Question

-Questão original em anexo:

A integral de linha $\int\limits^._C \frac{3x y^{2} }{20} ds,$ tal que ,$C$ é o segmento de reta cujas equações parametricas são: x=3t-1; y=4t, com $0 \leq t \leq 1,$ vale?


a. 4,67
b. 6,33
c. 5,00
d. 1,67
e. 2,33

Answer 1

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Calcular a integral de linha de uma função sobre uma curva parametrizada:

     $\mathsf{\displaystyle\int_C \dfrac{3xy^2}{20}\,d\overrightarrow{\mathbf{s}}}$


sendo  C  o segmento de reta descrito pelas equações paramétricas:

     $\mathsf{C:~~}\left\{ \!\begin{array}{l} \mathsf{x=3t-1}\\ \mathsf{y=4t} \end{array} \right.\qquad\quad\mathsf{0\le t\le 1.}$


Reescreva a integral de linha, substituindo  x  e  y  pelas expressões dadas nas equações paramétricas,  onde  ds  é a notação para o módulo do vetor tangente à curva:

     $\mathsf{\displaystyle\int_C \dfrac{3xy^2}{20}\,d\overrightarrow{\mathbf{s}}}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3xy^2}{20}\cdot \left\|C'(t)\right\|dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3xy^2}{20}\cdot \left\|\frac{d}{dt}\left\langle x,\,y\right\rangle\right\|dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3\cdot (3t-1)\cdot (4t)^2}{20}\cdot \left\|\frac{d}{dt}\left\langle 3t-1,\,4t\right\rangle\right\|dt}$

     $=\mathsf{\displaystyle\int_0^1 \frac{3\cdot (3t-1)\cdot 16t^2}{20}\cdot \left\|\left\langle 3,\,4\right\rangle\right\|dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3\cdot (3t-1)\cdot 16t^2}{20}\cdot \sqrt{9+16}\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3\cdot (3t-1)\cdot 16t^2}{20}\cdot \sqrt{25}\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{3\cdot (3t-1)\cdot 16t^2}{20}\cdot 5\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \frac{240\cdot (3t-1)\cdot t^2}{20}\,dt}\\\\\\ =\mathsf{\displaystyle 12\int_0^1 (3t-1)\cdot t^2\,dt}$

     $=\mathsf{\displaystyle 12\int_0^1 (3t^3-t^2)\,dt}\\\\\\ =\mathsf{12\cdot \left(\dfrac{3t^{3+1}}{3+1}-\dfrac{t^{2+1}}{2+1}\right)\bigg|_0^1}\\\\\\ =\mathsf{12\cdot \left(\dfrac{3t^4}{4}-\dfrac{t^3}{3}\right)\bigg|_0^1}\\\\\\ =\mathsf{12\cdot \left[\left(\dfrac{3\cdot 1^4}{4}-\dfrac{1^3}{3}\right)-\left(\dfrac{3\cdot 0^4}{4}-\dfrac{0^3}{3}\right)\right]}\\\\\\ =\mathsf{12\cdot \left(\dfrac{3}{4}-\dfrac{1}{3}\right)}\\\\\\ =\mathsf{12\cdot \left(\dfrac{9}{12}-\dfrac{4}{12}\right)}\\\\\\ =\mathsf{12\cdot \dfrac{5}{12}}\\\\\\ =\mathsf{5}\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Resposta:  alternativa  c.  5,00.


Bons estudos! :-)