Recently I got a problem that equated the time derivative of a cross product d/dt (P x Q) with a function of time (like t + t^2).
Ex. d/dt (P x Q) = 5t - 6t^2
My question is, how can you have an equation with a cross product derivative (which is itself a vector) with a function of time?
You are quite right. You can't sensibly equate a vector to a scalar. So the scalar function of $t$ on the right hand side of your equation can't be right. But you could have a vector function of $t$.
The cross product of two planar vectors is a scalar.
$$ \pmatrix{a \\ b} \times \pmatrix{x \\ y} = a y - b x $$
Also, note the following 2 planar cross products that exist between a vector and a scalar (out of plane vector).
$$ \pmatrix{a \\ b} \times \omega = \pmatrix{\omega\, b\\ -\omega\, a} $$
$$ \omega \times \pmatrix{x \\ y} = \pmatrix{-\omega\, y \\ \omega \, x} $$
All of the above are planar projections of the one 3D cross product.
Also note that the derivative is a linear operator which means the product rule applies
$$ \tfrac{\rm d}{{\rm d}t} ( \vec{a} \times \vec{b} ) = ( \tfrac{\rm d}{{\rm d}t} \vec{a} ) \times \vec{b} + \vec{a} \times (\tfrac{\rm d}{{\rm d}t} \vec{b}) $$
The vectors P and Q define a plane. The cross product is perpendicular to that plane. Your time function could describe the motion of an object in that perpendicular direction.
