8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before


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1 .1 Arc Length hat s the length of a curve? How can we approxmate t? e could do t followng the pattern we ve used before Use a sequence of ncreasngly short segments to approxmate the curve: As the segments get smaller (and the number of segments gets larger,) t seems reasonable that the approxmaton wll get closer to the actual length. But how long s each segment? P "#$ P " = (x " x "#$ ) + + (y " y "#$ ) + = ( x) + + ( y) + By the Mean Value Theorem, we know that, for every nterval, there exsts some x * such that f(x ) f(x 1 ) = f (x * ) (x x 1 ) or y = f (x * ) x so
2 P 1 P = ( x) + + ( y " ) + = ( x) + + [f 2 x " x] + = 1 + [f 2 x " ] + x So L = 1 + [f 2 x ] + dx Ex: Fnd the length of y 2 = x 3 between the ponts (1, 1) and (4, ). hat does the graph look lke? y = x 3/2 :; :< = = + x1/2 so :; :< + = >? x so L =? $ 1 + >? x dx
3 Sometmes, nstead of smply measurng a partcular arc, we want to set up an arc length functon that wll gve us a formula for the length n general. Ex 4: Fnd the arc length functon for the curve y = x 2 1/ lnx, wth P 0 (1,1) as the startng pont. Fnd f (x)= Then fnd 1 + [f (x)] 2 = So 1 + [f 2 x ] + = hch means that the arc length, s(x), s gven by < $ 1 + [f 2 x ] + = hat s the arc length along the curve from (1, 1) to (2, f(2))?
4 .2 Area of a Surface of Revoluton hat s a surface? Lateral surface of a crcular cylnder: Lateral surface area of a cylnder: A = 2πrh Lateral surface of a cone: Lateral surface are of a cone: A = πrl (from an earler result)
5 Surfaces of Revoluton If the surface s more complcated than a cone or a cylnder, how can we calculate ts surface area? Frst, mage the surface created when you revolve a curve around the xaxs: Then, magne breakng the curve down nto smaller peces, whch are revolved around the xaxs to make bands. hat s the area of ths band? The trck s to look at that band as beng smply a pece of a rght, crcular cone. On page 546, Stewart explans why the surface area of ths band s
6 A = 2πrl where l s the slant length (or wdth) of the band. So the surface has area S = S = S = 2πrl dx 2πf(x)(arc length) dx 2πf(x) 1 + [f 2 x ] + dx Ex: The arc of the parabola y = x 2 from (1, 1) to (2, 4) s rotated about the yaxs. Fnd the area of the resultng surface. The book gves 2 solutons one n x and one n y. e ll go through the frst one. S = 2πx ds + $ S = 2πx 1 + :; :< + dx y = x 2 :; :< = 2x
7 S = + 2πx 1 + 4x + dx $ e can use a u substtuton to evaluate the ntegral: u = du = (Look at Page 54 for the other verson of the soluton.)
8 .3 Applcatons Physcs and Teetertotters Queston: If you have an adult and a chld who want to play on a teetertotter, how should they st f the adult s much heaver than the chld? In fact, Archmedes dscovered that the teetertotter wll balance f m 1 d 1 = m 2 d 2 (Law of the Lever) where d 1 and d 2 are measured from the center of the teetertotter (more generally called the fulcrum): Each of the products m d s a measure of how much each person s pushng down on the end of ther sde of the teetertotter. Ths s called the torque, whch s one type of moment. Because the force s beng appled to the end of a lever connected to a sngle pont, the torque s a rotatonal force.
9 If, nstead of a lne, we have a thn plate of any gven shape, then we can talk about the center of mass as beng the pont where the plate wll balance horzontally, as shown: e can also consder the center of mass for our teetertotter example t would le at the fulcrum of the teetertotter. Le the teetertotter along the xaxs, where m 1 les at x 1 and m 2 les at x 2, and label the center of mass as x. Then d 1 = x  x 1 and d 2 = x 2  x e know: m 1 d 1 = m 2 d 2 so then also m 1 (x  x 1 ) = m 2 (x 2  x) and then m 1 x + m 2 x = m 1 x 1 + m 2 x 2 whch means that x = M NO N P M Q O Q M N P M Q
10 If you have a system wth n masses at dfferent ponts along the xaxs, you can extend ths result further: x = T SUN R S < S T SUN R S = T SUN R S < S R where m = "X$ m " s the total mass of the system and M = "X$ m " x " s the moment of the system about the orgn whch makes the equaton above nto mx = M Smlarly, we can extend these results to masses arranged on the plane to get M y = "X$ m " x " the moment of the system about the yaxs and M x = "X$ m " y " the moment of the system about the xaxs wth the correspondng equatons x = Y Z and y = Y [ R R
11 Ex: Fnd the moments and center of mass of the system of objects that have masses 3, 4, and at the ponts (1, 1), (2, 1), and (3, 2), respectvely. e need to fnd M y and M x M y = "X$ m " x " = 3(1) + 4(2) + (3) = 2 M x = "X$ m " y " and then we need to fnd x and y m = "X$ m " = so x = Y Z R = y = Y [ R = So, now that we have a formula for the center of mass of a lne segment, can we extend ths to fnd the center of mass for the thn plate of a gven shape (whch s also known as the centrod of the regon R)? To make t easer, assume the plate has a unform densty, p. (Hand wavng) As before, we ll break the regon down nto rectangles that are f(x \ ) by x, where the centrod of the th rectangle s gven by (x \, $ + f(x \)), where x \ s the mdpont of the subnterval n the x drecton.
12 The mass of each rectangle s then m = p f(x \ ) x (Densty * Area) and the moment of the th subnterval about the yaxs s gven by p f(x \ ) x * x \, whch means that the moment of the regon as a whole about the yaxs s: M y = lm b "X$ p x \ f(x \ ) x = p x f x dx Smlarly, we can fnd M x, whch s gven by M x = lm p $ "X$ f(x b + \) + x = p $ f x + dx + Fnally, we can calculate x and y: x = Y Z = f h < g < :< R h f g < :< = h < g < :< h g < :< = $ j x f x dx y = Y [ R = f hn Q g < Q :< h f g < :< = hn Q g < Q :< h = $ g < :< j $ f x + dx +
13 Ex #6: Fnd the centrod of the regon bounded by the lne y = x and the parabola y = x 2. Step 1: Fnd the area of the regon A = (x x + ) dx Step 2: Calculate x and y x = $ j x f x g(x) dx y = $ j $ + f x + g x + dx
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