5128_Ch07_pp378-433

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Section 7.5 Applications from Science and Statistics 419 7.5 What you’ll learn about • Work Revisited • Fluid Force and Fluid Pressure • Normal Probabilities . . . and why It is important to see applications of integrals as various accumulation functions. Applications from Science and Statistics Our goal in this section is to hint at the diversity of ways in which the definite integral can be used. The contexts may be new to you, but we will explain what you need to know as we go along. Work Revisited Recall from Section 7.1 that work is defined as force (in the direction of motion) times displacement. A familiar example is to move against the force of gravity to lift an object. The object has to move, incidentally, before “work” is done, no matter how tired you get trying. If the force F(x) is not constant, then the work done in moving an object from x a to x b is the definite integral W b a F(x)dx. 4.4 newtons 1 lb (1 newton)(1 meter) 1 N•m 1 Joule EXAMPLE 1 Finding the Work Done by a Force Find the work done by the force F(x) cos(px) newtons along the x-axis from x 0 meters to x 12 meter. SOLUTION 12 W cos(px) dx 0 1 1/2 sin(px) p 0 p 1 sin p 2 sin(0) p 1 0.318 Now try Exercise 1. 22 (N) Work 440 N. m EXAMPLE 2 Work Done Lifting A leaky bucket weighs 22 newtons (N) empty. It is lifted from the ground at a constant rate to a point 20 m above the ground by a rope weighing 0.4 N/m. The bucket starts with 70 N (approximately 7.1 liters) of water, but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done (a) lifting the bucket alone; (b) lifting the water alone; (c) lifting the rope alone; (d) lifting the bucket, water, and rope together. SOLUTION (a) The bucket alone. This is easy because the bucket’s weight is constant. To lift it, you must exert a force of 22 N through the entire 20-meter interval. (m) 20 Figure 7.39 The work done by a constant 22-N force lifting a bucket 20 m is 440 N • m. (Example 2) Work 22 N 20 m 440 N • m 440 J Figure 7.39 shows the graph of force vs. distance applied. The work corresponds to the area under the force graph. continued
420 Chapter 7 Applications of Definite Integrals 70 (N) y F(x) = 70 – 3.5x (m) 20 Figure 7.40 The force required to lift the water varies with distance but the work still corresponds to the area under the force graph. (Example 2) 8 (N) Work (m) 20 Figure 7.41 The work done lifting the rope to the top corresponds to the area of another triangle. (Example 2) x (b) The water alone. The force needed to lift the water is equal to the water’s weight, which decreases steadily from 70 N to 0 N over the 20-m lift. When the bucket is x m off the ground, the water weighs Fx 70 ( 20 x 20 ) ( 70 1 x 70 3.5x N. 20 The work done is (Figure 7.40) W b a 20 0 Fx dx 70 3.5x dx [ ) 20 70x 1.75x2] 1400 700 700 J. 0 (c) The rope alone. The force needed to lift the rope is also variable, starting at 0.420 8 N when the bucket is on the ground and ending at 0 N when the bucket and rope are all at the top. As with the leaky bucket, the rate of decrease is constant. At elevation x meters, the 20 x meters of rope still there to lift weigh Fx 0.4 20 x N. Figure 7.41 shows the graph of F. The work done lifting the rope is Fx dx 20 0.420 x dx 20 0 original weight of water proportion left at elevation x 0 20 [ 8x 0.2x2] 160 80 80 N • m 80 J. 0 (d) The bucket, water, and rope together. The total work is 440 700 80 1220 J. Now try Exercise 5. 10 y y 10 8 0 y Δy y 2x or x 1 y 2 1 y 2 5 (5, 10) Figure 7.42 The conical tank in Example 3. x EXAMPLE 3 Work Done Pumping The conical tank in Figure 7.42 is filled to within 2 ft of the top with olive oil weighing 57 lbft 3 . How much work does it take to pump the oil to the rim of the tank? SOLUTION We imagine the oil partitioned into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval 0, 8. (The 8 represents the top of the oil, not the top of the tank.) The typical slab between the planes at y and y Δy has a volume of about 2 Δy p 4 y2 Δy ft 3 . ΔV p(radius) 2 (thickness) p ( 1 2 y ) The force Fy required to lift this slab is equal to its weight, Fy 57 ΔV 57 p y 4 2 Δy lb. weight per Weight ( unit volume ) volume The distance through which Fy must act to lift this slab to the level of the rim of the cone is about 10 y ft, so the work done lifting the slab is about ΔW 57 p 10 yy 4 2 Δy ft • lb. The work done lifting all the slabs from y 0 to y 8 to the rim is approximately W 57 p 10 y y 4 2 Δy ft • lb. continued
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