Mathematical Immaturity

14.9 Exercises

• A particle of unit mass with position vector $\mathbf{r}(t)$ at time $t$ is moving in space under the actions of certain forces.
  (a) Prove that $\mathbf{r} \times \mathbf{r}'' = 0$ implies $\mathbf{r} \times \mathbf{r}' = \mathbf{c},$ where $\mathbf{c}$ is a constant vector.
  (b) If $\mathbf{r} \times \mathbf{r}' = \mathbf{c},$ where $\mathbf{c}$ is a constant vector, prove that the motion takes place in a plane. Consider both $\mathbf{c} \neq 0$ and $\mathbf{c} = 0.$
  (c) If the net force acting on the particle is always directed toward the origin, prove that the particle moves in a plane.
  (d) Is $\mathbf{r} \times \mathbf{r}'$ necessarily constant if a particle moves in a plane?
• (a) Recall the solution of 14.4.15 where we showed that $(F \times F')' = F \times F''$
  (b) Recall from Theorem (13.12.g) that for two vectors $A$ and $B,$ $A \times B = O$ if and only if $A$ and $B$ are linearly dependent.
  (c) If force is always directed towards the origin, then this implies that acceleration and position are linearly dependent.
• (a) We have shown previously that for some vector-valued function $F,$ $$ (F \times F')' = F \times F'' $$ Thus, if $\mathbf{r} \times \mathbf{r}'' = 0,$ then $(\mathbf{r} \times \mathbf{r}')' = 0$ and $\mathbf{r} \times \mathbf{r}'$ is constant. \begin{align*} \end{align*} (b) If $\mathbf{c} = 0,$ then $\mathbf{r}$ and $\mathbf{r}'$ are linearly dependent (Theorem 13.12). This means that the velocity vector is a scalar multiple of the position vector, implying that motion is linear, and thus in a plane. If $\mathbf{c} \neq 0,$ then $\mathbf{r}(t) \times \mathbf{v}(t)$ is constant for all $t,$ meaning that all motion is perpendicular to $\mathbf{c}.$ Thus, the motion defined by $\mathbf{r}$ and $\mathbf{v}$ must lie within a plane $($perpendicular to $\mathbf{c}).$ \begin{align*} \end{align*} (c) If the force is always directed towards the origin, then with unit mass, this implies that $\mathbf{a}(t) = -k\mathbf{r}(t)$ for some scalar $k.$ But this means that $\mathbf{a}$ and $\mathbf{r}$ are linearly dependent, and we know from before that this means $\mathbf{r} \times \mathbf{a} = O.$ But recall from part (a) that this implies that $\mathbf{r} \times \mathbf{v} = \mathbf{c}$ where $\mathbf{c}$ is constant, which means that the motion lies within a plane as shown in part (b). \begin{align*} \end{align*} (d) The direction of the vector normal to the plane will remain constant but can be scaled by some real constant, thus $\mathbf{r} \times \mathbf{v}$ need not be constant.