Instructor's solutions for Mathematical methods for physics by Riley, Hobson.

By Riley, Hobson.

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1 + t2 Substitution of these gives 1 π/2 (1 + sin x)1/2 dx = 0 1+ 0 2t 1 + t2 1/2 2 dt 1 + t2 1 2 + 2t dt (1 + t2 )3/2 1 2 1 dt + 2 − 2 3/2 (1 + t ) (1 + t2 )1/2 = 0 = 0 1 . 0 To evaluate the first integral we turn it back into one involving sinusoidal functions and write t = tan θ with dt = sec2 θ dθ. Then the original integral becomes π/2 π/4 2 sec2 θ 1 dθ + 2 1 − √ 3 sec θ 2 0 π/4 √ 2 cos θ dθ + 2 − 2 = 0 √ π/4 = 2[ sin θ ]0 + 2 − 2 √ √ = 2 − 0 + 2 − 2 = 2. (1 + sin x)1/2 dx = 0 38 PRELIMINARY CALCULUS An alternative evaluation can be made by setting x = (π/2) − y and then writing 1 + cos y in the form 2 cos2 (y/2).

Using first the half-angle identities, we have J= = 2 sin2 x2 1 − cos x dx = 1 + cos x 2 cos2 x2 x x x sec2 − 1 dx = 2 tan − x + c. 7) is 1 − t2 1 + t2 2 dt 1 − t2 1 + t2 1+ 1 + t2 2 2t dt 1 + t2 2 2 dt − dt 1 + t2 1− J= = = = 2t − 2 tan−1 t + c = 2 tan x − x + c. 2 (c) This integrand, containing only sinusoidal functions, can be converted to an algebraic one by writing t = tan(x/2) and expressing the functions appearing in the integrand in terms of it, 1 − t2 2t 2 2 1 + t2 1 + t2 cos x sin x 1 + t dx = dt 1 + cos x 1 − t2 1+ 1 + t2 2 2t(1 − t ) = dt (1 + t2 )2 A B = 2t + dt, 2 2 (1 + t ) 1 + t2 with A + B(1 + t2 ) = 1 − t2 , implying that B = −1 and A = 2.

A a y =a Therefore the curvature of the catenary at the point (x, y) is given by 1 = ρ x 1 cosh a a 2 x 1 + sinh a 3/2 x 1 cosh a a = = 2. x 3 a cosh y a To obtain this result we have used the identity cosh2 z = 1 + sinh2 z. We see that the curvature is maximal when y is minimal; this occurs when x = 0 and y = a. The maximum curvature is therefore 1/a. 20. 3). In this system a curve is defined by r, the distance from a fixed point O to a general point P of the curve, and p, the perpendicular distance from O to the tangent to the curve at P .

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