Answer
$v \cdot j =12; a \cdot j =26$
Work Step by Step
Since, $v=\dfrac{dx}{dt}i+\dfrac{dy}{dt}j$
$ v \cdot j=\dfrac{dy}{dt}=\dfrac{1}{3}x^2 \dfrac{dx}{dt}; (\dfrac{dx}{dt}=4)$
and $v \cdot j (3,3)=\dfrac{4}{3}x^2 \\ =\dfrac{(4)(9)}{3} \\=12$
$a(t)=\dfrac{2}{3}x(\dfrac{dx}{dt})^2+\dfrac{1}{3} x^2 \dfrac{d^2 x}{dt^2} $
$\implies a \cdot j=\dfrac{2}{3}x\times (\dfrac{dx}{dt})^2+\dfrac{1}{3} \times x^2 \dfrac{d^2 x}{dt^2} \\ =\dfrac{2}{3}x(4)^2+\dfrac{1}{3} (3)^2 (-2) \\ =26$
