Proof involving central acceleration and vector products

[Sho Kano]

Homework Statement

Suppose $r:R\rightarrow { V }_{ 3 }$ is a twice-differentiable curve with central acceleration, that is, $\ddot { r }$ is parallel with $r$.
a. Prove $N=r\times \dot { r }$ is constant
b. Assuming $N\neq 0$, prove that $r$ lies in the plane through the origin with normal $N$.

Homework Equations

The Attempt at a Solution

a. $\frac { d }{ dt } N=\frac { d }{ dt } r\times \dot { r } =r\times \ddot { r } +\dot { r } \times \dot { r } =\overrightarrow { 0 }$ because $r$ is parallel with $\ddot { r }$

b. $\dot { r }$ is in the same plane as $r$, then the equation of the plane through the origin is $\left< x,y,z \right> \cdot r\times \dot { r } =0$. If $r=\left< x,y,z \right>$, then $r\cdot r\times \dot { r } =r\times r\cdot \dot { r } =0$ which checks out

I'm really not sure if I'm right

 

[LCKurtz]

Homework Statement

Suppose $r:R\rightarrow { V }_{ 3 }$ is a twice-differentiable curve with central acceleration, that is, $\ddot { r }$ is parallel with $r$.
a. Prove $N=r\times \dot { r }$ is constant
b. Assuming $N\neq 0$, prove that $r$ lies in the plane through the origin with normal $N$.

Homework Equations

The Attempt at a Solution

a. $\frac { d }{ dt } N=\frac { d }{ dt } r\times \dot { r } =r\times \ddot { r } +\dot { r } \times \dot { r } =\overrightarrow { 0 }$ because $r$ is parallel with $\ddot { r }$

b. $\dot { r }$ is in the same plane as $r$, then the equation of the plane through the origin is $\left< x,y,z \right> \cdot r\times \dot { r } =0$. If $r=\left< x,y,z \right>$, then $r\cdot r\times \dot { r } =r\times r\cdot \dot { r } =0$ which checks out

I'm really not sure if I'm right

It looks correct.

 

[Sho Kano]

It looks correct.

Awesome, I can turn in my homework at ease now. Thanks.