Parametric Equations Question 2

For time $t\gt 0$ , the position of a particle moving in the $xy$ -plane is given by the parametric equations $x=4t+t^2$ and $y=\dfrac{1}{3t+1}$ . What is the acceleration vector of a particle at time $t=1$ ?

\left(2,\dfrac{1}{32}\right)

\left(2,\dfrac{9}{32}\right)

\left(5,\dfrac{1}{4} \right)

\left(6,-\dfrac{3}{16} \right)

Approach

x=4t+t^2

x'(t)=4+2t

x''(t)=2

x''(1)=2

y=\dfrac{1}{3t+1}

y'(t)=\frac{-3}{(3t+1)^2}

y''(t)= \frac{18}{(3t+1)^3}

y''(1)=\frac{18}{64} = \frac{9}{32}