Get A Second Course in Calculus PDF

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By Harley Flanders

ISBN-10: 148323102X

ISBN-13: 9781483231020

Moment direction in Calculus

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A ) . = (0,0,0). )*) = Wyx — Vzx + UZy — WXy>+ VXZ — UyZ = 0. 26. If / = / ( p ) , then g r a d / = /'(p) (p*, py, pz) = p" 1 /'(p)x. By the same formula, applied to p - 1 /'(p), and by Ex. 19, d i v [ g r a d / ( P ) ] = divCp-Tx] = grad(p" 1 /0 x + p" 1 /' divx = [p- 1 (p- 1 /0 , x]-x + 3p-T = p(p" 1 / , ) / + 3 p - r = p-Kpf" - / ' ) + 3P"1/' = / " + 2p~y = p-i(pf)" (or p - W ) ' ) . Section 5, page 306 2. 6, 2, - 6 4. -12, - 4 , -12 6. Vs 8. 6V2 Section 6, page 309 2. F = grad(#32/22), 1 4.

There is none. 18. Yes. By the Mean Value Theorem, \e-«2 - eri2\ = 20e"*2|s - t\, where 6 is between s and t, for s > 0, t > 0. By elementary calculus, 26e~d2 < \/2/e< 1 for 6 > 0, hence \e~s2 - e~t2\ < \s - t\ for s, t > 0. Set s = |x|, t = |y|. Then |e"ixi2 - e-i y ' 2 | < I |x| - |y| I < |x - y|, etc. of 20 Chapter 6 20. The composite of continuous functions (x,x) >f(x9x). is continuous: x > Section 4, page 212 16. the straight line x = (3, t, 9t) 18. The graph has no interior points since each line parallel to the 2-axis intersects the graph in at most one point.

Let g(x) be the piecewise linear function with vertices (zi, 2/1), (0,0), (x2,2/2), and let h(x) be the linear function with vertices (xi, 2/1), (x2,2/2). Then either g(x) < h(x) for all X\ < x < x2 or h(x) < g{x). Thus the area is rx2 (h — g) dx / ro h - J xi J xi r%% g - J x\ g «/0 = li(2/i + 2/2) (x2 - xi) + \xxyi — \x2y2\ = \\x\y% - x22/i|The argument is easily modified in case xi = 0 or x2 = 0. 6. A diagonal splits the parallelogram into two triangles of equal area, so we can use 5. Let the vertices of the base be Xi = (xh 2/1, 0) and x2 = (x2) y2) 0) and another vertex be x3 = (xs,2/3, 0).

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A Second Course in Calculus by Harley Flanders


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