Let $\mathbf { F } = \nabla f$, where $f ( x , y ) = \sin ( x - 6 y )$
Which of the following equations does the line segment from $( 0,0 )$ to $( 0 , \pi )$ satisy?
Select the correct answer.
A) $\int _ { C } \mathbf { F } \cdot d \mathbf { r } = 1$
B) $\int _ { C } \mathbf { F } \cdot d \mathbf { r } = 0$
C) none of these

Question

Quizplus · Accepted Answer

First, we compute $\mathbf{F}=
abla f =(\cos(x-6y),-6\cos(x-6y))$.
Then, we parameterize the line segment from $(0,0)$ to $(0,\pi)$ as $\mathbf{r}(t)=(0,t)$ for $0\leq t\leq \pi$.
We can now evaluate the line integral as follows:
\begin{align*}
\int_C\mathbf{F}\cdot d\mathbf{r}&=\int_0^{\pi} \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt\
&=\int_0^{\pi}(\cos(0-6t),-6\cos(0-6t))\cdot (0,1)\,dt\
&=\int_0^{\pi}-6\cos(0-6t)\,dt\
&=6\int_0^{\pi}\cos(6t)\,dt\
&=6\left[\frac{1}{6}\sin(6t)ight]_0^{\pi}\
&=0.
\end{align*}
Therefore, the line integral is equal to 0, which means that the line segment from $(0,0)$ to $(0,\pi)$ satisfies $\int_C\mathbf{F}\cdot d\mathbf{r}=0$. Therefore, the correct answer is B.

Let F=∇f\mathbf { F } = \nabla fF=∇f , Where f(x,y)=sin⁡(x−6y)f ( x , y ) = \sin ( x - 6 y )f(x,y)=sin(x−6y)

Let $\mathbf { F } = \nabla f$ , Where $f ( x , y ) = \sin ( x - 6 y )$