DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

ANSWERS TO SELECTED EXERCISES1.1.1 α(t) = ( 1−t 21+t 2 ,2t1+t 2 ).1.1.4 We parametrize the curve by α(t) = (t, f(t)), a ≤ t ≤ b, and so length(α) =∫ ba ‖α′ (t)‖ dt = ∫ ba√1+(f ′ (t)) 2 dt.1.1.6 β(s) = ( 12 (√ s 2 +4+s), 1 2 (√ s 2 +4− s), √ 2 ln(( √ s 2 +4+s)/2) ) .1.2.1 c. κ =12 √ 2 √ 1−s 21.2.3 a. T = 1 2 (√ 1+s, − √ 1 − s, √ 12), κ =2 √ 2 √ , N =1/√ 2( √ 1 − s, √ 1+s, 0), B =1−s 212 (−√ 1+s, √ 1 − s, √ 12), τ =2 √ 2 √ ; c. T = √ √1(t, √1−s 2 2 1+t1+t 2 , 1), κ = τ =21/2(1 + t 2 ), N = √ 1 (1, 0, −t), B = √ √1(−t, √1+t 2 2 1+t1+t 2 , −1)21.2.5 κ =1/ sinh t (which we see, once again, is the absolute value of the slope).1.2.6 B ′ =(T×N) ′ = T ′ ×N+T×N ′ =(κN)×N+T×(−κT+τB) =τ(T×B) =τ(−N),as required.1.2.9 b. If all the osculating planes pass through the origin, then there are scalar functionsλ and µ so that 0 = α + λT + µN for all s. Differentiating and using the Frenetformulas, we obtain 0 = T + κλN + λ ′ T + µ ( −κT + τB ) + µ ′ N; collecting terms, wehave 0 = ( 1+λ ′ − κµ ) T + ( κλ + µ ′) N + µτB for all s. Since {T, N, B} is a basisfor R 3 ,weinfer, in particular, that µτ =0. (We could also just have taken the dotproduct of the entire expression with B.) µ(s) =0leads to a contradiction, so wemust have τ =0and so the curve is planar.1.2.11 We have α ′ × α ′′ = κυ 3 B, so α ′ × α ′′′ = (α ′ × α ′′ ) ′ = (κυ 3 B) ′ = (κυ 3 ) ′ B +(κυ 3 )(−τυN), so (α ′ × α ′′′ ) · α ′′ = −κ 2 τυ 6 . Therefore, τ = α ′ · (α ′′ × α ′′′ )/(κ 2 υ 6 ),and inserting the formula of Proposition 2.2 gives the result.1.2.23 a. Consider the unit normal A s,t to the plane through P = 0, Q = α(s), andR = α(t). Choosing coordinates so that T(0) = (1, 0, 0), N(0) = (0, 1, 0), andB(0) = (0, 0, 1), we apply Proposition 2.6 to obtainα(s) × α(t) =st(s − t) (−κ212 0 τ 0 st + ...,2κ 0 τ 0 (s + t)+...,−6κ 0 +2κ ′ 0 (s + t) − κ3 0 st + ...) ,α(s) × α(t)so A s,t =→ A =(0, 0, −1) as s, t → 0. Thus, the plane through P‖α(s) × α(t)‖with normal A is the osculating plane.114