By Harley Flanders

ISBN-10: 148323102X

ISBN-13: 9781483231020

Moment direction in Calculus

**Read Online or Download A Second Course in Calculus PDF**

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**Extra resources for A Second Course in Calculus**

**Sample text**

8. 0, for the same reason: the function is odd with respect to the line y = x. 10. V5/2 12. f 18. | ( 6 4 In 2 - 15) 0, by symmetry in the line x = y 14. 20. 0, same reason as in Ex. 8 16. ff }e(e - l)(c* - 1) Section 3, page 380 2. 12. iln5 4. (1-e-1)2 6. 0 8. f 10. 2 0 if n is odd, 2/(n + 1) (n + 2) if n is even 13. Geometrically, the contribution f(x, y) dx dy is cancelled by the con tribution / ( — x9 —y) dxdy. Analytically, / / = / _ \ ( / - \ / f o 2/) ^ ) Likewise JJ = Jjf(-X> d * = / - 1 ( / - i ^ > -2/) d » ) 2/) da; efy.

1 4 1 4 1 4 (2" ' , 2 ' , 2- / ) 4. (±1,0,0) 6. 2 8. Each plane z = b cuts the paraboloid in an ellipse. If b > 1, the point (1, 0, 4), directly above (1, 0, 0), is closer to (1, 0, 0) than is any point on this ellipse; if 0 < 6 < 1, the point ( 2 V o , 0, b) where the ellipse meets the (first quadrant of) the x, z-plane is closest. Minimize f(b) = ( 2 v 6 — l ) 2 + b2 for 0 < b < 1. Section 6, page 35 7 2. min 4. neither 6. neither 8. neither 10. neither 12. min 14. none 16. none 18. min 20. neither 22.

A = *, fl = 1 14. Solve the two linear systems by Cramer's Rule: FzZx + Fwwx = -Fx FZZy + FwWy = Gzzx + Gwwx = —Gx GZZy -{- GWWy = -Fy Gy. There is a unique solution whenever the determinant FZGW — FWGZ at (x, y) z(x, y), w(x, y)) is not 0. Section 8, page 318 2. 4. 6. dz = (-c 2 /z) (x dx/a2 + y dy/b2) dz = {e~xy - xye-yz)-l\_(yze~xv — e~yz) dx + zx(e~xy + e~yz) dy~] dz = p dx + q dy, dp = r dx + s dy, dq = s dx + t dy 8. /(l - X2)/x2Z 10. 1. 5. ) SAMPLE TEST 1. Show that /(x, 1/) = \xy\* is differentiable at 0 by finding a suitable linear approximation.

### A Second Course in Calculus by Harley Flanders

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