0 (10/20)xdx = 1 2 (x2 /2) 20

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1 Work eample: Leak bucket Suppose ou lift a bucket of water straight up using a rope attached to a pulle. But as ou lift the bucket, it leaks water at a constant rate.the bucket weights 2lbs, the rope is 2 ft long and weights a total of lbs. The rope is wound around the pulle at a rate of 2 ft/s. The bucket starts out holding 5 lb of water and leaks at a rate of / lb/s. Howmuchworkisrequiredtoliftthe bucket to the top? Answer: We do this problem in three parts, () the bucket, (2) the rope, and (3) the water. pulle () Bucket. The bucket eerts a force of 2 lbs, and is lifted 2 ft, so W bucket = 2(2) ft-lbs. 2 bucket (2) Rope. Break the rope into vertical segments of length. Each segment eerts a force of (/2 lb/ft) ft, and the segment of rope at height gets lifted ft. W rope = R 2 (/2)d = 2 (2 /2) 2 = (2)2. Work eample: Leak bucket The bucket weights 2lbs,theropeis2 ft long and weights a total of lbs. The rope is wound around the pulle at a rate of 2ft/s.Thebucketstartsoutholding5 lb of water and leaks at a rate of / lb/s. How much work is required to lift the bucket to the top? W bucket = 2(2) W rope = (2)2 (3) Water. The work done to lift the water from height to height is f(), wheref() =weight of water remaining at position, sothatw water = R 2 f()d. As a function of time, starting from when the bucket begins to be lifted, the position of the bucket is = 2 2t. So time, as a function of position, is t = 2. Also as a function of time, the weight of the bucket is (/)t. So So f() = (/)t() = (/)( 2)= W water = Z d=(9 + 2 ) 2 = 9(2) + (2)2.

2 Work eample: Leak bucket The bucket weights 2lbs,theropeis2 ft long and weights a total of lbs. The rope is wound around the pulle at a rate of 2ft/s.Thebucketstartsoutholding5 lb of water and leaks at a rate of / lb/s. How much work is required to lift the bucket to the top? So in total, W bucket = 2(2) W rope = (2)2 W water = 9(2) + (2)2 W = W bucket +W rope +W water = 2(2) + (2)2 + 9(2) + (2)2. 9. arametric curves In the water portion of the previous problem, position and weight started out as functions of time: (t) = 2 2t and f(t) = (/)t. These are called parametric equations, with parameter t. Separatel, the re just two functions of time. But together, the are coupled b their common parameter. We can thus graph f versus b varing t. To find the equation for f as a function of, we solved for t, and plugged that into f: t = 2, so f = (/)( 2 )=9+ 2. f(t) t= t=5 t= ()=2, f()= (5)=, f(5)=9.5 ()=, f()= (t)

3 Eample: Define the parametric curve b (t) =t 2 2t, (t) =t +. lotting the curve: ick a sample of values for t: t t=2 t= t= (, ) t=3 t=_ t= 8 t=_2 This curve is suited best writing as a function of, so solve for t in terms of and plug in: t =, so =( ) 2 2( ) =( 2) 2. Eample: Define the parametric curve b (t) =t 2 2t, (t) =t +. t=2 For all t: t=3 t= For apple t apple : (8, 5) t= t= (, ) t=_ 8 t=_2 (, ) Writing the function just in terms of and loses some information. If we re thinking about the parametric function as a particle traveling on the - plane over time, we calculated that it traces the curve =( 2) 2,butitdoesn ttelluswhat direction or how fast. Further, we have put no restriction on t.

