By Einar Hille, John T. Anderson, Gordon D. Prichett, S. L. Salas
This is often the scholar complement for the 3rd version of S.L. Salas' "Calculus of 1 and several other Variables" textual content.
Uploader comment -- even though it's nonetheless readable, this was once released prior to TeX used to be round, so the typesetting is atrocious.
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Additional info for Salas Student Supplement for Calculus
Example text
We shall later be interested in those vector fields V(x), for which the tangential line integral only depends on the initial and end points of the curve K. ) We have here an example in which this ideal property is not satisfied. 7. 7) We get K V(x) · dx = π = π 2 π = π 2 {sin4 t + cos t + 3 cos2 t + 3 cos3 t + cos4 t}dt sin4 t + cos4 t + 2 cos2 t · sin2 t − π 2 π = (sin2 t + cos2 t)2 − π 2 = t− = {−y 3 dx + x3 dy} {− sin3 t · (− sin t) + (1 + cos t)3 cos t}dt π = K 1 3 3 sin2 2t + cos t + + cos 2t + 3 cos3 y dt 2 2 2 3 1 1 + cos 4t + cos t + cos 2t + 3 cos t − 3 sin2 t cos t dt 4 4 2 1 3 3 t + sin 4t + sin t + t + sin 2t + 3 sin t − sin3 t 4 16 2 4 1− 1 3 + 4 2 π π 2 9π π −4+1= − 3.
D Find r (t) . 5 1 Figure 47: The curve K. I It follows from r (t) = (t2 − 1, t2 + 1, 2t), that r (t) 2 = (t2 −1)2 +(t2 +1)2 +4t2 = 2t4 +2+ 4t2 = 2(t2 + 1)2 , hence 1 (K) = −1 r (t) dt = 2 1 √ 0 2 √ 2(t + 1) dt = 2 2 1 +1 3 √ 8 2 = . 9 A space curve K is given by the parametric description √ r(t) = 6t2 , 4 2 t3 , 3t4 , t ∈ [−1, 1]. Explain why the curve is symmetric with respect to the (X, Z)-plane. Then find the arc length of K. A Arc length. D Replace t by −t. Then find r (t). com 54 Calculus 2c-7 Arc lengths and parametric descriptions by the arc length –4 3 2 –2 1 0 2 1 2 4 3 4 5 6 Figure 48: The curve K.
5 for a = 1 and b = 2. Alternatively we get by using the parametric description (x, y) = a (cos t, sin t), t ∈ [0, 2π], that K V(x) · dx = 2π = 0 2π = 0 =− K x+y y−x dx + 2 dy x2 + y 2 x + y2 a2 {(cos t+sin t)(−sin t)+(sin t−cos t) cos t}dt a2 {−cos t · sin t−sin2 t+cos t · sin t−cos2 t}dt 2π 0 dt = −2π. 5) Here K V(x) · dx = π 2 = 0 = 0 π 2 K {(x2 − y 2 )dx − (x + y)dy} {(a2 cos2 t−b2 sin2 t)(−a sin t)−(a cos t+b sin t)b cos t}dt {−a[(a2 +b2 ) cos2 t−b2 ] sin t−ab cos2 t−b sin t cos t}dt 1 1 1 ab = +a(a +b ) cos3 t−ab2 cos t− (t+ sin 2t)− b2 sin2 t 3 2 2 2 2 2 π 2 0 a(a2 + b2 ) ab π b2 a b − + ab2 = (2b2 − a2 ) − (2b + aπ).