Business Calc Homework w answers_Part_22

Business Calc Homework w answers_Part_22 - 106 Section 3.7...

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Unformatted text preview: 106 Section 3.7 6x 2 31. d (6x 2) dx d (3xy) dx dy 12x 3x dx dy 3x dx 2y 2 3xy 17y 6 0 34. x sin 2y d d d (17y) (6) (0) dx dx dx dy dy (3y)(1) 4y 17 00 dx dx dy dy 4y 17 12x 3y dx dx dy (3x 4y 17) 12x 3y dx dy 12x 3y dx 3x 4y 17 y cos 2x d (2y 2) dx d (x sin 2y) dx (x)(cos 2y)(2) dy dx (sin 2y)(1) dy dx dy (cos 2x) dx dy (y)( sin 2x)(2) (2x cos 2y) d (y cos 2x) dx (cos 2x) dy dx 2y sin 2x sin 2y 2y sin 2x 2x cos 2y sin 2y cos 2x dx 6 (x 7 (a) Tangent: y 1) 7 (x 6 (b) Normal: y 2x 23 3 (a) Tangent: y d (sin y) dx dy cos y dx dy cos y) dx dy dx 2y(1) (2x 2 d (5) dx (b) Normal: y y 3 : 2 2 (x (x 1 2 4 0 2 cos ( x Slope at (1, 0): d (2 ) dx x 4 2 2 (x or y 2x 1 x 2 or y 2 2 sin ( x 2 cos ( x 5 8 y) d 2 sin ( x dx y) dy dx y) 2 cos ( x y) 2 cos ( x y) 1 2 cos ( x y) 2 cos 2 cos (a) Tangent: y 1 cos 2 dy dx dy dx dy y)] dx dy dx 4y [1 sin ( ) 2 cos ( ) y 2 1 1) 2 ( 1) 2( 1) 0 or y 2 2x 2 0 1 (x 2 (b) Normal: y 2y 1) x 2 0 or y 1 2 2y 2x 1) 1) sin 2x 2x x3 4 35. 2x 3 y 2 2 0 0 2( /2) 2(1) cos( /2) (a) Tangent: y (b) Normal: y 0 ( 1) (a) Tangent: y 4(2) sin y d 2 (xy) dx Slope at 1, 2 2 3 2xy dy dx 3 42 ( )(1) 7 6 : 2 (b) Normal: x 33. 23 6 7 , 5 d 2 (y 2 ) dx dy 3(y)(1) 4y dx dy ( x 3 4y) dx dy dx Slope at ( 3, 2): 2x 2y 2 3xy 2 Slope at 7 x 6 0 or y d 3 (xy) dx dy 3(x) dx 6 7 6 x 7 0 or y 1) x2 32. d2 (x ) dx 12 14 12( 1) 3(0) 3( 1) 4(0) 17 Slope at ( 1, 0): 2 2 2 2 x sin y 0 d2 (x cos2 y) dx d (sin y) dx dy (x 2)(2 cos y)( sin y) dx 2 or y or y 36. x 2 cos2 y cos y x (2x 2 cos y sin y 2 2 dy dx cos y) 2 cos y d (0) dx (cos2 y)(2x) dy dx 2x cos y 2x 2 cos y sin y Slope at (0, ): 1 2(0) cos 2(0)2 sin (a) Tangent: y (b) Normal: x 0 (cos y) 2x cos2 y 1 2x cos y 2x2 sin y 0 dy dx 0 107 Section 3.7 y4 37. (a) y2 d4 (y ) dx dy 4y3 dx dy 3 (4y 2y) dx dy dx 3 3 At , : 4 2 x2 39. (a) ( 1)3(1)2 d2 d2 (y ) x dx dx dy 2y 2x dx (x 3)(2y) 2x 2y x (2x 3y 2y3 y 2 2 2 4 3 4 3 4 At 33 4 2 3 y3 2y 1 3 4 13 2 1 2 1 4 4 4 1 5 t y (2 Tangent: y Normal: y x) dy dx dy x) dx dy dx 1 (x 2 1) 1) xy . Hence, 5 2 , and 1 xy (y)(1) 0 x)y y (3y 2 y 3y 2 x y d dx 3y 2 x (3y 2 x)(y ) (y)(6y y (3y 2 x)2 y xy 3y 2 y 2 (3y x)2 y 3x 2 2 y y2 2 3x 2y(2 1 or y 5 2 1 3 3x 2 4 2 1). The d ( 1) dx d3 (x ) dx x)] 3(1)2 (1)2 2(1)(2 1) 2(x 1 d (xy) dx 3y 2y x x)(2y) 2y(2 Slope at (1, 1): 