Let $F(x; y) = (-g(t)y; g(t)x)$, where $t = x^{2} + y^{2}$ and $g(t)$ is a function (of single variable) which is differentiable (class $C^1$) for $t > 0$.

I have to calculate the line integral on the unit circle, $F \cdot dr$, counter clockwise.

How does the information that $g(t)$ is differentiable help us ?

Thanks in advance

  • $\begingroup$ you may use Stokes theorem, for which differentiability is needed $\endgroup$ – vidyarthi Nov 26 '16 at 9:14
  • $\begingroup$ thanks but we are not supposed to do that because we did not go through it yet , this is on the problem set of green's theorem , line integrals and conservative vector fields. $\endgroup$ – Kasmir Khaan Nov 26 '16 at 9:16
  • $\begingroup$ Then use Green's theorem with $\vec{r}=x\bar{i}+y\bar{j}$ $\endgroup$ – vidyarthi Nov 26 '16 at 9:18
  • $\begingroup$ Can you please show me how ? i got stuck when i did that , because i end up taking double integral of 2 g(t) witch depend on the function g . $\endgroup$ – Kasmir Khaan Nov 26 '16 at 9:20

Hint. If $(x,y)$ is on the unit circle centred at the origin then $$\mathbf{F}(x; y) = (-g(t)y; g(t)x)=g(1)(-y,x).$$ Hence $$\int_{C}\mathbf{F}\cdot d\mathbf{r}=g(1)\int_0^{2\pi}\left(-\sin t,\cos t\right) \cdot\left(-\sin t,\cos t\right)dt.$$ P.S. If $g(1)\not=0$ then the vector field $\mathbf{F}$ is not conservative in a simply connected region which contains the origin. To be conservative we need that $$-g'(t)t_y\cdot y-g(t)=(-g(t)y)_y=(g(t)x)_x=g'(t)t_x\cdot x+g(t)$$ that is $-g(t)=g'(t)t$ which implies $g(t)=\frac{C}{t}$.

  • $\begingroup$ brilliant answer in minimum steps! $\endgroup$ – vidyarthi Nov 26 '16 at 9:30
  • $\begingroup$ Very elegent solution , can you give me a hint how to show that F is a conservative field in a simply connected region if and only if g(t) = k/t , k is constant. Thanks again $\endgroup$ – Kasmir Khaan Nov 26 '16 at 10:32
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    $\begingroup$ To be conservative you need that $(-g(t)y)_y=(g(t)x)_x$. $\endgroup$ – Robert Z Nov 26 '16 at 10:46
  • $\begingroup$ Thanks alot ! i did prove the implication but only for 1 direction , can you please post to both ( no full solution just highlight ) so i can see if my argument is correct and see how to think about these questions . $\endgroup$ – Kasmir Khaan Nov 26 '16 at 11:04
  • $\begingroup$ I started with g(t) = k/t and showed that it is valid , but not knowing g (t) = k/t how would I proceed ? Thanks so much again $\endgroup$ – Kasmir Khaan Nov 26 '16 at 11:07

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