Parametric Curve Acceleration

This is a parametric equation of $y=x^2$ on $x\in [1,1]$ . So we have $y'' = 2$ . And the parametric curve, $\textbf{r}(t)$ , is given by $\begin{aligned} \text{r}(t) = &\langle \sin(t), \sin^2(t)\rangle\\ \text{r}''(t) = &\langle -\sin(t), 2\cos(2t) \rangle \end{aligned}$ So we know $2 = 2\cos(2t) \Rightarrow t = 0, \frac{\pi}{2}$ where $|\textbf{r}''(t)| = \sqrt{\sin^2(t) +4\cos^2(2t)}$ so $|\textbf{r}''(0)|= 2$ and $|\textbf{r}''(\frac{\pi}{2})| = \sqrt{5}$ . $\frac{A}{\sqrt{B}} = \frac{2}{\sqrt{5}} \Rightarrow \boxed{A+B = 7}$

$~~~~~~\\ X=Sin(t),~~~~~Y=Sin^2(t).\\ \implies~X'=Cos(t),~~~~~Y'=Sin(2t).\\ \therefore~a_x=X"=-Sin(t),~~~~~a_y=Y"=2Cos(2t).\\ \therefore~{|a|}=\sqrt{a_x^2+a_y^2}............(A)\\ ~~~~~ \\ When~X=0,~~t=0.\\ When~X=1,~~t=\dfrac \pi 2.\\ Substituting~in~(A),\\ A=\sqrt{a_x^2+a_y^2}=\sqrt{(-Sin(0))^2+(2*Cos(2*0))^2},\\ \therefore~A=\sqrt{0+4}=2.\\ \sqrt B=\sqrt{ \left (-Sin(\dfrac \pi 2) \right )^2+ \left (2*Cos(2*\dfrac \pi 2) \right )^2}, \\ \sqrt B=\sqrt{1^2+2^2} =\sqrt 5.\\ A+B=2+5=\Large ~~~\color{#D61F06}{7}.$

Unable to understand why part of the Latex version also has come up !!!!!