1 y 1 = 0 are d3 (y ) dx 2 d2 [y (2 dx y y3 [ 1.8, 1.8] by [ 1.2, 1.2] 1 1)(y 2 2y + 1 as (y . (b) Parameter interval: 3 2 1 is one solution, and there are three possible y-values: 1, 2 (2 3 2 1, or (1) 4(1)( 1) 2(1) 3 1 (y2)( 1) 3x 2y 2 sin y 2y 2 1 y 3 (b) 38. (a) 2x 3y 0. Clearly, y solutions of y 2 2 3x 2y 2 1 we may factor y 3 3 4 1 2 dy dx dy dx 2, we have y 3 40. (a) When x 1 31 ,: 42 Slope sin y) dy dx 3( 1)2(1) 2( 1)3(1) sin 3 The slope of the tangent line is . 2 33 3 3 2 ( sin y)( ) Slope at ( 1, 1): 4 3 d cos ( y) dx (y 2)(3x 2) dy dx 1. cos( y) d 32 (x y ) dx 3 Slope x 3y 2 (b) 2x 4y3 cos ( ) is true since both sides equal y x) 2 2x 1 or y (b) One way is to graph the equations y Since we are working with numerical information, there is no need to write a general expression for y in terms of x and y. 1 1 x 2 3 2 x3 2 1) x . To evaluate f (2), evaluate the expression for y using x 2 and y f (2) 1: 1 3(1)2 2 1 1 1 To evaluate f (2), evaluate the expression for y using x 2, y f z (2) 1, and y (1) 2(1) [3(1)2 1: 3(1)2(1) 2]2 4 1 4 108 Section 3.7 41. Find the two points: The curve crosses the x-axis when y 0, so the equation becomes x 2 0x 0 7, or x 2 7. The solutions are x 7, so the points are ( 7, 0). Show tangents are parallel: x2 y2 xy (2x y) 0 2( 7) 2x 2 d (2x 2) dx 2 0 2 2(0) d2 (x ) dx 2x x d (xy) dx dy dx y2 xy 2y d (7) dx dy dx 0 d (3y 2) dx dy 4x 6y dx dy dx 5 d (5) dx 0 4x 6y (2x dy dx y) 3x 2 3x 2 2y the slopes are 2x y x 2y 2x 3y d3 x dx At (1, 1), the slopes are dy 2y) dx (x 3y 2 x3 d2 y dx dy 2y dx dy dx 7 d2 (y ) dx (y)(1) 7 . 3 Second curve: y2 x2 2 43. First curve: 2x y x 2y The tangents at these points are parallel because they have the same slope. The common slope is 2. 42. 7 7 and 2 , 3 3 Note that these are the same points that would be obtained by interchanging x and y in the solution to part (a). 2(0) 7 7 3 The points are 0 27 7, 0): 7 , 3 7 7 7 7 d (7) dx 7 Slope at ( ( 2y) y2 y2 y2 3y 2 y 7 d2 d d2 (x ) (xy) (y ) dx dx dx dy dy 2x x (y)(1) 2y dx dx dy (x 2y) dx dy dx Slope at ( 7, 0): x 2 xy ( 2y)(y) 4y 2 2y 2 2 2 and 3 2 3 and respectively. At (1, 3 2 1), 3 respectively. In both cases, the 2 tangents are perpendicular. To graph the curves and normal lines, we may use the following parametric equations for (a) The tangent is parallel to the x-axis when dy dx t 2x y x 2y Substituting 0, or y : 2x. First curve: x 2x for y in the original equation, we have Second curve: x 2 x2 2 x xy y (x)( 2x) ( 2x)2 x 2 2x 2 4x 2 3x 2 x The points are 7 7 7 7 7 7 ,2 and 3 3 5 cos t, y 2 32 t ,y 7 3 7 , 3 2 7 . 