Math F412: Homework 4 Solutions February 20, κ I = s α κ α

Transcription

1 All prts of this homework to be completed in Mple should be done in single worksheet. You cn submit either the worksheet by emil or printout of it with your homework. 1. Opre Let α be not-necessrily unit speed curve. Let I α be its involute. Show tht κ I = κ 2 α + τ 2 α s α κ α. Recll tht I(t) = α(t) s α (t)t α (t) where T α is the unit tngent to α. forth, ll symbols such s s, κ, etc. will men the ssocited quntities for α, nd we will drop the explicit dependence on t. Using the Frenet equtions nd the fct tht s = ν we compute I = α s T st = νt νt sνκn = sνκn. I = sνκ. From our expression for I nd the Frenet equtions gin we compute I = (sνκ) N (sνκ)n = (sνκ) N (sνκ)ν( κt + τb). Reclling tht N N =, T N = B nd N B = T we hve I I = (sνκ) 2 ν(κb + τt). Since B nd T re perpendiculr nd unit length, I I = (sνκ) 2 ν κ 2 + τ 2. s dvertised. κ I = I I = (sνκ2 )ν κ2 + τ 2 = I 3 (sνκ) 3 For the unit circle we hve κ = 1 nd τ = so κ2 + τ 2 sκ κ I = 1 s.

2 If we did this computtion by directly, we would strt with the formul I(s) = (cos(s), sin(s)) s( sin(s), cos(s)). Then I (s) = ( sin(s), cos(s)) ( sin(s), cos(s)) s( cos(s), sin(s)) = s(cos(s), sin(s)) nd the unit tngent is the signed unit norml is T(s) = (cos(s), sin(s)). N s,i = ( sin(s), cos(s)). Also, where ν I (s) = I (s) = s. Now T I (s) = κ s,i (s)ν I (s)n s (s) κ s,i (s)ν(s) = 1 for ll s. s dvertised. T I (s) = ( sin(s), cos(s)) = N s (s). κ s,i (s) = 1 ν(s) = 1 s 2. Opre (Mple ok, but not required) Show tht the evolute of the ellipse (α(t) = cos(t), b sin(t)) is the steroid ( 2 b 2 cos 3 (t), b2 2 sin 3 (t)). b 2

3 3. Opre (You ll hve to red the prgrph just prior to understnd the question.) Let β be unit speed cylindricl helix with xis u. Let h(s) = (β(s) β()) u. Let γ(s) = β(s) h(s)u. Show tht κ γ = κ b et/ sin 2 θ where θ is the constnt ngle such tht β (s) u = T u = cos(θ). We compute nd lso γ (s) = β (s) h (s)u However, T u = (T u) = (cos(θ)) =. Now = β (s) (β (s) u)u = T(s) (T(s) u)u. γ (s) = T (s) (T (s) u)u. γ (s) = T (s) = κ(s)n(s). γ (s) 2 = γ (s) γ (s) = β (s) 2 2(β (s) u) 2 since u u = 1. Since β (s) u = T u = cos(θ) we hve Also, γ (s) 2 = (1 cos 2 θ) = sin 2 θ. γ γ = (T (T u)u) (κn) = κt N (T u)u N = κb κ cos(θ)(u N). Now we cn write u = (u T)T + (u N)N + (u B)B = cos(θ)t + N + (u B)B. Since u hs unit length, we must hve (u B) 2 + cos(θ) 2 = 1 nd hence (u B) = ± sin(θ). So nd using the fct tht T N = B nd N B = T. So u = cos(θ)t ± sin(θ)b u N = cos(θ)b sin(θ)t γ γ = κ(1 cos(θ) 2 )B κ sin(θ) cos(θ)t = κ sin(thet) 2 B κ sin(θ) cos(θ)t. Since B nd T re orthonorml, γ γ 2 = κ 2 sin(θ) 4 + κ 2 sin(θ) 2 cos(θ) 2 = κ 2 sin(θ) 2. κ γ = γ γ / γ 3 = κ sin(θ) sin(θ) 3 = κ sin(θ) 2. 3

4 4. Opre Show β is circulr helix if nd only if tu nd κ re constnt. Suppose β is circulr helix, so γ is circle nd κ γ is constnt. Since κ β = κ γ sin(θ) 2, we conclude tht κ β is constnt. Since τ β = (τ β /κ b et)κ β is product of constnts, it is lso constnt. Suppose β is curve such tht κ nd τ re both constnts. Then the rtio τ/κ is constnt, so β is cylindricl helix. Then κ γ = κ β / sin(θ) 2 is constnt. γ is circle. 5. Write procedure in Mple tht tkes two rguments, curve α(t) nd the nme of the prmeter (t) nd returns list contining the following: [ν(t), κ(t), τ(t), T(t), N(t), B(t)]. The procedure should ttempt to crete expressions tht re s simple s possible. Demonstrte tht your procedure works by pplying it to the circulr helix α(t) = ( cos(t), sin(t), bt). 6. Use your procedure from the previous problem to ssist you in nswering Opre nd In clss we stted Green s theorem: if α is smooth, simple, positively oriented, closed plne curve nd V = (P, Q) is smooth vector field, int(α) Q x P y dxdy = α P dx + Q dy. The theorem ctully holds for piecewise C 1 curves, nd the point of this exercise is to show tht it holds when the domin is polygon. Demonstrte tht the theorem holds in the cse where α prmeterizes tringle. Then explin why Green s theorem holds for rbitrry polygons. Consider the cse where the vector field V is perpendiculr to one side of the tringle T. After pplying trnsltion nd rottion we cn ssume tht V = (P, ), one vertex of the tringle lies t the origin, nd one side of the tringle is the verticl line between (, y ) nd (, y 1 ) where y < y 1 nd >. Let S, S 1 nd S 2 denote the bottom, right, nd top sides of this tringle. We cn prmeterize S by α (t) = (t, ty /) for t. Then α V α = 4 P(t, ty /) dt.

5 The right side cn be prmeterized by α 1 (t) = (, y + t(y 1 y )) for t 1. Clerly α1 V α 1 =. The top side cn be prmeterized by α 2 (t) = ( t, ( t)y 1 /). Then α2 V α = P( t, ( t)y 1 /) dt = P(t, ty 1 /) dt. Combining the pth integrls into single pth round the boundry we find α V α = = P(t, ty /) P(t, ty 1 /) dt ty 1 / ty / y P(t, y) dy dt = y P(x, y) dy dx T = Q x P y dy dx. T This proves the result for vector fields tht re perpendiculr to one side. For vector fields tht re prllel to one side, we cn cut the tringle from one vertex in line perpendiculr to the vector field to decompose the tringle into two tringles, ech with side perpendiculr to the vector field. The result then follows by dding the contributions of the sides nd interiors, noting tht there is cncelltion on the common side of the two smller tringles. The result holds for rbitrry tringles nd vector fields by linerity. The result holds for polygons by decomposition rgument similr to the one outlined bove for cutting single tringle into two subtringles. 5