Velocity vector

[CellCoree]

If $$r^{\rightarrow} = bt^2i+ct^3j$$, where $$b$$ and $$c$$ are positive constants, when does the velocity vector make an angle of $$45.0^o$$ with the x- and y-axes?


i & j are vectors.

i have no clue on how to start it, can anyone give me a hint?

 

[Tide]

There are several approaches you could take. Here's one:

A vector that is at right angles to the the 45 degree direction is [itex]\vec a = \hat i - \hat j[/tex]. Now form the scalar product [itex]\vec a \cdot \vec r[/tex] and set it equal to zero. The desired result will follow!

 

[arildno]

Alternatively, require that the x and y components of the velocity vector are equal.

EDIt:
Tide, I believe he's been given the position vector, not the velocity vector..

 

[Tide]

Alternatively, require that the x and y components of the velocity vector are equal.

EDIt:
Tide, I believe he's been given the position vector, not the velocity vector..

Thanks, arildno! Change that to $\vec a \cdot \vec v$ with $\vec v \frac{d\vec r}{dt}$.