I want to find the length of the smooth curve $γ(t)$ from $t = 0$ to $t = \pi/4$
I'm using the formula for γ: $$l(γ) = \int_a^b|\gamma'(t)|\,dt$$ apologies I can't work out how to format that correctly.
I have $ \gamma'(t) $ I am now aware this is incorrect but am struggling to amend it myself I'm not sure how to apply this into the formula without it just integrating back into its original form?
Any help or identification of areas where I've made a mistake would be really useful.
Note that if $$\gamma(t)=e^{it}+t(\sin(t)-i\cos(t))=e^{it}+t\sin(t)-it\cos(t)$$ so, you have that $$\gamma'(t)=\color{blue}{ie^{it}}+\sin(t)+t\cos(t) +it\sin(t)-i\cos(t). \quad [1]$$ Now, since that for all $t\in \mathbb{R}$, we have that $$e^{it}=\cos(t)+i\sin(t)$$ The, you can re-write $[1]$ as $$\gamma'(t)=\sin(t)+t\cos(t)\color{blue}{-\sin(t)}+it\sin(t)-i\cos(t)\color{blue}{+i\cos(t)}.$$ Using that for all $z\in \mathbb{C}$ we have that $$|z|=\sqrt{\operatorname{Re}(z)^{2}+\operatorname{Im}(z)^{2}}$$ So, you have that \begin{eqnarray*} |\gamma'(t)|&=&\sqrt{(\sin(t)+t\cos(t)-\sin(t))^{2}+(t\sin(t)-\cos(t)+\cos(t))^{2}}\\ &=&\sqrt{t^{2}\cos^{2}(t)+t^{2}\sin^{2}(t)}\\ &=&\sqrt{t^{2}(\cos^{2}(t)+\sin^{2}(t))}\\ &=&\sqrt{t^{2}}, \quad \text{since that} \quad \forall x\in \mathbb{R}: \quad \cos^{2}(x)+\sin^{2}(x)=1. \end{eqnarray*}
Finally, you need calculate $$\ell(\gamma(t))=\int_{0}^{\frac{\pi}{2}}|\gamma'(t)|dt.$$