# What happens to $\frac{d}{dt}\left(\hat{v}\right)$ at the highest point a projectile reaches when launched vertically upwards?

Acceleration is given by $$\dot{\vec{v}} = \frac{d}{dt}\left( v \hat{v}\right) = \dot{v} \hat{v} + v \dot{\hat{v}}$$.

What happens to $$\dot{\hat{v}}$$ when the direction of velocity flips by $$180^o$$?

E.g. when a ball is thrown straight up in the air; when it reaches it's highest point, it goes from having an upwards velocity at the instant before to having a downwards velocity the instant afterwards.

Would $$\dot{\hat{v}}$$ be undefined in this case? But how come $$\dot{\vec{v}}$$ isn't? (We know it would be $$\vec{g}$$ in my example.) Is it because at the point when velocity flips by $$180^o$$ (if we assume that velocity as a function of time is 'smooth and continuous'), $$v$$ has to be zero, so $$\vec{v} = \vec{0}$$, which has no direction vector, so $$\dot{\hat{v}}$$ can't be defined at this point in time?

• What's the direction of a vector with norm = 0? It's not defined. Once you know it, it's better to avoid non-regular description of a problem, if you need to do calculations on it, or implement model in a program Commented Jun 2 at 9:59
• While the unit vector may be ill-defined at the apex, thankfully it's being multiplied by $v = 0$ there to ensure its value doesn't matter. Commented Jun 2 at 13:35
• @eyeballfrog Not necessarily, in the left term, $\dot{v}\mathbf{\hat{v}}$, you have two undefined quantities being multiplied and in the right term, $v\dot{\mathbf{\hat{v}}}$, you have an undefined term multiplied by zero which is again undefined. Even if we take it to be zero, the left term makes the entire expression undefined unless we make "handwavy" $0/0$ assumptions. Commented Jun 2 at 13:40

This is a fun question, you can answer your own question by looking at a simple example. Let's look at vertical motion under the influence of gravity: $$v(t)=v_0-gt\rightarrow v(t)=-gt$$ Where $$g>0$$. Here I chose $$t=0$$ to be the turning point. In vector notation, if we define our unit vector using the standard convention of vector divided by its magnitude: $$\mathbf{v}(t)=|gt|\frac{-gt}{|gt|}\mathbf{\hat{z}}=v\mathbf{\hat{v}}$$ Here we see we introduced the identity to write it in terms of its unit vector, where: $$v=|gt|$$ $$\mathbf{\hat{v}}=-\frac{gt}{|gt|}\mathbf{\hat{z}}=-\frac{t}{|t|}\mathbf{\hat{z}}=-\mathrm{sgn}(t)\mathbf{\hat{z}}$$ Let's take the derivative of $$\mathbf{v}(t)$$ written: $$\frac{d}{dt}\mathbf{v}=\frac{d}{dt}\left(-gt\mathbf{\hat{z}}\right)=-g\mathbf{\hat{z}}$$ And now written we take the derivative when it is written as $$v\mathbf{\hat{v}}$$: $$\frac{d}{dt}\left(v\mathbf{\hat{v}}\right)=\dot{v}\mathbf{\hat{v}}+v\dot{\mathbf{\hat{v}}}$$ Using: $$\dot{v}=\frac{d}{dt}|gt|=g\frac{d}{dt}|t|=\frac{gt}{|t|}=g\,\mathrm{sgn}(t)$$ $$\dot{\mathbf{\hat{v}}}=-\frac{d\,\mathrm{sgn}(t)}{dt}\mathbf{\hat{z}}$$ Combining everything both terms from the expanded form ($$v\mathbf{\hat{v}}$$) and equating it to the unexpanded form ($$\mathbf{v}$$): $$-g\,\mathrm{sgn}(t)^2-g|t|\frac{d\,\mathrm{sgn}(t)}{dt}=-g$$ For $$t\neq0$$, $$\mathrm{sgn}(t)^2=1$$ and the derivative of the sign function is 0, so everything works. When $$t=0$$ we want to return to when we wrote $$\dot{v}=g\,\mathrm{sgn}(t)$$ and instead write this as: $$-\frac{t}{|t|}\frac{d|t|}{dt}-|t|\frac{d}{dt}\left(\frac{t}{|t|}\right)\stackrel{?}{=}-1$$ So we can't really evaluate either derivative at $$t=0$$ without getting a divided by zero or into $$0/0$$ territory, but this gives you (assuming I didn't accidentally divide by zero anywhere) a "balancing" expression for 2 expressions that you can't evaluate. The only way we can resolve this is if we undo our introduced unity identity in: $$\mathbf{v}(t)=|gt|\frac{-gt}{|gt|}\mathbf{\hat{z}}$$ Where at $$t=0$$ we have introduced a $$0/0$$ expression. So If we just never applied that, we have no ambiguity. You may find a better discussion in a textbook discussing the theory of distributions.