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mrs_skeptik [129]
3 years ago
7

Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

Mathematics
1 answer:
horsena [70]3 years ago
5 0

Answer:

a.

\mathbf{r_1 = (t,0)  \implies  t = 0 \ to \ 1}

\mathbf{r_2 = (2-t,t-1)  \implies  t = 1 \ to \ 2}

\mathbf{r_3 = (0,3-t)  \implies  t = 2 \ to \ 3}

b.

\mathbf{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}

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:

C_1 = (0,0) \\ \\  C_2 = (1,0) \\ \\ C_3 = (0,1)

Also:

\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }

\mathbf{r_1 = (t,0)  \implies  t = 0 \ to \ 1}

\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }

\mathbf{r_2 = (2-t,t-1)  \implies  t = 1 \ to \ 2}

\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }

\mathbf{r_3 = (0,3-t)  \implies  t = 2 \ to \ 3}

b Evaluate :

Integral of (x+2y^1/2)ds

\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}}

\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 } }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds = \int  \limits 2- t + 2\sqrt{t-1}  \ \sqrt{1+1}  }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2} \int  \limits^2_1  2- t + 2\sqrt{t-1} \ dt }

\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}

\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} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} 2 -\dfrac{3}{2} + \dfrac{4}{3} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} \dfrac{12-9+8}{6} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =  \sqrt{2}  }  \ \begin {pmatrix} \dfrac{11}{6} \end {pmatrix} }

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =   \dfrac{ \sqrt{2}  }{6} \  (11 )}

\mathtt{\int  \limits _{c2} (x+ 2 \sqrt{y}) ds =   \dfrac{ 11 \sqrt{2}  }{6}}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds =  \int  \limits ^3_2 0+2 \sqrt{3-t}   \ \sqrt{0+1} }

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds =  \int  \limits ^3_2 2 \sqrt{3-t}   \ dt}

\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 }

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [(0)-(1)]}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = -\dfrac{4}{3} [-(1)]}

\mathtt{\int  \limits _{c3} (x+ 2 \sqrt{y}) ds = \dfrac{4}{3}}

\mathtt{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}}{6}+\dfrac{1}{2}+ \dfrac{4}{3}}

\mathtt{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+3+8}{6}}

\mathbf{\int  \limits _{c} F \ dr =\dfrac{11 \sqrt{2}+11}{6}}

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