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

`(\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

`(\mathrm d\vec r(t))/(\mathrm dt)\cdot\vec\imath=2bt=√(4b^2t^2+9c^2t^4)\cos\theta`

and

`(\mathrm d\vec r(t))/(\mathrm dt)\cdot\vec\jmath=3ct^2=√(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=√(4b^2t^2+9c^2t^4)\\3\sqrt2\,ct^2=√(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=(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=(2b)/(3c)}`