MATH 255 Applied Honors Calculus III Winter 2011 Homework 5 Solutions Note: In what follows, numbers in parentheses indicate the problem numbers for users of the sixth edition. A * indicates that this problem is not in the sixth edition and you should look in the Michigan edition. Section 15.1, pg. 933: 16 (14), 32, 53 58(*), 60 (62). Section 15.2, pg. 944: 6 (*), 8 (10), 28 (30), 36 (38). Section 15.3, pg. 955: 32 (36), 44 (48), 64 (68), 77 (81), 83 (87). Section 15.4, pg. 966: 4, 14, 18 (20), 36 (38). 15.1: #16 We have two restrictions. From y x we have that y x, and from ln(x + y) we have y > x. The domain of the function is where both these inequalities hold. That is: D {(x, y) y x and y > x} In figure (1) The gray areas are where each inequality is true, and the darker gray is where both are true. Figure 1. Plot for 15.1 #16. 15.1: #32 If we start at the origin and move along the x-axis, the z-values of a cone centered at at the origin increase at a constant rate, so we would expect its level curves to be equally spaced. A paraboloid with vertex the origin, on the other hand, has z-values which change slowly near the origin and more quickly as we move farther away. Thus, we would expect its level curves near the origin to be spaced more widely apart than those farther from the origin. Therefore the contour map I corresponds to the paraboloid, and contour map II to the cone. 15.1: #53-58 53. The function is constant on any circle centered at the origin. a) B; b) III 54. The function is symmetric about the plane x y. It is equal to 0 along the x- and y-axes. Moreover, it vanishes for x,y big enough. a) C; b) II 55. Level curves are ellipses, and for large x, y the the function becomes almost 0. a) F; b) V 56. Along the lines y ± 1 3 x and x 0, this function is 0 a) A; b) VI 57. This function is periodic in both x and y, with period 2π in each variable. a) D; b) IV 58. This function is periodic along x-axis, and increases as y increases. a) E; b) I 1
15.1: #60 Level surfaces of this function have equations x 2 + 3y 2 + 5z 2 k These are ellipsoids for k > 0 and the origin for k 0. 15.2: #6 Easy one, since the function is defined and well behaved there. lim xy cos(x 2y) 6 3 cos(6 6) 18 (x,y) (6,3) 15.2: #8 Try two lines: x 0 and y 0. f(x, y) x2 + sin 2 y 2x 2 + y 2 lim x 0 x 2 + 0 2x 2 + 0 1/2 lim f(x, y) y 0 lim y 0 0+sin2 y 0+y 2 lim y 0 ( sin y y )2 (lim y 0 sin y y )2 (lim y 0 cos y 1 )2 1 1/2 The limit doesn t exist. 15.2: #28 This is a rational function, so it is continuous on the whole domain. The denominator is never zero, so the domain is all x, y. So, the function is continuous for all x and y. 15.2: #36 For this function to be continuous at (a,b), its limit must equal its value at (a,b). This is a rational function, so it is automatically continuous when the denominator polynomial isn t equal to zero, which occurs only when (x,y) 0. Let s check if it is continuous at (0,0). { f(x, y) xy x 2 +xy+y 2 0 if (x,y) (0,0) otherwise
Following example 2, let s take y mx, for any slope m, and take the limit of x going to 0. x(mx) lim f(x, mx) lim x 0 x 0 x 2 + x(mx) + (mx) 2 mx 2 lim x 0 (1 + m + m 2 )x 2 m lim x 0 (1 + m + m 2 ) m (1 + m + m 2 ) As we can see, the limit depends on the slope of the line of approach we take. Therefore, it depends on our avenue of approach (whether or not it s a line), and the limit doesn t exist, much less equal 0. So f(x,y) is not continuous at (0,0), and is therefore continuous for all (x, y) (0, 0). 15.3 #32 y2. x t+2z 2xy. y t+2z 2xy2. z (t+2z) 2 xy2. t (t+2z) 2 15.3 #44 F (x, y, z) sin (xyz) x 2y 3z. z x F/ x F/ z (cos (xyz) (yz) 1)) cos (xyz) (xy) 3 and z y F/ y F/ z (cos (xyz) (xz) 2)) cos (xyz) (xy) 3 15.3 #64 u x a y b z c. 6 u x y 2 z 3 a b(b 1) c(c 1)(c 2)xa 1 y b 2 z c 3 Notice that this formula works even if a 0, b 0, 1 c 0, 1, 2 (the derivative is 0). 15.3 #77 R For starters, let s rewrite R(R 1, R 2, R 2 ) 1 R 2 R 3 R 1 R 2 +R 2 R 3 +R 1 R 3. Then, by the quotient rule R R 1
