As the title states, I'm trying to show that $$\left|\left|\frac{d\hat{v}}{dt}\right|\right| = \frac{a\cdot\sin(\theta)}{s}, $$ where
$\hat{v}$ is the unit velocity vector of a particle
$a = \left|\left|\vec{a}\right|\right| = $ the magnitude of the acceleration vector
$s = \left|\left|\vec{v}\right|\right| = $ the magnitude of the velocity vector
$\theta$ is the angle between $\vec{v}$ and $\vec{a}$.
I've been trying a wide variety of manipulations but the closest I've come is with:
$\left|\left|\frac{d\hat{v}}{dt}\right|\right| = \frac{a\sqrt{\cos(\theta)}}{s}$. But here, $\theta$ represents the angle between the acceleration vector and itself... Plus there's the square root.
Any thoughts on this?
Thank you very much for your time,
Eric
The acceleration is defined as $$ \mathbf{a} = \frac{d \mathbf{v}}{dt}$$ and the unit norm vector as $$ \hat{\mathbf v} = \frac{\mathbf{v}}{|\mathbf{v}|}=\frac{\mathbf{v}} {\sqrt{\mathbf{v}\cdot \mathbf{v}}}$$ Now the task is to calculate $|d\hat{\mathbf v}/dt|.$
We have with the chain rule $$\frac{d \hat{\mathbf v}}{dt} = |\mathbf{v}|^{-1}\mathbf{a} +\mathbf{v} \frac{d}{dt}(\mathbf v \cdot \mathbf{v})^{-1/2} = \frac{\mathbf{a}}{|\mathbf{v}|} - \frac{\mathbf{v}}{|\mathbf v|^3} \mathbf{v}\cdot\mathbf{a}.$$ We additionally know that $\mathbf{v}\cdot \mathbf{a} =|\mathbf v| \,|\mathbf a| \cos\theta$. We obtain $$ \frac{d \hat{\mathbf v}}{dt} \cdot \frac{d \hat{\mathbf v}}{dt} =\frac{|\mathbf{a}|^2}{|\mathbf{v}|^2} + \frac{|\mathbf{v}\cdot \mathbf{a}|^2}{|\mathbf{v}|^4}- 2 \frac{|\mathbf{v}\cdot \mathbf{a}|^2}{|\mathbf{v}|^4} =\frac{|\mathbf a|^2}{|\mathbf v|^2} - \frac{|\mathbf a|^2 \cos^2\theta}{|\mathbf v|^2} = \frac{|\mathbf a|^2 \sin^2\theta}{|\mathbf v|^2} ,$$ from which the result follows (with $|\sin \theta|$).