3 dx can be obtained by interchanging x and y dy dy dx 2y x in the expression for . That is, . The dx dy x 2y dx tangent is parallel to the y-axis when 0, or dy equation, 2y. Substituting equation, we have: t Tangents at (1, 1): x 1 3t, y x 1 2t, y 1 3t Tangents at (1, 1): x 1 3t, y x 1 2t, y 1 3t (b) Since x and y are interchangeable in the original x 5 sin t 3 2y for x in the original [ 2.4, 2.4] by [ 1.6, 1.6] 1 2t 1 2t Section 3.7 s (t) a(t) d (4 dt 9(4 44. v(t) 3 (4 2 6t)1/2 At t (6) 27(4 2, the velocity is v(2) 6t) 1/2 dv dt t) 4(s t) 4(s t) 1/2 (v 4(s t) 1/2 1] [(8(s 1/2 x 3 0 x6 27 3x 3 0 54) 0 9x x 1) t)1/2 1) 1] At x 1/2 (s 0 2 x 3 1 dt 9xy 23 x3 3 (x 27 x3 1/2 1/2 ds t) y3 x3 d [8(s dt 32(s 3y y2 x3 36 m/sec and the 27 m/sec2. 4 acceleration is a(2) 45. Acceleration x2 x2 0, or y . 3x 3 x2 Substituting for y in the original equation, we have: 3 6t)1/2] 1/2 6t) (b) The tangent is horizontal when 6t)1/2(6) dy dx d [9(4 dt v (t) 9 (4 2 6t)3/2 109 t) 3 0 or x 02 3 0, we have y 54 32 ft/sec2 y 13 (9 4) 3 3 2 0, which gives the point (0, 0), which is the origin. At x 132 (3 2) 3 3 3 3 3 3 3 2, we have 4, so the point other 3 than the origin is (3 2, 3 4) or approximately 46. y 4 4y 2 x d4 (y ) dx d (4y 2) dx dy dy 8y 4y 3 dx dx 4 9x d4 (x ) dx 4x 3 4y 3 47. (a) 2): (c) The equation x 3 9(3) 4(2) 2(3)3 2( 2)3 x3 d3 d3 (x ) (y ) dx dx dy dy 3x 2 3y 2 9x dx dx (3y 2 3(2) Slope at (4, 2): 2 (2) 3(4) Slope at (2, 4): 2 (4) 2x 3 2y 3 the line y 9x 4y 3 9(3) 4( 2) d dx x2 48. 27 8 2) or 2x d (0) dx 9(y)(1) 0 dy dx dy dx 9y 3y2 0 d d d 2 (xy) (3y2) (0) dx dx dx dy dy 2x 2(y)(1) 6y 0 dx dx dy (2x 6y) 2x 2y dx dy 2x 2y xy dx 2x 6y 3y x 0 9 (xy) 2xy d2 (x ) dx 27 8 9xy (4)2 3(4) (2)2 3(2) 3 approximately (4.762, 3.780). 27 8 9( 3) 4( 2) 9x) x and we may find the desired point by to part (b). The desired point is (3 4, 3 9( 3) 4(2) y3 9xy is not affected by interchanging the x-value and the y-value in the answer 27 8 2( 3)3 2( 2)3 y3 interchanging x and y, so its graph is symmetric about 18x 8y 2( 3)3 Slope at ( 3, 2): 2(2)3 2): d (9x 2) dx 18x 2(3)3 Slope at (3, 2): 2(2)3 Slope at (3, (3.780, 4.762). 4x 3 dy dx Slope at ( 3, 2 At (1, 1) the curve has slope normal line is y 3x 2 9y 3x 2 3y 2 9x 10 5 8 4 4 8 5 10 3y y2 x2 3x Substituting x 1(x 11 3(1) 1 1) + 1 or y 2 2 1, so the x 2. 