By Nico M. Temme
This publication provides an advent to the classical, famous particular features which play a job in mathematical physics, specifically in boundary price difficulties. Calculus and complicated functionality thought shape the foundation of the ebook and various formulation are given. specific cognizance is given to asymptomatic and numerical features of certain capabilities, with a number of references to fresh literature supplied.
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Extra info for Special Functions: An Introduction to the Classical Functions of Mathematical Physics
Sample 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).