7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?


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1 7.1 Integrl s Net Chnge Clculus 7.1 INTEGRAL AS NET CHANGE Distnce versus Displcement We hve lredy seen how the position of n oject cn e found y finding the integrl of the velocity function. The chnge in position is displcement. To see the difference etween distnce nd displcement, consider the following sying: "Two steps forwrd nd one step ck" Wht is the totl distnce trveled? Wht is the totl displcement? To find displcement, we only need to find vt () dt. In order to find your new loction, we sy tht your new position = initil position + displcement. To find totl distnce we use vt () dtor find when the oject is moving in the negtive direction, rek the integrl into pieces nd sutrct the vlue of the integrl for the re under the curve. Exmple: Suppose the velocity of prticle moving long the x xis is given y t. vt () = 6t 18t+ 1 when ) When is the prticle moving to the right? When is it moving left? When is it stopped? ) Find the prticle's displcement for the time intervl. c) Find the prticle's totl distnce trveled y setting up ONE integrl nd using your clcultor. d) Find the prticle's totl distnce trveled without using solute vlue. 111
2 7.1 Integrl s Net Chnge Clculus Exmple: ) Integrting velocity gives ) Integrting the solute vlue of velocity gives. Consumption over Time Velocity is not the only rte in which you cn integrte to get totl. In fct if you were given function tht gve the numer of tickets per hour tht the police wrote ech dy, nd you wnted to find the totl numer of tickets in 4hour period, you could integrte. Exmple: [5 AP Clculus AB # Clcultor Allowed] The tide removes snd from Sndy Point Bech t rte modeled y the function R given y 4πt Rt () = + 5sin 5. A pumping sttion dds snd to the ech t rte modeled y the function S, given y 15t S() t = t Both R (t) nd S (t) hve units of cuic yrds per hour nd t is mesured in hours for < t < 6. At time t =, the ech contins 5 cuic yrds of snd. ) How much snd will the tide remove from the ech during this 6hour period? Indicte units of mesure. ) Write n expression for Y (t), the totl numer of cuic yrds of snd on the ech t time t. c) Find the rte t which the totl mount of snd on the ech is chnging t time t = 4. d) For < t < 6, t wht time t is the mount of snd on the ech minimum? Wht is the minimum vlue? Justify your nswers. 11
3 7.1 Integrl s Net Chnge Clculus Exmple: [4 AP Clculus AB (Form B) # Clcultor Allowed] For < t < 31, the rte of chnge of the t numer of mosquitoes on Tropicl Islnd t time t dys is modeled y Rt () = 5 tcos 5 mosquitoes per dy. There re 1 mosquitoes on Tropicl Islnd t time t =. ) Show tht the numer of mosquitoes is incresing t time t = 6. ) At time t = 6, is the numer of mosquitoes incresing t n incresing rte, or is the numer of mosquitoes incresing t decresing rte? Give reson for your nswer. c) According to the model, how mny mosquitoes will e on the islnd t time t = 31? Round your nswer to the nerest whole numer. d) To the nerest whole numer, wht is the mximum numer of mosquitoes for < t < 31? Show the nlysis tht leds to your conclusion. 113
4 7.1 Integrl s Net Chnge Clculus Exmple: [3 AP Clculus AB # Clcultor Allowed] A prticle moves long the x xis so tht its velocity t time t is given y t vt () = ( t+ 1sin ). At time t =, the prticle is t position x = 1. ) Find the ccelertion of the prticle t time t =. Is the speed of the prticle incresing t t =? Why or why not? ) Find ll times t in the open intervl < t < 3 when the prticle chnges direction. Justify your nswer. c) Find the totl distnce trveled y the prticle from time t = until time t = 3. d) During the time intervl < t < 3, wht is the gretest distnce etween the prticle nd the origin? Show the work tht leds to your nswer. 114
