We divide the interval [a, b] into subintervals of equal length x = b a n

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1 Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl: We divide the intervl [, b] into subintervls of equl length x = b n. Let s i be the length of the stright line connecting the two points (x i 1, y i 1 ) nd (x i, y i ), the concept of limit tells us to define the length of the curve L to be: n L = lim s i n i=1 According to the Pythgoren theorem, s i = ( x) + ( y i ), where y i = f(x i ) f(x i 1 ) From the Men Vlue Theorem we know tht in ech intervl [x i 1, x i ], there exists x i such tht f (x i ) = f(x i) f(x i 1 ) x i x i 1 The eqution for s i becomes: s i = ( x) + ( y i ) = f(x i ) f(x i 1 ) = f (x i )(x i x i 1 ) y i = f (x i ) x ( x) + (f (x i ) x) = ( x) + (f (x i )) ( x)

2 [ = 1 + (f (x i ))] ( x) = 1 + [f (x i )] x Tking n gives: n L = lim s i = lim n n i=1 n i=1 1 + [f (x i )] x The limit on the right is Riemnn, which by definition, is represented by b 1 + [f (x)] We hve obtined the following formul: Arc Length Formul: Suppose f is function of x nd tht f (x) is continuous on [, b], then the length of the grph of f(x) from to b is given by: b L = 1 + [f (x)] Using the dy/ nottion, letting f (x) = dy, we cn write the bove formul s: b [ ] dy L = 1 + Notice the expression involved inside the rdicl is the derivtive of the originl function, not the originl function. Also notice tht the rdicl is involved.

3 Exmple: Find the length of the curve y = 1 x for 1 x 1 Ans: Using the formul for rc-length, nd the fct tht dy = the integrl: [ x L = 1 + = 1 x ] = x 1 x = 1 1 x = rcsin(x) ] 1 1 = / ( /) = 1 x 1 x 1 1 x You my hve noticed tht the curve is the upper hlf of the unit circle. we setup

4 Exmple: Find the length of the curve y = x, 0 x 1 Ans: dy = x, we get the integrl: L = = [x] = [ ln ( 1 + ) ] x = x 1 + x + 1 ln ( x x ) ] 1 [ ln ( 0 + 1) ] = + ln(1 + ) 0 In the formul for rc length, if we let x b, nd define x s(x) = 1 + [f (t)] dt This is function whose vlue t x represents the rc length of the curve f from to x. We cll this the rc length function. The Fundmentl Theorem of Clculus tells us tht: [ ] ds dy = 1 + [f (x)] = 1 + If we look t ds,, dy s differentils, then we get the eqution:

5 ds = 1 + [ ] dy If we squre both sides nd simplify we get the symmetric form for the differentil of the rc length function in dy nd : (ds) = () + (dy)

6 Are of Surfce of Revolution: Given continuous function f(x) between x = nd x = b, we wnt to find the re of the surfce of revolution obtined by rotting the grph of f bout the x xis. We first strt by finding the lterl surfce re of cone: r l r θ l Imgine tht we cut the cone long lterl side nd ly it flt, the re of the resulting plne is prt of circle. The re of this piece is clculted by: ( ) ( θ r ) A = l = l l = lr The lterl surfce re of cone with bse rdius of r nd slnt height l is defined to be: A = rl We now wnt to find the lterl re of truncted cone. The lterl re of truncted cone is the lterl re of the bigger cone subtrcted the lterl re of the smller cone.

7 l 1 r 1 l r In the picture bove, the lterl re of the lrger cone is: A = r (l + l 1 ) The lterl re of the smller (top) cone is: A 1 = r 1 (l 1 ) Therefore, the re of the truncted cone is given by A = A A 1 = r (l + l 1 ) r 1 l 1 = [(r r 1 )l 1 + r l ] Using similr tringles we see tht l 1 = l 1 + l r 1 r cross multiply gives: r l 1 = r 1 l 1 + r 1 l (r r 1 )l 1 = r 1 l The formul for A now simplifies to: A = A A 1 = r (l + l 1 ) r 1 l 1 = [(r r 1 )l 1 + r l ] = (r 1 l + r l) If we let r = r 1 + r be the verge rdius of the top nd bottom circle of the truncted cone, then the formul for the lterl surfce re of truncted cone becomes: A = rl We cn now derive formul for generl surfce of revolution.

8 In the picture bove, if we let s i be the distnce between the points P i 1 nd P i, then the formul for the surfce re of truncted cone we just derived tells us tht the re of the green strip is given by: ( ) f(xi 1 ) + f(x i ) A i = s i From the derivtion of the formul for the rc length, we hve s i = where x i is number inside the intervl [x i 1, x i ]. 1 + [f (x i )] x, If f is continuous, then for smll x, f(x i 1 ) f(x i ) nd f(x i) f(x i ), so the formul for the re of the strip becomes: ( f(xi 1 ) + f(x i ) A i = = f(x i ) 1 + [f (x i )] x ) s i Adding up the re of ll the strips gives: n A n f(x i ) 1 + [f (x i )] x i=1 ( f(x i ) + f(x i ) ) 1 + [f (x i )] x Our intuition tells us tht this pproximtion will be exct if we tke n, so we hve: n S = lim f(x i ) 1 + [f (x n i )] x i=1 This limit is nother Riemnn sum which by definition is the definite integrl:

9 b S = f(x) 1 + [f (x)] Formul for Surfce of Revolution: The re of the surfce obtined by rotting the curve y = f(x), x b, bout the x xis, is given by: b S = f(x) 1 + [f (x)] Exmple: Find the re of the surfce obtined by rotting the curve y = 4 x, x, round the x xis. Ans: By the formul, f(x) = 4 x nd f (x) = x, we hve: 4 x S = [ 4 x x 1 + = 4 x ] 4 x 1 + x 4 x = ] 4 4 x 4 x = () = 4x = 8 ( 8) = 16 You my hve noticed tht the curve is the upper hlf of the circe with rdius, nd the surfce generted is sphere with rdius.

10 E.g. Find the re of the surfce obtined by rotting the curve y = sin(x), 0 x 1 bout the x xis. Ans: f(x) = sin(x), so f (x) = cos(x), we hve: S = sin(x) 1 + [ cos(x)] = sin(x) 1 + () cos (x) 0 Using the substitution u = cos(x) we get: du = ( sin(x)) 1 du = sin(x). Also, when x = 0, u =, nd when x = 1, u =, the bove integrl chnges to: = 1 + u du Switching the upper nd lower limits of integrtion nd switching the sign gives: = 1 + u du Using tble or integrtion by tri-substitution gives: ( = u 1 + u + 1 ( ln u + ) ] ) 1 + u = [ 1 + () + 1 ( ln + ) ( 1 + () = [ 1 + () + 1 ( ln + ) 1 + () + = [ 1 + () ( ) + 1 ln ( ( ) ) )] 1 + () 1 ln ( () ) ] ( ln ( + ) 1 + () ln ( + )) ] 1 + () Using property of logrithm gives: = [ ( 1 + () + 1 ln + )] 1 + () () Rtionlize the denomintor inside ln gives: ( = 1 + () ) 1 + () ( + ) 1 + () ln ( + ) 1 + () ( + ) 1 + () [ 1 + () + 1 ln ( () ) ] = = [ 1 + () + 1 ( ln + ) ] 1 + ()

11 = [ 1 + () + ln ( + )] 1 + () Distribute gives clener looking expression: = 1 + () + ( ln + ) 1 + ()

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