1. Seja \gamma: I \rightarrow \mathbb{R}^{2} uma curva derivável até a 2 .^{\text {a }} ordem, com \left\|\gamma^{\prime}(t)\right\| \neq 0 no intervalo I. Seja
    s=\int_{t_{0}}^{t}\left\|\gamma^{\prime}(u)\right\| d u, t \in I, \operatorname{com} t_{0} fixo em I. Sejam, ainda, \vec{T}(t)=\frac{\gamma^{\prime}(t)}{\left\|\gamma^{\prime}(t)\right\|} o versor de \gamma^{\prime}(t) e \vec{t}(s) dada por \vec{t}(s)=\vec{T}(t), onde t=t(s). Mostre que
    a) \frac{d \vec{T}}{d t}(t)=\frac{\gamma^{\prime \prime}(t)\left\|\gamma^{\prime}(t)\right\|^{2}-\gamma^{\prime}(t)\left(\gamma^{\prime \prime}(t) \cdot \gamma^{\prime}(t)\right)}{\left\|\gamma^{\prime}(t)\right\|^{3}}.
    b) \frac{\overrightarrow{d t}}{d s}(s)=\frac{\gamma^{\prime \prime}(t)\left\|\gamma^{\prime}(t)\right\|^{2}-\gamma^{\prime}(t)\left(\gamma^{\prime \prime}(t) \cdot \gamma^{\prime}(t)\right)}{\left\|\gamma^{\prime}(t)\right\|^{4}}, t=t(s).
    c) \left\|\frac{\overrightarrow{d t}}{d s}(s)\right\|_{\text {onde } t=t(s) .}=\sqrt{\frac{d \vec{t}}{d s}(s) \cdot \frac{d \vec{t}}{d s}(s)}=\frac{\sqrt{\left(\left\|\gamma^{\prime \prime}(t)\right\|\left\|\gamma^{\prime}(t)\right\|\right)^{2}-\left(\gamma^{\prime \prime}(t) \cdot \gamma^{\prime}(t)\right)^{2}}}{\left\|\gamma^{\prime}(t)\right\|^{3}},

Passo 1

Resumindo o enunciado, temos uma curva \gamma:I\to\mathbb{R}^2 duas vezes diferenciável com \|\gamma'(t)\|\neq0. Definimos o versor tangente
\vec T(t)=\frac{\gamma'(t)}{\|\gamma'(t)\|},
e o parâmetro de comprimento de arco
s=\int_{t_0}^t\|\gamma'(u)\|\,du,
de modo que, em termos de s, \vec t(s)=\vec T\bigl(t(s)\bigr).
Para resolver precisamos aplicar a regra do quociente em vetores, a regra da cadeia e usar
\frac{d}{dt}\|\gamma'(t)\|=\frac{\gamma''(t)\cdot\gamma'(t)}{\|\gamma'(t)\|}.

Passo 2

Queremos \dfrac{d\vec T}{dt}(t). Usando a regra do quociente em \vec T(t)=\gamma'(t)\,/\,\|\gamma'(t)\|, temos
\frac{d\vec T}{dt} =\frac{\|\gamma'\|\,\gamma''-\gamma'\,\frac{d}{dt}\|\gamma'\|}{\|\gamma'\|^2}.
Como
\frac{d}{dt}\|\gamma'(t)\| =\frac{\gamma''(t)\cdot\gamma'(t)}{\|\gamma'(t)\|},
segue
\frac{d\vec T}{dt} =\frac{\|\gamma'\|\,\gamma''-\gamma'\,\bigl(\gamma''\cdot\gamma'/\|\gamma'\|\bigr)}{\|\gamma'\|^2} =\frac{\gamma''\,\|\gamma'\|^2-\gamma'\,(\gamma''\cdot\gamma')}{\|\gamma'\|^3}.

Passo 3

Para obter \dfrac{d\vec t}{ds}(s) usamos que \vec t(s)=\vec T\bigl(t(s)\bigr). A regra da cadeia dá
\frac{d\vec t}{ds} =\frac{d\vec T}{dt}\,\frac{dt}{ds}.
Mas ds/dt=\|\gamma'(t)\|, logo dt/ds=1/\|\gamma'(t)\|. Assim
\frac{d\vec t}{ds} =\frac{1}{\|\gamma'\|}\,\frac{d\vec T}{dt} =\frac{\gamma''\,\|\gamma'\|^2-\gamma'\,(\gamma''\cdot\gamma')}{\|\gamma'\|^4},
avaliado em t=t(s).

Passo 4

Por fim, para a norma de \dfrac{d\vec t}{ds} usamos a identidade \|A-\lambda B\|^2=\|A\|^2-2\lambda(A\cdot B)+\lambda^2\|B\|^2, mas aqui reconhecemos diretamente
\bigl\|\gamma''\,\|\gamma'\|^2-\gamma'\,(\gamma''\cdot\gamma')\bigr\|^2 =\|\gamma'\|^4\!\bigl(\|\gamma''\|^2\|\gamma'\|^2-(\gamma''\cdot\gamma')^2\bigr).
Dividindo por \|\gamma'\|^8 e tirando a raiz fica
\left\|\frac{d\vec t}{ds}\right\| =\frac{\sqrt{\bigl(\|\gamma''\|\|\gamma'\|\bigr)^2-(\gamma''\cdot\gamma')^2}}{\|\gamma'\|^3}.

Resposta

a) \displaystyle \frac{d\vec T}{dt}(t) =\frac{\gamma''(t)\,\|\gamma'(t)\|^2-\gamma'(t)\,\bigl(\gamma''(t)\cdot\gamma'(t)\bigr)}{\|\gamma'(t)\|^3}. b) \displaystyle \frac{d\vec t}{ds}(s) =\frac{\gamma''(t)\,\|\gamma'(t)\|^2-\gamma'(t)\,\bigl(\gamma''(t)\cdot\gamma'(t)\bigr)}{\|\gamma'(t)\|^4},\quad t=t(s). c) \displaystyle \left\|\frac{d\vec t}{ds}(s)\right\| =\frac{\sqrt{\bigl(\|\gamma''(t)\|\|\gamma'(t)\|\bigr)^2-\bigl(\gamma''(t)\cdot\gamma'(t)\bigr)^2}}{\|\gamma'(t)\|^3}.

Exercícios relacionados