Let \(\displaystyle \mathbf{F} = \bigtriangledown f\), where \(\displaystyle f(x, y) = \sin(x - 3y)\). Find curves \(\displaystyle C_1\) and \(\displaystyle C_2\) that are not closed and satisfy the equation.
(a)
\(\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = 0\)
\(\displaystyle C_1: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1\)
(b)
\(\displaystyle \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = 1\)
\(\displaystyle C_2: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1\)
I know that \(\displaystyle \mathbf{F} = \bigtriangledown f = \ <f_x, \ f_y> \ = \ <\cos(x - 3y), \ -3\cos(x - 3y)>\) and \(\displaystyle d\mathbf{r} = \ <dx, \ dy>\)
\(\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy\) = 0
What should I do now?
(a)
\(\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = 0\)
\(\displaystyle C_1: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1\)
(b)
\(\displaystyle \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = 1\)
\(\displaystyle C_2: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1\)
I know that \(\displaystyle \mathbf{F} = \bigtriangledown f = \ <f_x, \ f_y> \ = \ <\cos(x - 3y), \ -3\cos(x - 3y)>\) and \(\displaystyle d\mathbf{r} = \ <dx, \ dy>\)
\(\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy\) = 0
What should I do now?