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Pre-Calc Homework Solutions 232

# Pre-Calc Homework Solutions 232 - 232 Section 6.1(V max)2...

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Unformatted text preview: 232 Section 6.1 (V max)2 sin2 (120 t) 5. Again h k 1. The y-coordinate of the midpoint of each 7 2 57. (a) V 2 Using NINT: av(V ) (Vmax) Vrms (b) Vmax 2 2 line segment is . The x-coordinates of the midpoint of each line segment are , , , ..., 2 2 1 1 1 0 1 (Vmax) sin (120 t) dt 2 1 3 5 2 2 2 2 19 . From left to right the 2 slopes of the line segments are 21 sin (120 t) dt 0 (Vmax) (Vmax)2 2 1 2 2 2 (Vmax)2 2 Vmax 2 , 3 1 2 2 , 5 1 2 2 , ..., 19 1 2 1 The 10 graphs are graphs of the functions. 240 2 339.41 volts y1 2 1 2 x 1 Chapter 6 Differential Equations and Mathematical Modeling s Section 6.1 Antiderivatives and Slope Fields (pp. 303315) Exploration 1 Constructing a Slope Field 1. As i and j vary from 1 to 10, 100 ordered pairs are produced. Each ordered pair represents a distinct point in the viewing window. 2. The distance between the points with j fixed and i r and i r 1 is the distance between their x-coordinates. Xmin (Xmin 2(r 1) 1 h 2 1 2 7 ,0 2 x 1, y2 3 3 2 x 1 3 2 7 ,1 2 x 2, y3 2 5 2 x 1 5 2 7 ,2 2 x 3, y10 2 19 2 x 1 19 2 7 ,9 2 x 10. Xmin 2 1 2r (2r 1) h 2 1) h 2 Xmin) (2r h [0, 10] by [0, 10] 3. The distance between the points with i fixed and j r and j r 1 is the distance between their y-coordinates. Ymin (Ymin 4. Here h 4 7 6. For each line segment in part (5), make a column of parallel line segments as in part (4). 7. WL (2(r 1) 1) k 2 Ymin 2 1 2r (2r 1) k 2 1) k 2 Quick Review 6.1 k 1. 100(1.06) 2. 100 1 \$106.00 \$106.14 \$106.17 \$106.18 (cos 3x)(3) sec 2 x (Ce 2x)(2) 2) 1 x 2 5 2 5 2 0.06 4 4 0.06 12 12 0.06 365 365 Ymin) k (2r 1. Each line segment in the third column has 3. 100 1 4. 100 1 5. 6. 7. 8. dy dx dy dx dy dx dy dx slope , because the x-coordinate of the midpoint of each line segment is 2.5. The y-coordinates are , , , ..., 1 3 5 2 2 2 19 . 2 The 10 graphs are graphs of the functions y 4 (x 7 2.5) n ,2 2 x 3, for n 1, 3, 5, ..., 19. d sin 3x dx 5 d tan x 2 dx d Ce 2x dx d ln (x dx 3 cos 3x 5 5 sec 2 x 2 2 The length of the line segment can be increased or decreased by adjusting the restriction 2 x 3. 2Ce 2x [0, 10] by [0, 10] ...
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