6.4 Variation 323 Solution a) Since d varies directly as the square of t, we start with the equation d kt .2 = Next, to determine k, substitute 144 for d and 3 for t into the variation equation and then solve for k. = = = = = d kt k k k k 144 (3) 144 9 144 9 9 9 16 2 2 Thus, the constant of proportionality, k, is 16. b) To determine the distance an object falls after 2 seconds, we start with the variation equation, d kt ,2 = and substitute k 16 = and t 2. = = = = = d kt d d d 16(2) 16(4) 64 2 2 Thus, the object falls 64 feet after 2 seconds. 7 Now try Exercise 41 Example 4 Using the Constant of Proportionality The area, a, of a picture projected on a movie screen varies directly as the square of the distance, d, from the projector to the screen. If a projector at a distance of 25 feet projects a picture with an area of 100 square feet, what is the area of the projected picture when the projector is at a distance of 40 feet? Solution We begin with the formula a kd .2 = Since the constant of proportionality is not given, we must determine k, using the given information. = = = = = a kd k k k k 100 (25) 100 (625) 100 625 0.16 2 2 We now use k 0.16 = to determine a when d 40. = = = = = = a kd a d a a a 0.16 0.16(40) 0.16(1600) 256 2 2 2 Thus, the area of a projected picture is 256 ft2 when the projector is at a distance of 40 ft. 7 Now try Exercise 47 Inverse Variation A second type of variation is inverse variation. When two quantities vary inversely, as one quantity increases the other quantity decreases, and vice versa. To explain inverse variation, we use the formula distance rate time. = ⋅ If we solve for time, we get time distance/rate. = Assume the distance is fixed at 100 miles; then time 100 rate = WithGod/Shutterstock
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