Respuesta :
Answer:
a.
[tex]\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}[/tex]
[tex]\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}[/tex]
[tex]\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}[/tex]
b.
[tex]\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}[/tex]
Step-by-step explanation:
Given that:
C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)
a. Find a piecewise smooth parametrization of the path C.
r(t) = { 0
If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),
Then:
[tex]C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)[/tex]
Also:
[tex]\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }[/tex]
[tex]\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}[/tex]
[tex]\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }[/tex]
[tex]\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}[/tex]
[tex]\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }[/tex]
[tex]\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}[/tex]
b Evaluate :
Integral of (x+2y^1/2)ds
[tex]\mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \int \limits ^1_{0} \ (t + 0) \sqrt{1} } \\ \\ \mathtt{ \int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \begin {pmatrix} \dfrac{t^2}{2} \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_{c1} (x+ 2 \sqrt{y}) ds = \dfrac{1}{2}}[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits (x+2 \sqrt{y} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 \ dt } }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \int \limits 2- t + 2\sqrt{t-1} \ \sqrt{1+1} }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} \int \limits^2_1 2- t + 2\sqrt{t-1} \ dt }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2t - \dfrac{t^2}{2}+ \dfrac{2(t-1)^{3/2}}{3} (2) \end {pmatrix} ^2_1}[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{1}{2} (4-1)+\dfrac{4}{3} (1)^{3/2} -0 \end {pmatrix} }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \sqrt{2} } \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ \sqrt{2} }{6} \ (11 )}[/tex]
[tex]\mathtt{\int \limits _{c2} (x+ 2 \sqrt{y}) ds = \dfrac{ 11 \sqrt{2} }{6}}[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 0+2 \sqrt{3-t} \ \sqrt{0+1} }[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits ^3_2 2 \sqrt{3-t} \ dt}[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \int \limits^3_2 \begin {pmatrix} \dfrac{-2(3-t)^{3/2}}{3} (2) \end {pmatrix}^3_2 }[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}[/tex]
[tex]\mathtt{\int \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}[/tex]
[tex]\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}[/tex]
[tex]\mathtt{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}[/tex]
[tex]\mathbf{\int \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}[/tex]