5 7.1 Integrl s Net Chnge Clculus Exmple: [ AP Clculus AB # Clcultor Allowed] The rte t which people enter n musement prk on given dy is modeled y the function E defined y 156 E() t = t 4t The rte t which people leve the sme musement prk on the sme dy is modeled y function L defined y 989 Lt () = t 38t Both E (t) nd L (t) re mesured in people per hour nd time t is mesured in hours fter midnight. These functions re vlid for 9 < t < 3, the hours during which the prk is open. At time t = 9, there re no people in the prk. ) How mny people hve entered the prk y 5: pm (t = 17)? Round your nswer to the nerest whole numer. ) The price of dmission to the prk is $15 until 5: pm (t = 17). After 5: pm, the price of dmission to the prk is $11. How mny dollrs re collected from dmissions to the prk on the given dy? Round your nswer to the nerest whole numer. t c) Let H () t = ( E( x) L( x) ) dx for 9 < t < 3. The vlue of H ( 17 ) to the nerest whole numer is 375. Find 9 the vlue of H '17 ( ) nd explin the mening of H ( 17) nd H '17 ( ) in the context of the prk. d) At wht time t, for 9 < t < 3, does the model predict tht the numer of people in the prk is mximum? 115
6 7. Ares in the Plne Clculus 7. AREAS IN THE PLANE Let s Review the concept of re s it reltes to clculus! Recll the re under curve cn e pproximted through the use of Riemnn sums: We cn rek the re into rectngles. Consider the one rectngle drwn. It s height is given y the function vlue of the curve t the right endpoint nd the width is given s Δ x. The re under the curve then is pproximtely the sum of the res of ALL the rectngles just like this one. f ( c) Are n k= 1 f ( c ) Δx k k Δx As the numer of rectngles, n, increses, the pproximted re gets closer to the ctul re, so we sy Are under the curve = lim f ( ck) Δ xk = f ( x) d n = k n x We cn pply this sme concept to the re etween curves. Consider the two functions f nd g elow. First: Drw rectngulr strip. Wht is the height nd width of your rectngle? Would the height nd width of the rectngle strip e different if you drew it in different plce? 1 g f Second: The re etween the curves is pproximtely the sum of ll of these rectngles. We cn write this s Third: How cn we get closer to the ACTUAL re etween the curves? Fourth: If we let the numer of rectngles pproch infinity, then we hve 116
7 7. Ares in the Plne Clculus Are of Region Between Two Curves If f nd g re continuous on [, ] nd g( x) f ( x) grphs of f nd g nd the verticl lines x = nd x = is for ll x in [, ], then the re of the region ounded y the ( ) ( ) A = f x g x dx Exmple: Find the re of the region ounded y the grphs of y= x +, y = x, x =, nd x = 1. Step 1: Drw picture nd shde the desired region. Step : Drw n ritrry rectngulr strip. Step 3: Using the re of the rectngulr strip s guide, set up nd solve n integrl to find the re etween the curves. Exmple: Find the re of the region ounded y the grphs of f ( x) = x nd g( x) = x. 117
8 7. Ares in the Plne Clculus 3 Exmple: Find the re of the region etween the grphs of f ( x) = 3x x 1x nd g ( x) = x + x. Exmple: Find the re of the region ounded y the grphs of x = 3 y nd x= y Exmple: Find the re of the region ounded y the grphs of g( x) = x, y = 4, nd x =. Exmple: The re of the region ounded y the grphs of 1 1 y 3 ( x ) x dx. 3 = x nd y = x cnnot e found y the single integrl Explin why this is so. Use symmetry to write single integrl tht does represent the re. 118
