Begin2.DVI

∫∞ 0 sinax^2 cos 2bx dx=^12 √ π 2 a ( cosb 2 a −sin b^2 a ) ∫∞ 0 cosax^2 cos 2bx dx=^12 √ π 2 a ( cosb 2 a + sin b^2 a ) ...

∫ 1 0 ln(1−x) x dx=− π^2 6 ∫ 1 0 (ax^2 +bx+c) ln^1 x 1 −xdx= (a+b+c) π^2 6 −(a+b)− a 4 ∫ 1 0 √lnx^1 1 −x^2 dx=π 2 ln ...

∫∞ 0 e−axsinxbxdx= tan−^1 ab ∫∞ 0 e−ax−e−bx x dx= ln b a ∫∞ 0 e−a^2 x^2 cosbx dx= √π 2 ae −b^2 / 4 a^2 ∫∞ 0 e−(a ...

∫∞ 0 x^4 e−x^2 cosax dx= √π 8 ( 3 − 3 a^2 +a 4 4 ) e−a^2 /^4 ∫∞ 0 (lnx x− 1 ) 3 dx=π^2 ∫∞ −∞ xsinrx dx (x−b)^2 +a^2 ...

∫∞ 0 sinhpx sinhqxdx= π 2 qtan( πp 2 q), |p|< q ∫∞ 0 coshax−coshbx sinhπx dx= ln ∣∣ ∣∣ ∣ cos 2 b cosa 2 ∣∣ ∣∣ ∣, −π & ...

Appendix D Solutions to Selected Problems Chapter 6 I6-1. (a)A~+B~= 9ˆe 1 +eˆ 2 + 3ˆe 3 (b) 6A~− 3 B~= 15ˆe 2 (c)A~+ 2B~= 15ˆe 1 ...

I6-4. By construction A~+^1 2 B~=C,~ B~+^1 2 A~=D,~ A~+E~=B~ All these vectors are coplaner so that there exists scalar constant ...

I6-7. (a) A~·A~=|A~||A~|cos 0 =|A~|^2 =C^2 (b) d dt (A~·A~) = d dt C^2 =⇒ A~·d A~ dt +d A~ dt ·A~= 0 =⇒ A~·d A~ dt = 0which show ...

I6-12. (c) 23 / 9 (d) 23 / 3 I6-13. (a) ˆeC=ˆe^2 √+ ˆ 2 e^3 also−ˆeC (b)− 1 / √ 3 I6-14. (b)A~·ˆeα=−cosα+ √ 3 sinα (c)α=π/ 6 or ...

I6-20. (d) (~r−~r 1 )·N~= 0 is equation of plane I6-21. T~=ˆe 1 −ˆe 2 is tangent to line. x= 3 +λ, y= 4−λ, z= 2 I6-22. N~ =12ˆe ...

I6-28. (a) If ˆe1 = cosα 1 ˆe 1 + cosβ 1 ˆe 2 + cosγ 1 ˆe 3 and ˆe 2 = cosα 2 ˆe 1 + cosβ 2 ˆe 2 + cosγ 2 ˆe 3 , then ˆe` 1 ·ˆe` ...

I6-35. (a) t 22 ˆe 1 +tˆe 2 −t 33 ˆe 3 +~c I6-36. ~a=d~vdt= costˆe 1 + sintˆe 2 ~v=~v(t) = sintˆe 1 −costˆe 2 +~c ~v(0) =−ˆe 2 + ...

I6-50. If dtd(B~×C~) =B~×d C~ dt + dB~ dt × C~, then d dt [ A~×(B~×C~) ] =A~× d dt( B~×C~) +dA~ dt×( B~×C~) =A~× [ B~×dC~ dt + d ...

I6-57. On circlex= cosθ, y= sinθ, dx=−sinθdθ, dy= cosθdθ ∫ C F~·d~r= ∫ C yz dx+ 2xdy+y dz = ∫ 2 π 0 −2 sin^2 θdθ+ 2 cos^2 θ dθ= ...

I6-68. If~r 0 is center of sphere ,~r 1 is point on sphere where tangent plane is con- structed and~ris a general point on the t ...

Chapter 7 I7-1. I7-2. I7-3. I7-4. ~r=αcosωteˆ 1 +αsinωtˆe 2 +βtˆe 3 d~r dt=−αωsinωtˆe^1 +αωcosωteˆ^2 +βˆe^3 ds dt =|d~r dt |= √ ...

I7-4 (Continued) κ^2 =dˆet ds ·dˆet ds = α (^2) ω 4 (α^2 ω^2 +β^2 )^2 Curvature κ= αω 2 α^2 ω^2 +β^2 radius of curvature ρ= 1 κ= ...

I7-7. d~r dt=~r ′, d~r ds ds dt= d~r dt, ds dt=s ′= (~r′·~r′) 1 / 2 d~r ds=ˆet= ~r′ s′, d^2 s dt^2 =s ′′= ~r′·~r′′ (~r′·~r′)^1 / ...

I7-10. d~r ds ds dt= d~r dt=~r ′ or ˆetds dt=~r ′, |~r′|=ds dt d dt ( ˆetds dt ) =d (^2) ~r dt^2 =~r′′ ˆetd (^2) s dt^2 +ds dt d ...

I7-14. Use the vector identityA~×(B~×C~) =B~(A~·C~)−C~(A~·B~)and show A~×(B~×C~) =B~(A~·C~)−C~(A~·B~) B~×(C~×A~) =C~(B~·A~)−A~(B ...