# Is s 4 ln 2 1 3 ln 2 2 t dt ds

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Unformatted text preview: ′ (t) dt = ds = 1 1 1 = 2t + (t ≥ 1). The t t 2 1 2t + dt = t2 + ln | t | = t 1 = 2t + 1 4. As suggested in the question, we can put A = (0, 0, 0, 0), C = (0, 0, 0, Ct), on the t–axes, and B = (Bx , 0, 0, Bt), in the xt– plane. We now parametrize the paths by γ AB (λ) = ′ (1 − λ)(0, 0, 0, 0) + λ(Bx , 0, 0, Bt ) = (λ Bx , 0, 0, λ Bt), 0 ≤ λ ≤ 1, so (γ AB ) (λ) = (Bx , 0, 0, Bt); γ BC (λ) = (1 − λ)(Bx , 0, 0, Bt) + λ(0, 0, 0, Ct) = (1 − λ) Bx , 0, 0, Bt + ′ λ(Ct − Bt ) , 0 ≤ λ ≤ 1, so (γ BC ) (λ) = (−Bx , 0, 0, Ct − Bt ), and γ AC (λ) = (1 − ′ λ)(0, 0, 0, 0) + λ(0, 0, 0, Ct) = (0, 0, 0, λ Ct), 0 ≤ λ ≤ 1, so (γ AC ) (λ) = (0, 0, 0, Ct). 1 11 1 2 2 2 2 Now the proper times are τAB = −Bx + c2 Bt dλ = −Bx + c2 Bt λ = c0 c 0 1 1 1 1 2 2 2 2 −Bx + c2 Bt , τBC = −Bx + c2 (Ct − Bt )2 dλ = −Bx + c2 (Ct − Bt )2 ) c c0 c 11 1 1 and τAC = c2 Ct2 dλ = c2 Ct2 = cC t . c0 c c MATB42H page 3 Solutions # 3 To show that τAB + τBC &lt; τAC we need to show that 2 2 2 −Bx + c2 Bt + −Bx + c2 (Ct − Bt )2 ) &lt; cCt 2 2 2 −Bx + c2 (Ct − Bt )2 ) &lt; cCt − −Bx + c2 Bt which holds ⇐⇒ 2 2 2 2 2 ⇐⇒ −Bx + c2 (Ct − Bt )2 &lt; c2 Ct2 − Bx + c2 Bt − 2cCt −Bx + c2 Bt 2 2 2 2 2 2 ⇐⇒ −Bx + c2 Ct2 − 2c2 Ct Bt + c2 Bt &lt; c2 Ct2 − Bx + c2 Bt − 2cCt −Bx + c2 Bt 2 2 ⇐⇒ −2c2 Ct Bt &lt; −2cCt −Bx + c2 Bt 2 2 ⇐⇒ cBt &gt; −Bx + c2 Bt which is trivially true. Hence τAB + τBC &lt; τAC . 3 5. (a) f ds = γ t7 t5 + 7 5 3 f γ (t) 0 3 0 3 t2 γ ′ (t) dt = 2 t2 + 1 dt = t6 + t4 dt = 0 0 2187 243 12636 = + = . 7 5 35 2 2 1 (b) f ds = f (γ (t)) γ (t) dt = (t + 8t) 1 + dt = t γ 1 1 2 2 2 3 45 t =. +t =9 4− 9 (t + 1) dt = 9 2 2 2 1 1 2 ′ 9t 1 + 1 1 t dt = (c) We ﬁrst note that γ = γ 1 + γ 2 where γ 1 (t) = (2t, t), 0 ≤ t √ 1 and γ 2 (t) = ≤ (t + 1, 5 − 4t), 1 &lt; t ≤ 3. Now γ 1 ′ (t) = (2, 1), γ 1 ′ (t) = 5 and γ 2 ′ (t) = √ √ f ds = (1, −4), γ 2 ′ (t) = 1 + 16 = 17. Since γ is piecewise C 1 , we have γ 1 f ds = f ds + f (γ 2 (t)) γ 2 ′ (t) dt = f (γ 1 (t)) γ 1 ′ (t) dt + γ2 γ1 γ2 γ1 1 3 √ √ √ √ t dt + 17 2t) 5 dt + (5 − 4t) − (t + 1) 17 dt = − 5 0 1 √ 3 √ √ t2 1 √ 5 t2 5 + 17 − + 4t = − − 12 17. −5 20 2 2 1 1 (−5t + 4) dt = f (γ (t)) γ ′ (t) dt (sin3 t + t + cos2 t) √ 2 dt = γ 1 0 =0 2π &gt;  2π √ t2 √ :   =0 t 1 1  =0 :  2  t = cos dt = 2 + sin −  t sin t cos + t + +  2t 2 2 2 2 20 √0 2 π 2π + 1 . f ds = (t − 3 2π 2 (d) 0 = π 6. We can parametrize the quarter circle with γ (t) = 3 cos t, 3 sin t , 0 ≤ t ≤ . Hence, 2 √ √ γ ′ (t) = − 3 sin t, 3 cos t and γ ′ (t) = 9 sin2 t + 9 cos2 t = 9 = 3. If the density function is ρ, the mass m is given by m = ρ ds. Here the density is given γ π 2 π 2 by ρ(x, y ) = x + y so the mass is m = 0 9 1 − (−1) = 18. 3 cos t + 3 sin t (3) dt = 9 sin t − cos t = 0 MATB42H Solutions # 3 page 4 7. Using Example 7 on page 146 of the text with R = 3, we have the path vt vt , 3 − 3 cos and the velocity is given by γ (t) = v t − 3 sin 3 3 vt vt , v sin . The velocity is horizontal when the y – γ ′ (t) = v − v cos 3 3 vt vt component, v sin = 0. This occurs whenever is an integer multiple of π . 3 3 If it is an even multiple, the x–component is also zero and γ (t) also has a zero y – 3π component, so the critical points occur when t = 2k + 1 for integers k . The v speed is (v − v cos(2k + 1)π, 0) = (v + v, 0) = 2 v . (For even multiples, the point touches the ground and the speed is 0.)...
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