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Sa sk The length of the curve (x(t), y(t)) as t varies from t0 to t1 is given by L= t=t1 t=t0 DEO PAT- ET RIE n U Arc Length of versity ni atc he w a 2 2 (x (t)) + y (t) dt If we wish to ﬁnd the length of a curve which is the graph of a function y = f (x), as x runs from a to b, we let x(t) = t, y(t) = f (x(t)) = f (x) and we get x (t) = 1, and y (t) = f (x(t))x (t) = f (x), so we have a simple formula for the length: L= x=b x=a 2 1 + (f (x)) dx = b a 2 1 + (f (x)) dx = b Similarly, if we have a curve x = g(y) with y running from c to d we get L= y=d y=c Example: 2 1 + g (y) dy = d c 1 + g (y) dy = Consider the curve given by x(t) = cos t, y(t) = sin t , 0 ≤ t ≤ π . Its length is L= t=π t=π t=0 t=0 x (t)2 + y (t)2 dt t=π = (− sin t)2 + (cos t)2 dt = t=0 1dt = t|π 0 = π 1 2 a 2 1 + y dx d c 1 + (x )2 dy 2 1 a b 3 1 2 2 x 1+ dx = 2 a b 9 1 + xdx = 4 a 0 3 b 9 2 (1 + x) 4 4 = 3 9 2 a b 32 b 32 b 3 9 8 4 + 9x 1 2 1+ x = = 27 (4 + 9x) = 4 27 4 a a sk 3 b 2 We have L = 1 + y dx = 8 27 Sa 3 y = x2, 0 ≤ a ≤ x ≤ b y a 3 3 1 (4 + 9b) 2 − (4 + 9a) 2 27 2 0 1 2 x DEO PAT- ET RIE n U Find the length of the curve of Example: versity ni atc he w a x = y 2, 0 ≤ c ≤ y ≤ b 1 0 x -1 0 -1 1 2 3 4 -2 y=d y=d y=d 2 2 We have L = 1 + (x ) dy = 1 + 2y dy = 1 + 4y 2 dy y=c y=c y=c Making the substitution y = 12 tan θ, we have 1 dy = 2 sec2 θdθ, 1 + 4y 2 = 1 + tan2 θ = sec2 θ, θc = arctan 2c when y = c and θd = arctan 2d when y = d, so L= 1 2 θ=θd θ=θc sec2 1 1 θ sec2 θdθ = 2 2 θ=θd θ=θc sec3 θdθ = θ=θd 1 = (sec θ tan θ + ln | sec θ + tan θ|) 2 θ=θc 1 1 [sec θd tan θd + ln | sec θd + tan θd |] − [sec θc tan θc + ln | sec θc + tan θc |] = 4 4 1 1 1 + 4d2 2d + ln | 1 + 4d2 + 2d| − 1 + 4c 2 2c + ln | 1 + 4c 2 + 2c| = 4 4 √ √ √ d 1 + 4d2 − c 1 + 4c 2 1 1 + 4d2 + 2d + ln √ 2 4 1 + 4c 2 + 2c Note that if we let c = 0, we get the formula for the distance along the parabola to the point (d2 , d): √ d 1 + 4d2 1 L= + ln( 1 + 4d2 + 2d) 2 4 √ and if we let b = d2 , we get the (equivalent) formula for the distance from (0,0) to (b, b): √ √ b 1 + 4b 1 L= + ln 1 + 4b + 2 b 2 4 3 sk DEO PAT- ET RIE n U 2 Find the length of the curve Sa Example: of y versity ni atc he w a U 2 1 2 x +2 , 0≤x ≤3 3 Example: 1 2 Find the length of the curve y = Sa 12 13 2 x + 2 2x = x x 2 + 2 , so 32 2 2 2 1+ y = 1 + x 2 (x 2 + 2) = x 2 + 2x 2 + 1 = x 2 + 1 , and 3 3 3 3 x 2 2 L= + x = x + 1dx = (x 2 + 1) dx = 3 0 0 0 03 33 +3 − + 0 = 12 3 3 y = 1 2 sk 6 3 x 0123 3 4 3 2 x 3 − x 3 + 5, 1 ≤ x ≤ 8 y 4 8 16 Solution: 12 1 1 3 4 1 3 2 −1 1 x3 − x 3 = x 3 − x − 3 , so 43 83 4 2 2 2 1 1 1 1 1 1 8 1 + y = 1 + x 3 − x− 3 = x 3 + x − 3 , and 4 4 4 2 8 8 8 2 3 3 4 1 1 1 1 x 1 1 1x L= x 3 + x − 3 dx = x 3 + x − 3 dx = 4 + 2 = 4 4 4 3 1 1 3 y = 8 3 4 3 2 3 4 3 2 3 4 3 2 3 + 3 3 + 3 L = x3 + x3 8 8 1 1 = − = 4 8 1 4 8 4 8 3 4 3 2 3 3 3 9 2 + 2 − + = 12 + − = 4 8 4 8 2 8 12 4 3 8 1 0 x 012345678 PAT- ET RIE atc he w 9 0 DEO n 3 Find the length of the curve y = y Solution: of Example: versity ni a