If $T$ be the smallest positive number such that the tangent to $$\cos t \hat{i}+\sin t\hat{j}+\frac {t}{\sqrt{2}}\hat{k}$$ at $t=T$ is orthogonal to the tangent at $t=0$, then line integral of $\vec{F}=x\hat{j}-y\hat{i}$ along the above vector from $t=0$ to $t=T$ is?
I am stuck regards the question. The answer seems to be $2$, but am unable to derive it. Specifically, I used the formulation $\int_{\Gamma}\vec{F}\cdot d\vec{r}=\int_{\Gamma}\vec{F(\vec{r}(t))}\cdot \vec{r}'(t) dt$, but I get the answer to be $\frac{2\pi}{3}$, because, I think $T=\frac{2\pi}{3}$. Any hints? Thanks beforehand.
Let $\vec{R}(t)=\cos (t) \hat{i}+\sin( t)\hat{j}+\frac {t}{\sqrt{2}}\hat{k}$
And $\int_{\Gamma} \vec{F}\cdot d\vec{r}=\int_{t=0}^{t=T} (xdy-ydx) $
$\implies \int_{\Gamma} \vec{F}\cdot d\vec{r}=\int_{t=0}^{t=T} (\sin^2t+\cos^2t) dt=T$ (*)
Now , we have to find $T$ . As given ,
$\{\frac{d\vec{R}}{dt}\}_{t=0} \cdot \{\frac{d\vec{R}}{dt}\}_{t=T}=0$ (**)
(since, tangent at $t=0$ is orthogonal i.e. perpendicular to the tangent at $t=T$ )
So , we get from (**) ,
$\cos T=\frac{-1}{2}$ .
And $T$ is the smallest $+ve$ real number....
Hence, $T=\frac{2\pi}{3}$ .