Question

If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 with the x-and y-axes?

Answer 1

$\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath$

The velocity at time $t$ is

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath$

Take two vectors that point in the positive $x$ and positive $y$ directions, such as $\vec\imath$ and $\vec\jmath$. The dot products of the velocity vector with $\vec\imath$ and $\vec\jmath$ are

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

and

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

We want the angles between these vectors to be 45º, for which we have $\cos45^\circ=\frac1{\sqrt2}$. So

$\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0$

$\implies t(3ct-2b)=0$

$\implies t=0\text{ or }t=\dfrac{2b}{3c}$

When $t=0$, the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

$\boxed{t=\dfrac{2b}{3c}}$