推 asdf322505:感謝大大的幫忙 05/08 09:41
※ 引述《asdf322505 ()》之銘言：
: → → → →
: r =e^t cos2ti+e^t sin2tj+ e^tk 在t=拍/4 之 曲率 k(s)
: ans:√5/3(e^-拍/4)
: 這提該怎嚜算?
．
→ → → →
r(t) = e^t * (cos2t - 2sin2t)i + e^t * (sin2t + 2cos2t)j + e^t k
..
→ → → →
r(t) = e^t * (-3cos2t - 4sin2t)i + e^t * (4cos2t - 3sin2t)j + e^t k
． ..
→ → → →
r × r = e^2t * (4sin2t - 2cos2t)i + e^2t * (-2sin2t - 4cos2t)j
+ e^2t * [10 * (cos2t)^2 + 10 * (sin2t)^2]
當 t = π/4 時
．
→ → → →
r(π/4) = e^(π/4) * (-2 i + j + k )
．
→
=> ｜r(π/4)｜ = √6 * e^(π/4)
． ..
→ → → → →
r × r (π/4) = e^(π/2) * (4i - 2j + 10k )
． ..
→ →
=> ｜r × r (π/4)｜ = 2√30 * e^(π/2)
． ..
→ →
｜r × r (π/4)｜ √5
故 κ = -------------------- = ----- * e^(-π/4)
． 3
→ 3
｜r｜
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