9 7.3 Volumes Clculus 7.3 VOLUMES Just like in the lst section where we found the re of one ritrry rectngulr strip nd used n integrl to dd up the res of n infinite numer of infinitely thin rectngles, we re going to pply the sme concept to finding volume. The key Find the volume of ONE ritrry "slice", nd use n integrl to dd up the volumes of n infinite numer of infinitely thing "slices". We will first pply this concept to the volume of solid with known cross section, then we will find the volumes of solids formed y revolving region out horizontl or verticl line. We will discuss three different methods of finding volumes of solids of revolution, ut first Dy 1: Volumes of Solids with Known Cross Sections First Question Wht is cross section? Imgine lof of red. Now imgine the shpe of slice through the lof of red. This shpe would e cross section. Techniclly cross section of three dimensionl figure is the intersection of plne nd tht figure. It would e like cutting n oject nd then looking t the fce of where you just cut. The cross sections we will e deling re lmost entirely perpendiculr to the x xis. Here's the sic ide You will e given region defined y numer of functions. We will grph tht region on n x nd y xis. Then we will ly they region flt nd uild upon tht region solid which hs the sme cross section no mtter where you slice it. To see some nimted views of this go to (We will wtch few of them in clss) Second question How do we find the volume of this solid tht hs een creted to hve similrly shped cross section, even though ech cross section my hve different size? We get to use clculus, of course! But first, we need to know how to find the Volume of prism. Even though every shpe my e different, we cn find the volume of prism y finding the re of the se times the "height". The "height" of our prisms will e the thickness of the slices. Once you know the volume of one slice, you just use n integrl to dd the volumes of ll the slices to get the volume of the solid. Exmple: Find the volume of the following squre "slice". Since most of the "slices" we will e deling with will hve thickness of dx, we will use tht sme thickness here. dx x Exmple: Find the volume of the following semicirculr "slice". x dx 119
10 7.3 Volumes Clculus You will lso need to e le to find the volume of equilterl tringle cross sections, isosceles right tringle cross sections, nd others. Rememer, wht you relly need is formul for the re of the se, which is just the cross sectionl shpe. Exmple: The se of solid is the region enclosed y the grph of y = e x, the coordinte xes, nd the line x = 3. If ll plne cross sections perpendiculr to the x xis re squres, then its volume is A) 1 e 6 B) 1 6 C) e 6 e D) 3 e E) 1 e 3 Exmple: The se of solid is the region in the first qudrnt enclosed y the prol y = 4x, the line x = 1, nd the x xis. Ech plne section of the solid perpendiculr to the x xis is squre. The volume of the solid is A) 4π 3 B) 16 π 5 C) 4 3 D) 16 5 E)
11 7.3 Volumes Clculus Exmple: The se of solid S is the region enclosed y the grph of y= ln x, the line x = e, nd the xxis. If the cross sections of S perpendiculr to the x xis re squres, then the volume of S is A) e B) 3 3 C) 1 D) E) ( 1) Exmple: The se of solid is region in the first qudrnt ounded y the xxis, the yxis, nd the line x+ y=8. If the cross sections of the solid perpendiculr to the x xis re semicircles, wht is the volume of the solid? A) B) C) D) 67.1 E) Exmple: The se of solid is the region in the first qudrnt enclosed y the grph of y = x nd the coordinte xes. If every cross section of the solid perpendiculr to the yxis is squre, the volume of the solid is given y A) π ( y) dy B) ( y) dy D) ( x ) dx E) ( ) C) π ( ) x dx x dx 11
12 7.3 Volumes Clculus Exmple: Let R e the region enclosed y the grphs of y= ln( x + 1) nd y = cos x. ) Find the re of R. (Clcultor ok) ) The se of solid is the region R. Ech cross section of the solid perpendiculr to the xxis is n equilterl tringle. Write n expression involving one or more integrls tht gives the volume of the solid. Do not evlute. 1
