- #1
karush
Gold Member
MHB
- 3,269
- 5
$\tiny{243.13.0113}$
$\textsf{The vector $r(t)$ is the position vector of a particle at time $t$.}]$
$\textsf{Find the angle between the velocity and the acceleration vectors at time $t=0$}\\$
\begin{align*} \displaystyle
r_{13}(t)&=sin^{-1}(4t)\large\textbf{i}+\ln(7t^2+1)\large\textbf{j}+\sqrt{8t^2+1}\large\textbf{k}\\
v_{13}(t)&=\frac{4}{\sqrt{1 - 16 t^2}}\large\textbf{i}
+ \frac{14 t}{(7t^2+1)}\large\textbf{j}
+ \frac{8 t}{\sqrt{8 t^2 + 1}}\large\textbf{k}\\
a_{13}(t)&=a_{13}(t)=\frac{64t}{(1- 16t^2)^{3/2}}\large\textbf{i}
+\frac{14-98t^2}{(7t^2+1)^2}\large\textbf{j}
+\frac{8}{(8t^2+1)^{3/2}}\large\textbf{k}\\
\textit{book answer}&=\color{red}{\frac{\pi}{2}}
\end{align*}
ok this looks kinda hefty
can we plug in the $t=0$
before the leap of faith into dot product or no?(drink)
$\textsf{The vector $r(t)$ is the position vector of a particle at time $t$.}]$
$\textsf{Find the angle between the velocity and the acceleration vectors at time $t=0$}\\$
\begin{align*} \displaystyle
r_{13}(t)&=sin^{-1}(4t)\large\textbf{i}+\ln(7t^2+1)\large\textbf{j}+\sqrt{8t^2+1}\large\textbf{k}\\
v_{13}(t)&=\frac{4}{\sqrt{1 - 16 t^2}}\large\textbf{i}
+ \frac{14 t}{(7t^2+1)}\large\textbf{j}
+ \frac{8 t}{\sqrt{8 t^2 + 1}}\large\textbf{k}\\
a_{13}(t)&=a_{13}(t)=\frac{64t}{(1- 16t^2)^{3/2}}\large\textbf{i}
+\frac{14-98t^2}{(7t^2+1)^2}\large\textbf{j}
+\frac{8}{(8t^2+1)^{3/2}}\large\textbf{k}\\
\textit{book answer}&=\color{red}{\frac{\pi}{2}}
\end{align*}
ok this looks kinda hefty
can we plug in the $t=0$
before the leap of faith into dot product or no?(drink)
