Mult Variate Calc: Proof for N(t) vector

[opus]

Homework Statement

Part (d):
Use parts (a), (b), and (c) to argue that $$\vec{N}(t) = \frac{\vec{a}_{v\perp}}{|\vec{a}_{v\perp}|}$$

Relevant Equations

$$\frac{d}{dt}(\vec{v}\cdot\vec{v}) = 2\vec{a} \cdot \vec{v}$$
$$\frac{d}{dt}v(t) = \frac{\vec{v}\cdot\vec{a}}{v(t)}$$
$$v(t)\vec{T'}(t) = \vec{a}(t) - \frac{v'(t)\vec{v'}(t)}{v(t)}$$

I need to prove this using the given equations.
$$\vec{N}(t) = \frac{\vec{a}_{v\perp}}{|\vec{a}_{v\perp}|}$$

Here is the entirety of my work up to this point. So far I've wanted to use what I have to find something that is perpendicular to the velocity vector and maybe show that with the dot product but no luck so far and I'm pretty stuck. Any hints to get me started?

 

[andrewkirk]

It will be easier to help with this if some context is provided. Presumably $t$ is time and $\vec a(t),\vec v(t)$ are the acceleration and velocity of a particle, but what do $\vec T(t), \vec N(t),\vec a_{v\bot}$ and the scalar $v(t)$ represent?

 

[opus]

Ok so $\vec{T}(t)$, $\vec{N}(t)$, and $\vec{a}_{v\perp}$ arent ever mentioned in any previous problem for this homework, but in the text they are given as:

$\vec{T}(t) = \frac{\vec{v}(t)}{v(t)}$ so a unit vector in the direction of the velocity vector.

$\vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|}$ which I'm a little unclear on but I think its a unit vector perpendicular to the plane created by the velocity and acceleration vectors.

$v(t)$ is just the speed or magnitude of the velocity vector $\vec{v}(t)$.

And I THINK $\vec{a}_{v\perp}$ is a vector perpendicular to the velocity and acceleration vectors but the homework nor the book goes over this.

 

[Mark44]

Ok so $\vec{T}(t)$, $\vec{N}(t)$, and $\vec{a}_{v\perp}$ arent ever mentioned in any previous problem for this homework, but in the text they are given as:

$\vec{T}(t) = \frac{\vec{v}(t)}{v(t)}$ so a unit vector in the direction of the velocity vector.

AKA the unit tangent vector, thus the name T.

$\vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|}$ which I'm a little unclear on but I think its a unit vector perpendicular to the plane created by the velocity and acceleration vectors.

The unit normal vector, thus the name N. The N vector is perpendicular to the tangent plane at a particular point. The acceleration could be in the same direction as the velocity vector, so it (acceleration) doesn't necessarily define a normal vector.

$v(t)$ is just the speed or magnitude of the velocity vector $\vec{v}(t)$.

And I THINK $\vec{a}_{v\perp}$ is a vector perpendicular to the velocity and acceleration vectors but the homework nor the book goes over this.

Maybe or maybe not, as the acceleration and velocity could be in the same direction.

 

[opus]

Thanks Mark that is more clear! Going to keep plugging at this to see what I can get

 

[Orodruin]

And I THINK →av⊥a→v⊥\vec{a}_{v\perp} is a vector perpendicular to the velocity and acceleration vectors but the homework nor the book goes over this.

It is the component of acceleration orthogonal to the velocity. In other words, the acceleration minus any component aceleration has in the velocity direction.

 

[WWGD]

It seems like it may be a setup for the Frenet-Serret frames for space curves? Any chance they are mentioned on this or upcoming chapters ( maybe check the index?)?

 

[Antarres]

You may notice from it's definition that $\vec{N}$ is in the direction of $\vec{T'}$. So you just have to figure out the magnitude of $\vec{T'}$ and combine it with the third formula you derived.

So analyze the third formula. Notice that $\vec{a}_{v\perp}$ is the component of acceleration that is perpendicular to velocity, as Orodruin pointed out. So this component is just full acceleration minus the component of acceleration that is tangential to velocity(usually called tangential acceleration).

Now tangential acceleration has magnitude $v'(t)$, because it is the component that actually varies the magnitude of velocity, while the perpendicular component changes its direction. So we have that:
$$\vec{a}_{v\perp} = \vec{a} - \vec{a}_{v}$$
$$\vec{a}_{v} = v'(t)\vec{T}(t)$$
in your notation.
Compare this to the formulas you have, and the solution should be pretty obvious.