2 for y in the original equation, we have: x 2 2xy 3y 2 0 x 2x( x 2) 3( x 2)2 0 2 x 2x 2 4x 3(x 2 4x 4) 0 4x 2 16x 12 0 4(x 1)(x 3) 0 x 1 or x 3 Since the given point (1, 1) had x 1, we choose x and so y (3) 2 1. The desired point is (3, 2 3 1). 110 Section 3.7 49. xy d (xy) dx dy x dx 2x y 0 d (2x) dx d (y) dx dy (y)(1) 2 dx dy (x 1) dx dy dx 0 b 2x 2 0 2 y 2 1 2, we wish to 2, that is, where 1 2y 1 0 2b 2x 2a 2y b 2x1 a 2y1 2y Substituting 2y d (x) dx 1 dy dx y b 2x1x But a 2y12 3 . b 2x1 y1 a 2y1 (x x1). This gives: 2 a 2y12 b 2x12 b 2x12. a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives x2 a2 (b) b 2x 2 b2 a 2b 2 d 22 (b x ) dx d 22 (a y ) dx dy 2b 2x 2a 2y dx dy dx d 22 (a b ) dx 0 2b 2x 2a 2y b 2x 1 a 2y 1 y1 a 2y1y a2y12 b2x1x b2x12 a 2y12 b 2x1x a 2y1y hyperbola. Therefore, b 2x1x 2b3 b 2b b2 1 greater than if b 2 2b(x 1 or b 2 1 12 2 2 b2 ) b. This line intersects the x-axis at 1 , which is the value of a and must be 2 b) will be perpendicular when 1, which gives 2y 1 or 1 . The corresponding value of a is 2 1 3 b2 . Thus, the two nonhorizontal 2 4 3 normals are perpendicular when a . 4 a 2y12 a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives y2 b2 y2 b2 y1y 2 b2 a y 2 y 1 x2 a2 a 2y1y x1x 52. (a) Solve for y: x2 a2 0. The two normals at (b 2, they have slopes x1). b2x12 b, or x (x This gives: line at (b 2, b) is y 2b b 2x1 a 2y1 But b 2x12 2bx b 2x a 2y . 2y. Thus, the normal y 1. 1 The normal line at (x, y) has slope 3 a 2b 2, and y1y a2 a 2y 2 The slope at (x1, y1) is b 2x1x x1x y2 b2 The tangent line is y d2 (y ) dx dy 2y dx 1 2y b 2x12 b 2x1x ellipse. Therefore, a 2y1y 3 in the original equation, we have: 3)y x a 2y12 a 2y1y xy 2x y 0 2( 2y 3) y 0 2y 2 8y 6 0 2(y 1)(y 3) 0 y 1 or y 3 At y 1, x 2y 3 2 3 1. At y 3: x 2y 3 6 3 3. The desired points are ( 1, 1) and (3, 3). Finally, we find the desired normals to the curve, which are the lines of slope 2 passing through each of these points. At ( 1, 1), the normal line is y 2(x 1) 1 or y 2x 3. At (3, 3), the normal line is y 2(x 3) 3 or y 2x 3. ( 2y y b 2x a 2y x a 2y1y x 50. d 22 (a b ) dx The tangent line is y x 4 a 2b 2 The slope at (x1, y1) is 1 2 y) a 2y 2 dy dx 1 the tangent has slope . Thus, we have 2 2(2 1 d 22 (a y ) dx dy 2b 2x 2a 2y dx y x 0 is find points where the normal has slope y x y2 b2 d 22 (b x ) dx y 2y x1 Since the slope of the line 2x 2 1 x2 a2 51. (a) 1 b2 2 (x a 2) a2 b x2 a2 a 1. a 2b 2, and ...
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