Answer
When $t = \frac{2b}{3c}$, the velocity vector makes an angle of $45.0^{\circ}$.
Work Step by Step
$x(t) = bt^2$
$v_x(t) = \frac{dx}{dt} = 2bt$
$y(t) = ct^3$
$v_y(t) = \frac{dy}{dt} = 3ct^2$
If the velocity vector makes an angle of $45.0^{\circ}$, then:
$v_x(t) = v_y(t)$
$2bt = 3ct^2$
$t = \frac{2b}{3c}$.
When $t = \frac{2b}{3c}$, the velocity vector makes an angle of $45.0^{\circ}$.
