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.