Reparameterize the curve r(t)= cos(t)i without (t)j (t)k by the arc length.



Answer to a math question Reparameterize the curve r(t)= cos(t)i without (t)j (t)k by the arc length.

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Para reparametrizar a curva \mathbf{r}(t) = \cos(t)\mathbf{i} + \sin(t)\mathbf{j} + t\mathbf{k} pelo comprimento de arco, primeiro precisamos calcular o comprimento de arco s(t) da curva.

O comprimento de arco s(t) é dado por:
s(t) = \int_{a}^{t} |\mathbf{r}'(\tau)| \ d\tau

Onde \mathbf{r}'(\tau) é a derivada de \mathbf{r}(t) em relação a \tau, e a é uma constante de integração que podemos escolher. Vamos escolher a = 0 para simplificar o cálculo.

A derivada de \mathbf{r}(t) em relação a t é dada por:
\mathbf{r}'(t) = -\sin(t)\mathbf{i} + \cos(t)\mathbf{j} + \mathbf{k}

Então, podemos calcular |\mathbf{r}'(t)|:
|\mathbf{r}'(t)| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + 1} = \sqrt{2}

Agora, podemos escrever a expressão para s(t):
s(t) = \int_{0}^{t} \sqrt{2} \ d\tau = \sqrt{2}t

Para reparametrizar a curva pelo comprimento de arco, precisamos inverter a fórmula anterior para resolver em termos de t. Temos:
t = \frac{s}{\sqrt{2}}

Agora, substituímos t novamente na expressão original da curva \mathbf{r}(t):
\mathbf{r}(s) = \cos\left(\frac{s}{\sqrt{2}}\right)\mathbf{i} + \sin\left(\frac{s}{\sqrt{2}}\right)\mathbf{j} + \frac{s}{\sqrt{2}}\mathbf{k}

Portanto, a curva \mathbf{r}(t) = \cos(t)\mathbf{i} + \sin(t)\mathbf{j} + t\mathbf{k} reparametrizada pelo comprimento de arco é:
\mathbf{r}(s) = \cos\left(\frac{s}{\sqrt{2}}\right)\mathbf{i} + \sin\left(\frac{s}{\sqrt{2}}\right)\mathbf{j} + \frac{s}{\sqrt{2}}\mathbf{k}

Resposta: \mathbf{r}(s) = \cos\left(\frac{s}{\sqrt{2}}\right)\mathbf{i} + \sin\left(\frac{s}{\sqrt{2}}\right)\mathbf{j} + \frac{s}{\sqrt{2}}\mathbf{k}

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