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微分積分II 補助演習問題 No. 9 解答例

1 曲線

r(t) =et i+e−t j+√

2t k, (0 ≤ t≤ 1) に対して,

∥r′(t)∥= √

e2t+ e−2t + 2 = et +e−t となるので, その弧長は

∫ 1 0

∥r′(s)∥ds =

∫ 1 0

(es+ e−s)ds= [

es−e−s ]1

0

= e−e−1.

2 C : r(t) = cost i+ sint j+t k (0≤ t ≤2π) に対して, I =

∫

C

y dx+z dy+ x dz =

∫ 2π 0

(y(t)x′(t) +z(t)y′(t) + x(t)z′(t))dt

=

∫ 2π 0

(sint(−sint) + tcost+ cost)dt = −π.

3 ベクトル場a = (cosy −ycosx) i+ (−xsiny −sinx+ 1) jを考える. (1) 曲線C : r(t) = (1−t)i (0 ≤ t≤ 2) に沿った線積分は

∫

C

a·dr =

∫ 2 0

cos 0 ×(−1)dt = −2

となる.

(2) ϕ(x, y) = xcosy −ysinx+ y に対して,

a = grad ϕ

が成り立つことに注意する. 従って, 点P(1,0,0)を始点とし, 点Q(−1,0,0)を終 点とするいかなる曲線Cに対しても,

∫

C

a·dr =

∫

C

grad ϕ ·dr = ϕ(Q)−ϕ(P) = (−1)−1 = −2 となる.

参照

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