determine o comprimento de arco da curva paramétrica.
x=\frac{1}{2} t, y=\frac{1}{3}(1-t)^{3 / 2},
z=\frac{1}{3}(1+t)^{3 / 2} ;-1 \leq t \leq 1

Passo 1

O enunciado pede o comprimento de arco da curva paramétrica
x(t)=\tfrac12\,t,\quad y(t)=\tfrac13\,(1-t)^{3/2},\quad z(t)=\tfrac13\,(1+t)^{3/2},
para -1\le t\le1.
O conteúdo é comprimento de arco de curva em \mathbb R^3, e vamos usar a fórmula
S=\int_a^b\sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2+\bigl(z'(t)\bigr)^2}\,dt.

Passo 2

Calculamos as derivadas:
x'(t)=\tfrac12,
y'(t)=\tfrac13\cdot\tfrac{3}{2}(1-t)^{1/2}\cdot(-1)=-\tfrac12\,(1-t)^{1/2},
z'(t)=\tfrac13\cdot\tfrac{3}{2}(1+t)^{1/2}=+\tfrac12\,(1+t)^{1/2}.

Passo 3

Somamos os quadrados das derivadas:
\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2+\bigl(z'(t)\bigr)^2 =\Bigl(\tfrac12\Bigr)^2 +\Bigl(-\tfrac12(1-t)^{1/2}\Bigr)^2 +\Bigl(\tfrac12(1+t)^{1/2}\Bigr)^2 = \tfrac{1}{4}+\tfrac{1-t}{4}+\tfrac{1+t}{4} = \tfrac{3}{4}.

Passo 4

A raiz é constante:
\sqrt{\bigl(x'\bigr)^2+\bigl(y'\bigr)^2+\bigl(z'\bigr)^2} =\sqrt{\tfrac{3}{4}}=\tfrac{\sqrt{3}}{2}.
Logo
S=\int_{-1}^{1}\tfrac{\sqrt{3}}{2}\,dt =\tfrac{\sqrt{3}}{2}\,(1-(-1)) =\tfrac{\sqrt{3}}{2}\cdot2 =\sqrt{3}.

Resposta

\sqrt{3}

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