13 7.3 Volumes Clculus Dy : Volumes of Solids of Revolution: The Disc Method In finding the re of region, we drew n ritrry representtive rectngle. Keeping with the sme ide, if we revolve rectngle round line, it forms cylinder, s shown elow. The key to using the disc method will e twofold: 1) The rectngulr strip must e connected to the xis of revolution (no mtter where you drw it), nd ) the rectngulr strip must e perpendiculr to the xis of revolution. Exmple: Wht is the volume of the cylinder shown if the height of the rectngle is considered R nd the width of the rectngle is dx? Just like we did in finding the re, s we increse the numer of rectngles to infinity, the width of ech rectngle ecomes infinitely smll nd we denote this dx (if it is verticl strip) or dy (if it is horizontl strip). We then use n integrl to sum the volume of every one of these infinitely thin cylinders. This concept leds to the following: The Disc Method To find the volume of solid of revolution with the disc method, use one of the following; HORIZONTAL AXIS OF REVOLUTION VERTICAL AXIS OF REVOLUTION V = π R( x) dx V = π R( y) dy where R (x) nd R (y) re the "heights" of your representtive rectngulr strips. Exmple: Drw n pproprite rectngulr strip nd find the volume of the solid formed y revolving the region out the x xis. y = 4 x 13
14 7.3 Volumes Clculus Exmple: Find the volume of the solid formed y revolving the region out the y xis. (Drw representtive rectngulr strip) y = 16 x Exmple: Find the volume of the solid generted y revolving the region ounded y the grphs of the equtions xy = 6, y =, y = 6, nd x = 6 out the indicted lines. Sketch the region formed, nd drw representtive rectngulr strip for ech solid. ) the line x = 6. ) the line y = 6. 14
15 7.3 Volumes Clculus Dy 3: Volumes of Solids of Revolution: The Wsher Method For the disc method, the re we revolved hd to e connected to the re of rottion nd the representtive rectngle hd to e perpendiculr to the xis of revolution. For the wsher method, the representtive rectngle will still e perpendiculr to the xis of revolution, ut no longer ttched to the xis of revolution. Exmple: Sketch the figure formed y rotting the rectngle round the given line. Do you see why it's clled the wsher method? 5 3 Exmple: Wht is the volume of the figure formed ove? We will cll the outer rdius R, nd we will cll the inner rdius r. The height of the cylinder formed is just the width of the strip. Just like efore, if we hve nd infinitely thin strip, this distnce will e denoted dx (if it is verticl strip) nd dy (if it is horizontl strip). The volume of the solid formed y revolving region round the xis using the wsher method is given y π R r dx Exmple: Set up nd integrl, ut do not solve to find the volume of the solid generted y revolving the region ounded y the grphs of the equtions out the indicted lines. y= x ; y= ; x= ) the y xis ) the x xis c) the line y = 8 d) the line x = 15
16 7.3 Volumes Clculus Dy 4: Volumes of Solids of Revolution: The Shell Method We hve now used two different methods to find the volume of solid formed y revolving region out line. As with the disc nd wsher methods we will egin our discussion of the shell method y considering rectngle hving width w nd length h. The mjor difference etween the shell method nd the previous methods is tht the rectngle will e prllel to the xis of revolution. w h p : Let p e the distnce etween the xis of revolution nd the CENTER of the rectngulr strip. To find the volume of this figure, we cn proceed exctly like we did with the wsher method using the vlues of w, h, nd p. Exmple: If p is defined s ove to e the distnce etween the xis of revolution nd the center of the rectngulr strip, then wht is the rdius of the outer cylinder? Exmple: Wht is the volume of the outer cylinder? Exmple: Wht is the rdius of the inner cylinder? Exmple: Wht is the volume of the shell? If we were to rotte the figure elow round the line given, we could estimte the volume of the solid formed y finding the volume of the solid formed y rotting ech rectngulr strip nd dding these volumes together. If we were to consider infinitely mny strips, then ech strip would e so incredily thin tht the verge rdius, p, would e the distnce etween the xis of revolution nd the strip. The height of the shell formed y ech strip is just the length of the strip, h, nd the thickness of ech strip is given y dy (if the strip is horizontl) or dx (if the strip is verticl). dy h p : if the strip is dy, then oth p nd h must e written s functions of y 16
