y). G s G t = F x x = F x s + F y y s = 2s sin(s2 t 2 )F x + 2s cos(s 2 t 2 )F y x t + F y y t = 2t sin(s2 t 2 )F x 2t cos(s 2 t 2 )F y Therefore t G s + s G t = 0 Yes G satisfies the differential equation.

6 Problem 6. Let f and g be one-dimensional twice differentiable functions. Let a and b be two constants, with b 0. Show that is a solution of the wave equation u(x,t) = f(ax + bt) + g(ax bt) 2 u x 2 = ( a 2 b 2 ) 2 u t 2. For u(x,t) = f(ax + bt) + g(ax bt), u x = d dx (f(ax + bt) + g(ax bt)) = af (ax + bt) + ag (ax bt) 2 u = d2 x 2 dx 2(f(ax + bt) + g(ax bt)) = a2 f (ax + bt) + a 2 g (ax bt) u = d t dt (f(ax + bt) + g(ax bt)) = bf (ax + bt) bg (ax bt) 2 u = d2 t 2 dt 2(f(ax + t) + g(ax t)) = b2 f (ax + bt) + b 2 g (ax bt) Since b 0, we have 2 u x 2 = a2 (f (ax + bt) + g (ax bt)) = a2 b 2 2 u t 2. Therefore u(x,t) = f(ax + bt) + g(ax bt) is a solution of 2 u x 2 = a2 b 2 2 u t 2. (Comment: You may also separate it into two cases: a = 0 and a 0. If a = 0, then u(x,t) = f(bt) + g( bt) is independent of x, therefore 2 u = 0 = a2 2 u and the equation is satisfied. x 2 b 2 t 2 If a 0, we have = 1 2 u a 2 x = 1 2 u 2 b 2 t 2 = f (ax + bt) + g (ax bt) which also satisfies 2 u x 2 = a2 b 2 2 u t 2. )

7 Problem 7. Let the surface be z = f(x,y) = ln(y 2 + cos(xy) + e 2). (a) Find the gradient vector of f f(x,y) = 1 y sin(xy), 2y x sin(xy) y 2 + cos(xy) + e 2 (b) Evaluate f(0, 1). Find the tangent plane to this surface at the point ( 0, 1,f(0, 1)). f(0, 1) = ln(e) = 1. The tangent plane of z = f(x,y) at (x 0,y 0,z 0 ) is z z 0 = f x (x 0,y 0 )(x x 0 ) + f y (x 0,y 0 )(y y 0 ) Plugging in (x 0,y 0 ) = (0, 1) into f(x,y), we get f x (0, 1) = 0, f y (0, 1) = 2 e. Therefore the tangent plane is z 1 = 0(x 0) + 2 (y 1), or 2y ez = 2 e. e (c) Find the symmetric equation of the normal line at the point ( 0, 1, f(0, 1))? The normal line is perpendicular to the tangent plane, therefore the direction of the normal line is the same as the normal direction of the tangent plane. From (b) we see the normal direction of the tangent plane at ( 0, 1, 1 ) is parallel to 0, 2 e, 1, Therefore the symmetric equation of the normal line at ( 0, 1, 1 ) is e(y 1) = z x = 0. Each equation represents a plane, and their intersection gives the normal line.

8 Problem 8. Let f(x,y,z) = e x3 +cos(y)+sin(z) (a) Find the gradient vector of f(x,y,z), and evaluate f(1, 0, 0). f(x,y,z) = f x,f y,f z = e x3 +cos(y)+sin(z) 3x 2, sin(y), cos z f(1, 0, 0) = e 0 3, 0, 1 = 3, 0, 1 (b) Find the directional derivative of f at (1, 0, 0) in the direction u = 1/3, 2/3, 2/3. D u f(1, 0, 0) = f(1, 0, 0) u = = 5 3 (c) What is the 3-d unit directional vector v that can maximize D v f(1, 0, 0)? What is the largest possible value of D v f(1, 0, 0)? Notice that for any unit vector v, D v f(x,y,z) = f(x,y,z) v = f(x,y,z) cos(θ), where θ is the angle between f(x,y,z) and v. So the maximum D v f(x,y,z) is obtained when θ = 0, i.e., when v is parallel to f(x, y, z). Therefore, max D v f(x,y,z) = f(x,y,z) v And the direction for this maximum is Plugging x = 1,y = z = 0 gives v = f(x,y,z) f(x,y,z) max v D v f(1, 0, 0) = f(1, 0, 0) = 10, v = , 0, 1.

9 Problem 9. We want to build a rectangular aquarium with given volume V. The base is made of marble, and the other four sides made of glass. Assume that marble costs 16 times as much as glass. Find the dimensions of the aquarium so that the cost of materials will be minimized. Let the edges of the base be x and y, and the height be z, then this problem can be modeled as { min f(x,y,z) = 16xy + 2yz + 2xz x,y,z>0 s.t. g(x,y,z) = xyz V = 0 Apply Lagrange multiplier method to find the optimal dimensions. By Lagrange multiplier, We get f = λ g xyz = V 16y + 2z = λyz 16x + 2z = λxz 2y + 2x = λxy Multiply respectively by x,y,z on both sides of the above three equations: λxyz = 16xy + 2xz = 16xy + 2yz = 2yz + 2xz. Therefore, x = y = z 8. Since xyz = V, we have 8x 3 = V, therefore the optimal dimensions are x = y = 3 V 2, z = 4 3 V.

