Week #15 - Word Problems & Differential Equations Section 8.2
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1 Week #1 - Word Problems & Differential Equations Section 8. From Calculus, Single Variable by Huges-Hallett, Gleason, McCallum et. al. Copyrigt 00 by Jon Wiley & Sons, Inc. Tis material is used by permission of Jon Wiley & Sons, Inc. SUGGESTED PROBLEMS. A parabola, upside down, sifted up by 4. We take only te part to te left of x = 0 and above te axes. In te second picture, we see tat area spun around te x-axis. y x If we take vertical slices of tis object, we ll get circles. Te circle at point x will ave radius r = 4 x. Tus te volume of te object can be approximated by Riemann Sum = (πr x = π(4 x x Converting tat sum to integral format, 1
2 0 Exact Volume = 0 π(4 x dx [16x 8x = 6 1 π (16 8x + x 4 dx + x ] 0 [ ( 0 (16( (8( 8 + ] y 1 1. x If we take vertical slices of tis object, we ll get circles. Te circle at point x will ave radius r = e x. Tus te volume of te object can be approximated by Riemann sum = πr x (e x x e x x Using integration to compute te exact volume,
3 1 Exact Volume 1 e x dx [ ] e x 1 1 [ e e ] 11.9 units cubed y 0 10 e (x e x x Cutting vertically troug tis object gives disc sapes. Te cross-sectional area of a disk is πr out πr in te area of te outer circle minus te area of te inner circle. See te example on page 76 for a nicer looking diagram. Te outer radius is given by r out = e x, wile te inner radius is r in = e x. Tis gives us
4 Riemann sum = (πr out πr in x (e x (e x x (e 6x e x x Te corresponding integral gives us te exact area, given tat we re adding up tese slices from x = 0 to x = 1: 1 Exact Volume (e 6x e x dx 0 [ e 6x ] 6 ex 1 0 ] [( e6 6 e (e0 6 e0 ( e 6 6 e units cubed. Te first difficulty in tis question is te wording. It sounds like a is someting you want to optimize, but really it is a fixed constant, and you are just trying to find te volume of te boat. Wit tat in mind, computing te volume becomes anoter example of slicing up an object to find an appropriate volume integral. For tis object, te easiest way to slice it would be parallel to te parabolic sape (perpendicular to L, so tat every slice is an identical parabola. Volume (Parabola Area x for x = 0 to x = L. Te area of te parabola is a 1D integration problem: H y H a 0 H a x Te area of tis parabola can be computed by taking vertical slices. Eac slice will be (H ax ig: te difference between te orizontal line y = H and te parabola, y = ax. Te widt of eac slice will be x. Tis area is given by te integral below: 4
5 H/a Parabola Area = (H ax dx H/a H/a = Hx ax ( H = H a a = H/a ( H a ( H / 1 H / a a = 4 H / a / ( H H a a ( H / a If tis is te area of te cross-section of te boat for its wole lengt, ten te total volume of te boat is (area (lengt: If L and H are in meters, Volume = ( 4 H / L a Buoyancy force = 40,000LH/ a newtons Tis is te maximum weigt supportable by te boat. Converting to a mass, te boat can old m = 1 g (Buoyancy force, were g is te acceleration due to gravity, 9.8 m/s. QUIZ PREPARATION QUESTIONS 4. (a Tere are several approaces possible to solving tis problem. We sow two possibilities below. Solution 1 - Using Integration If we take orizontal slices of te gutter, of tickness, we will get rectangular slabs. Tese slabs will be 100 cm long, and w( wide. To determine te widt of a given slice, we look at te details of te trapezoid:
6 w Te volume of any slice will be w 100. From te diagram, w = + /, so our Riemann sum for te volume is Riemann sum = 100 ( +. Converting to an integral form for te exact computation, we want to use as te upper limit of integration. Tis means we sould use a different variably inside te integral, say y: Volume up to = 100 ( + y dy 0 = 100 (y + y 0 = 100 ( + Solution - Witout Integration If te gutter as te same trapezoidal profile for its wole 100 cm lengt, ten its maximum volume will be (Area of trapezoid (100 cm. You can use te 60 o angle to find te ratio of te sides in te triangles. See te diagram below. Wen te eigt of te water in te gutter is at, we ave te following dimensions: 6
7 Cross-section area = Rectangle + Triangle ( 1 = w + b = + = + Total Volume = 100 cm (Cross-section area = 100 ( + (b From te second diagram above, te maximum value of is 4.. At tat eigt, te gutter is filled so tat te side lengt of is completely used. (c Te maximum volume te gutter can old can be computed using our general formula for volume, and setting to its largest possible value, Max volume = 100 ( + = 100 ( cm for = (d If te gutter currently olds alf of its maximum volume, ten te dept will be above alf te maximum dept. Tis is because te gutter starts narrow at te bottom. If one unit of volume were added, it would be spread around a smaller area, and so would be deeper. As te gutter fills, te cross-section gets larger, meaning tat te same volume could be spread tinner across a wider area. 7
8 (e We are given tat V = 1 max V, or V 16. We want to solve for eigt in te equation 16 = 100 ( = = 0 ± 4( 1 (16. Quadratic formula: = /., 11. =. makes sense in te problem, as it is a little over alf te maximum dept of 4.. Tus, wit alf of its maximum volume, te gutter will be filled to approximately. cm dept of water. 8
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