17 7.3 Volumes Clculus To find the volume of solid of revolution with the shell method, use one of the following: Horizontl Axis of Revolution Verticl Axis of Revolution V = π p( y) h( y) dy V = π p( x) h( x) dx dy h (y) p (y) dx h (x) Exmple: Let R e the region ounded y the grphs of y= x + 4, y = 8, nd x =, set up nd evlute the integrl tht gives the volume of the solid generted y revolving R out the y xis. ) Use the disc method p (x) ) Use the shell method 17
18 7.3 Volumes Clculus 1 Exmple: Let R e the region ounded y the grphs of y =, y =, x = 1, nd x = 4. In the lst exmple, we were x le to use oth the disc nd shell methods to rrive t the sme nswer. Explin why the volume formed y revolving R round the y xis is est found using the shell method insted of the disc nd wsher methods. Set up nd evlute the integrl tht gives the volume formed y revolving R round the y xis. 3 Exmple: Find the volume of the solid formed y revolving the region ounded y the grphs of y= x + x+ 1, y = 1, nd x = 1 out the line x =. Explin why it is necessry to use the shell method in this prolem. Exmple. Find the volume of the solid formed y revolving the region ounded y the grphs of y = x nd y= 4x x out the yxis. 18
19 7.3 Volumes Clculus π Exmple: [1973 AP Clculus AB #35] The region in the first qudrnt ounded y the grph of y = sec x, x =, 4 nd the xes is rotted out the xxis. Wht is the volume of the solid generted? A) π B) 1 4 π C) π D) π E) 8 π 3 Exmple: [1985 AP Clculus AB #45] The region enclosed y the grph of revolved out the yxis. The volume of the solid generted is y = x, the line x =, nd the xxis is A) 8 π B) 3 π C) 16 π D) 4π E) 16 π Exmple: [1985 AP Clculus BC #35] The region in the first qudrnt etween the xxis nd the grph of y = 6x x is rotted round the yxis. The volume of the resulting solid of revolution is given y 6 A) π( 6x x ) dx B) π ( 6x x ) dx C) π ( 6 ) 6 D) π( 3 9 y) dy E) π( ) y dy 6 x x x dx x Exmple: [1988 AP Clculus AB #3] A region in the first qudrnt is enclosed y the grphs of y = e, x = 1, nd the coordinte xes. If the region is rotted out the yxis, the volume of the solid tht is generted is represented y which of the following integrls? A) 1 x π xe dx B) 1 x π e dx C) 1 π e 4x dx D) e π yln y dy E) e π 4 ln ydy Exmple: [1988 AP Clculus AB #43] The volume of the solid otined y revolving the region enclosed y the ellipse x + 9y =9 out the xxis is A) π B) 4π C) 6π D) 9π E) 1 π 19
20 7.3 Volumes Clculus Exmple: [1988 AP Clculus BC #36] Let R e the region etween the grphs of y = 1 nd y = sin x from x = to π x =. The volume of the solid otined y revolving R out the xxis is given y A) π π xsinxdx B) π π C) π ( ) x cosxdx π 1 sinx dx D) π π π sin x dx E) π ( ) 1 sin x dx Exmple: [1993 AP Clculus AB #] Let R e the region in the first qudrnt enclosed y the grph of ( ) 1 3 y= x+ 1, the line x = 7, the xxis, nd the yxis. The volume of the solid generted when R is revolved out the yxis is given y 7 A) ( ) 3 π x + 1 dx B) ( ) 1 3 π x x+ 1 d C) ( ) 3 x π x dx 7 D) ( ) π x x+ 1 dx E) π ( y 1) dy Exmple: [1993 AP Clculus BC #19] The shded region R, shown in the figure elow, is rotted out the yxis to form solid who volume is 1 cuic inches. y y = kx x k x Of the following, which est pproximtes k? A) 1.51 B).9 C).49 D) 4.18 E)
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