Determine o triedro de Frenet, a curvatura e a torção em cada instante da curva:
alpha: R & R^3 & beta: R & R^3 t & mapsto(sen 4 t, 3 t, cos 4 t), & t & mapsto(raiz2 t, e^t, e^-t)

Passo 1

A questão pede o triedro de Frenet (T, N, B), a curvatura κ e a torção τ para duas curvas em ℝ³. Vamos usar as fórmulas:
– Vetor tangente unitário: T=r'|r'|
– Curvatura: kappa=|r'times r''||r'|^3
– Vetor normal: N=(r'times r'')times r'|r'times r''|,|r'|
– Binormal: B=r'times r''|r'times r''|
– Torção: tau=(r'times r'')cdot r'''|r'times r''|^2

Passo 2

Para alpha(t)=(sin4t,3t,cos4t): calcule alpha'=(4cos4t,3,-4sin4t) com velocidade |alpha'|=5. Logo
Talpha=15(4cos4t,;3,;-4sin4t)

Passo 3

Ainda em alpha:
alpha''=(-16sin4t,0,-16cos4t). Então
T'alpha=15(-16sin4t,0,-16cos4t),|T'alpha|=165
Curvatura
$$\kappa_\alpha=\frac{|T’_\alpha|}{|\alpha’|}=\frac{16/5}{5}=\frac{16}{25}.
kappaalpha=|T'alpha||alpha'|=16/55=1625 N_\alpha=\frac{T’\alpha}{|T’\alpha|}=(-\sin4t,0,-\cos4t).
$$

Passo 4

Ainda em alpha:
Binormal pelo produto vetorial
$$B_\alpha=T_\alpha\times N_\alpha=(-\tfrac{3}{5}\cos4t,;\tfrac{4}{5},;\tfrac{3}{5}\sin4t).
Balpha=Talphatimes Nalpha=(-tfrac35cos4t,;tfrac45,;tfrac35sin4t) çãó\tau_\alpha=\frac{(\alpha’\times\alpha’’)\cdot\alpha’’’}{|\alpha’\times\alpha’’|^2}=\frac{12}{25}.
$$

Passo 5

Para beta(t)=(raiz2,t,,e^t,,e^-t):
beta'=(raiz2,e^t,-e^-t) e |beta'|=2cosh t. Logo
$$T_\beta=\frac{1}{2\cosh t}(\sqrt2,;et,;-e{-t}).
$$

Passo 6

Ainda em beta:
beta''=(0,e^t,e^-t). O produto beta'timesbeta''=(2, -raiz2e^-t, raiz2e^t) tem norma 2raiz2cosh t. Curvatura
$$\kappa_\beta=\frac{2\sqrt2\cosh t}{(2\cosh t)3}=\frac{\sqrt2}{4\cosh2t}.
kappabeta=2raiz2cosh t(2cosh t)^3=raiz24cosh^2t áN_\beta=(-\tanh t,;\tfrac{1}{\sqrt2\cosh t},;\tfrac{1}{\sqrt2\cosh t}).
$$

Passo 7

Ainda em beta:
Binormal
$$B_\beta=\frac{\beta’\times\beta’’}{|\beta’\times\beta’’|}=(\tfrac{1}{\sqrt2\cosh t},;-\tfrac{e^{-t}}{2\cosh t},;\tfrac{e^t}{2\cosh t}).
Bbeta=beta'timesbeta''|beta'timesbeta''|=(tfrac1raiz2cosh t,;-tfrace^-t2cosh t,;tfrace^t2cosh t) çã\tau_\beta=\frac{(\beta’\times\beta’’)\cdot\beta’’’}{|\beta’\times\beta’’|2}=-\frac{\sqrt2}{4\cosh2t}.
$$

Resposta

Para α(t):
T_α(t)=\frac{1}{5}(4\cos4t,3,-4\sin4t),
N_α(t)=(-\sin4t,0,-\cos4t),
B_α(t)=(-\tfrac{3}{5}\cos4t,\tfrac{4}{5},\tfrac{3}{5}\sin4t),
κ_α=\tfrac{16}{25},
τ_α=\tfrac{12}{25}.
Para β(t):
T_β(t)=\frac{1}{2\cosh t}(\sqrt2, e^t, -e^{-t}),
N_β(t)=(-\tanh t,\tfrac{1}{\sqrt2\cosh t},\tfrac{1}{\sqrt2\cosh t}),
B_β(t)=(\tfrac{1}{\sqrt2\cosh t}, -\tfrac{e^{-t}}{2\cosh t}, \tfrac{e^t}{2\cosh t}),
κ_β=\tfrac{\sqrt2}{4\cosh^2t},
τ_β=-\tfrac{\sqrt2}{4\cosh^2t}.

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