R 2 R 3 (R 1 R 2 +R 2 R 3 +R 1 R 3 ) (R 1 R 2 R 3 )(R 2 +R 3 ) (R 1 R 2 +R 2 R 3 +R 1 R 3. Distribute out and cancel terms on the top to get R ) 2 R2 2R2 3 (R 1 R 2 +R 2 R 3 +R 1 R 3. Multiply and divide by R 2 ) 2 1 to get R R1 R 1 2R2 2 R2 3 R1 2(R 1R 2 +R 2 R 3 +R 1 R 3 R2. ) 2 R1 2 15.3 #83 No, don t believe them. f xy 4 and f yx 3 are continuous, but not equal. They violate Clairaut s Theorem. 15.4 #4 z y ln x. To get the plane at (1,4,0), we need the partial derivatives f x (1, 4) and f y (1, 4). R 1 f x (x, y) y x f x (1, 4) 1 4 f y (x, y) ln x f y (1, 4) 0 The equation for the tangent plane is z 1 (x 1). 4 15.4 #14 Both partial derivatives exist and are continuous at the point (3, 0), so it s differentiable at the point. f(x, y) x + e 4y. f x (x, y) 1 2 (x + e4y ) 1/2 f x (3, 0) 1 4 f y (x, y) 2(x + e 4y ) 1/2 (e 4y ) f y (3, 0) 1 f(3,0) 2, so the linearization at (3, 0)is 15.4 #18 f(x, y) ln (x 3y) f(7, 2) 0. L(x, y) 5 4 + 1 4 x + y f x (x, y) 1 x 3y f x (7, 2) 1 f y (x, y) 3 x 3y f y (7, 2) 3
The approximation near (7, 2) is and the approximate value at (6.9,2.06) f(x, y) (x 7) 3(y 2) f(x, y) ( 0.1) 3(0.06) 0.28 Graphs omitted. 15.4 #37 The goal here is to make the linear approximation, and then see how much change in P we get if we change V, T accordingly. The equation is The differential can be calculated via Calculate the partial derivatives: P 8.31 T V. dp P T dt + P V dv. P T 8.31 1 V P V 8.31 T V 2 Now V 12, T 310, dv 0.3 and dt 5. Plug in to the formula to find dp 8.31 1 ( 5) + 8.31310(0.3) 8.83 kilopascals. 12 144 Additional Problem 1: You have to remember the material from Chapter 14 in order to do this problem! The first thing you have to realize is that when Dave snaps a picture while on the roller coaster, he is going to capture what is along the tangent line to his motion. The tangent line to Dave s position is given by < t 2, 2t, 3t 1 > +s < 2t, 2, 3 > which you find by taking the partial derivatives. Now we are looking for a t value so that the point (12, 8, 11) is on this tangent line. This just means we are trying to solve the three equations t 2 + 2st 12, 2t + 2s 8, and 3t 1 + 3s 11. From the second equation, we see that t 4 s, and plugging into the first equation we see that (4 s)(4 + s) 12. This has two solutions, s 2 (and then t 2) and s 2 (and then t 6) but only the first solution makes sense. A negative value of s means that the target would be behind Dave and he wouldn t capture it on his camera. So Dave should snap the picture exactly at t 2. Additional Problem 2: Your company is in great shape! Notice that if you increase labor and capital by a factor of m, then P (ml, mk) m α L α )m 2 α K 2 α m 2 P (L, K). In other words, if you increase your inputs by a factor of m, your outputs are increased by a factor of m 2. This is amazing, you should expand as much as possible and become a generous philanthropist. In the real world, even if your company had a production function that resembled this one - it would not last long and you wouldn t be guaranteed infinite wealth.