4 Eample: Unit circle. t=π (t) = cos(t) (t) =sin(t), apple t apple 2 lotting points π This curve traces out a circle! t= 2 (Recall the unit circle) t (cos t, sin t) t= (, ) t=2π Converting to a function of just and : = cos 2 (t)+sin 2 (t) =. 3π t= 2 You tr: Graph and compare the following parametric curves to each other and the eample above. () (t) = cos(2t),(t) =sin(2t), apple t apple 2 ; (2) (t) = cos(t/3),(t) =sin(t/3), apple t apple 2. Graph transformations Since a parametric curve gives the and coordinates separatel, transformations are a little more straightforward. Eample: We saw that (t) = cos(t), (t) =sin(t), apple t apple 2 is the unit circle centered at the origin. If I want a circle centered at the point (2, 5), that sthesameas shifting all the -coordinates right b 3 and all the -coordinates up b 5: (t) = cos(t)+2 (t) =sin(t)+5, apple t apple 2. If instead I still want a circle centered at (, ), butiwantitsradius dilated to 3, I want to multipl the and coordinates all b 3: (t) = 3 cos(t) (t) =3sin(t), apple t apple 2. If I want a bigger circle that s also shifted, dilate first and then shift (just as before). A circle of radius r, centeredat(a, b) is given b (t) =r cos(t)+a (t) =r sin(t)+b, apple t apple 2. Check: ( a) 2 +( b) 2 =(rcos(t)) 2 +(rsin(t)) 2 = r 2 (cos 2 (t)+sin 2 (t)) = r 2.X

5 You tr: Sketch the following curves and give a formula for their shape just in terms of and. (Hint: Think about graph transformations, and scaling or shifting either or both coordinates to get them to fit the pthagorean identit.) Give an eample of a domain for t that would trace the curve eactl once.. (t) = 2 cos(t), (t) =sin(t). 2. (t) = cos(t), (t) =3sin(t). 3. (t) = cos(t), (t) =sin( t).. (t) = 5 cos(2t), (t) =3sin(2t). 5. (t) = cos(t), (t) =sin( t). 6. (t) =sin(t), (t) = cos(t). 7. (t) = 2 cos(t)+, (t) = 3(sin(t) ). 8. (t) = 5 cos( t)+, (t) =2sin(t)+5.

6 Eample: Sketch (t) =sin(t), (t) =sin 2 (t). 228_ch9_ptg_hr_5-5.qk_228_ch9_ptg_hr_5-5.qk /7/ :7 AM age 5 We could solve for t from one and plug it into the other. But as a shortcut, it s clear to see that =sin 2 (t) = 2. So this curve appears to be a parabola. 5 CHATER 9 ARAMETRIC EQUATIONS AND OLAR COORDINATES.5 3 Graphing devices are particularl useful for sketching complicated curves instance, the curves shown in Figures 9,, and would be virtuall impossib produce 2 b hand. t=-!/2+2k! - t=!/2+2k! t=, ±!, ±2!, But _.5 apple sin() apple, sothe.5 values _2 can t go outside 2these _.8 bounds. This is actuall a curve traced out b a particle bouncing back and forth between (, ) and (, ) along the curve = 2. _.5 _.8 _.8 FIGURE 9 =sin t+ cos 2 5t+ sin 3t r Take =cos t+ a wheel sin or radius r, and mark =cos onet 2 5t+ cos 3t point on its boundar. Now roll that wheel, and trace the path that the marked point takes: TEC An animation in Module 9.B shows how the ccloid is formed as the circle moves. This curve is called a ccloid. To calculate its formula, we ll use a FIGURE 2 param. curve: arameter:, the rot l angle of circle. Center: The edge of the circle has all touched it has rolled thefrom ground. the origin is So the distance of the center from the -ais is the arc C(r, r) r length OT of the circle with angle. Sothe Q center is at C =(r,r). C r, r Then from Figure 3 we see that Then O T = OT Q = r r sin, r = TC QC = r r cos. FIGURE 3 OT Q FIGURE =sin t-sin 2.3t THE CYCLOID EXAMLE 7 The curve traced out b a point on the circumference of a circle the circle rolls along a straight line is called a ccloid (see Figure 2). If the circ has radius r and rolls along the -ais and if one position of is the origin, find parametric equations for the ccloid. OT arc T r arc T r C r, r OT Q r r sin r sin r r sin r sin SOLUTION We choose as parameter the angle of rotation of the circle when is at the origin). Suppose the circle has rotated through radians. Becau the circle has been in contact with the line, we see from Figure 3 that the distan Therefore the center of the circle is. Let the coordinates of be,. Therefore parametric equations of the ccloid are FIGURE =sin t+ sin 2 5t+ cos 2.3t =cos t+ cos 2 5t+ sin 2.3t TC QC r r cos r cos r sin r cos,