10 Problem 10. (Bonus 10 points) Find the plane that passes through the point (1, 2, 1) and cuts off the smallest possible volume in the first octant. (Hint: Express the cut off volume as a function of the 3 components of the normal direction of the plane. Then solve the optimization problem.) The plane that goes though (1, 2, 1) may be written as where n = a,b,c is the normal direction. a(x 1) + b(y 2) + c(z 1) = 0, (1) To find the volume of the cutoff tetrahedron, we first find the intersection points of the plane with the axises. 1) Set y = z = 0 in (1), then x = a+2b+c a 2) Set x = z = 0 in (1), then y = a+2b+c b 3) Set x = y = 0 in (1), then z = a+2b+c c So the volume of the cutoff tetrahedron is V (a,b,c) = 1 (a + 2b + c) 3 6 abc, this is the intercept with the x-axis., this is the intercept with the y-axis., this is the intercept with the z-axis. Since the constraint that the plane passes (1, 2, 1) is already satisfied in (1), we only need to solve an unconstraint optimization problem min V (a,b,c). a,b,c>0 So we can find the critical point by solving V = 0. However, it is easier to apply the 1st order necessary condition to ln(6v ) (since ln() function is strictly increasing, the minimum critical point of ln(6v ) is the same as that of V ). Let f(a,b,c) = ln(6v ) = ln((a+2b+c) 3 ) ln(abc) = 3 ln(a+2b+c) ln(a) ln(b) ln(c). Then f = 0 gives a = 0 = 3 a + 2b + c 1 a = 0 b = 0 = 6 a + 2b + c 1 b = 0 c = 0 = 3 a + 2b + c 1 c = 0 Therefore, the critical point of f(a,b,c) must satisfy 3 a+2b+c = 1 a = 1 c = 1 2b, or a = c = 2b. Since the scaling of a normal direction as in (1) does not matter, we can set b = 1 and write the plane as 2(x 1) + y 2 + 2(z 1) = 0, or 2x + y + 2z = 6. The minimum is V (a,b,c) = 1 (3a) 3 = 9. The maximum of V does not exist 6 a 3 2 (E.g., V when a 0 + while b and c are set to 1).

11 Another solution: We can formulate the problem as a constrained optimization problem and apply Lagrange multiplier. Express the plane as x a + y b + z c = 1. (It simply means that the normal direction is n = 1, 1, 1 a b c.) This form of plane is convenient for this problem since it readily shows that the intercepts with the x-, y-, z- axises are a,b,c respectively. Therefore the volume of the cutoff tetrahedron is V (a,b,c) = 1 abc. Now we need to satisfy the constraint 6 that the plane passes a giving point. Plugging the point (1, 2, 1) into the plane we get = 1. a b c So the mathematical model is a constraint optimization problem { min f(a,b,c) = 6V (a,b,c) = abc a,b,c>0 s.t. g(a,b,c) = = 0 a b c By Lagrange multiplier, taking partial derivatives w.r.t. a,b,c as in We get f = λ g bc = λa 2 ac = 2λb 2 ab = λc 2 Multiplying respectively a, b, c on both sides of the above three equations, we get abc = λ 1 a = λ2 b = λ1 c. Since a,b,c are positive, we have λ 0. So we must have a = c, b = 2a. Plugging the above into the constraint = 1 we get a = c = 3,b = 6. a b c Therefore the plane that cuts off smallest volume is x 3 + y 6 + z 3 = 1, or 2x + y + 2z = 6. And the smallest cutoff volume is V (3, 6, 3) = = 9.

12 Problem 11. (Bonus 10 points) A central concept in statistical mechanics/thermodynamics and information science is Entropy. For n random variables p i > 0, (i = 1,...,n), where p i s are independent, the entropy can be defined as n S(p 1, p 2,..., p n ) = p i ln(p i ), with the constraint n p i = 1. i=1 Apply Lagrange multiplier method to find the extreme value of S. (In this case it is a maximum.) Here this objection function is and the constraint is i=1 S(p 1, p 2,..., p n ) = g(p 1, p 2,..., p n ) = n p i ln(p i ), i=1 n p i 1 = 0. i=1 The Lagrange multiplier satisfies S = λ g, that is This gives S p i = λ g p i, (i = 1, 2,...,n). ln(p i ) p i 1 p i = λ, = ln(p i ) = λ + 1, (i = 1, 2,...,n). Therefore at the extreme value of S all p i s are equal. Since they sum to 1, we must have p 1 = p 2 = = p n = 1 n. The point ( 1, 1,..., 1) in n n n Rn corresponds to a maximum of S, which is n ( ) 1 max p i ln(p i ) = ln = ln(n). p 1,...,p n > 0 n i=1 n i=1 p i = 1 (We ll have to skip the second order derivative test for the Lagrangian L = S λg in this higher dimension case, but it is not hard to see that 2 L = 1 p i p 2 i = 2 S p 2 i < 0 for all positive p i, which means the critical point of L cannot correspond to a minimum. Note also that 2 L p i p j = 2 S p i p j = 0 for i j, so the Hessian matrix of L is positive definite, that is why the critical point corresponds to a maximum. The result says that the highest unordered state (i.e., no difference among p i s) has the maximum entropy. )