Chain rule and division by zero

[jain_arham_hsr]

Homework Statement

The velocity of a particle moving along the x-axis is given as v=x^2-5x+4 (in m/s), where x denotes the x-coordinate of the particle in metres. Find the magnitude of acceleration of the particle when the velocity of the particle is zero.

Relevant Equations

Differential Calculus

My approach is as follows:
a = dv/dt
= (dv/dx) * (dx/dt)
= (dv/dx) * v
Putting v = 0:
a = (dv/dx) * 0 = 0 m s^(-2)

But, what I don't understand is this:
If v=0, then dx/dt must also be 0. Consequently, dx must also be 0 at that particular instant. But, we are writing acceleration as (dv/dx) * v. Won't this expression become undefined when dx=0?

 

[malawi_glenn]

dx is not a number

dv / dx is a shorthand for the derivative as a limit notation here.

 

[pasmith]

dx is not a number

dv / dx is a shorthand for the derivative as a limit notation here.

And the point of the limit process is that you never actually divide by zero.

 

[SammyS]

Homework Statement: The velocity of a particle moving along the x-axis is given as v=x^2-5x+4 (in m/s), where x denotes the x-coordinate of the particle in metres. Find the magnitude of acceleration of the particle when the velocity of the particle is zero.
Relevant Equations: Differential Calculus

My approach is as follows:
a = dv/dt
= (dv/dx) * (dx/dt)
= (dv/dx) * v
Putting v = 0:
a = (dv/dx) * 0 = 0 m s^(-2)

But, what I don't understand is this:
If v=0, then dx/dt must also be 0. Consequently, dx must also be 0 at that particular instant. But, we are writing acceleration as (dv/dx) * v. Won't this expression become undefined when dx=0?

In fact, it can be shown that $dv/dx \ne 0$ when $v=0$ for this velocity function.

Factoring the velocity function, we have: $\displaystyle \quad v=x^2-5x+4=(x-4)(x-1)$ .

Setting this equal to zero and solving for $x$ we see that $v=0$ when $x=1$ or $x=4$.

But $dv/dx=2x-5$ which is $-3$ when $x=1$ and is $3$ when $x=4$.

Also, notice that $dv/dx=0$ at $x=5/2$, so the acceleration is zero there too. (Yes, I realize that the problem did not ask for you to find this.)

 

[PeroK]

Homework Statement: The velocity of a particle moving along the x-axis is given as v=x^2-5x+4 (in m/s), where x denotes the x-coordinate of the particle in metres. Find the magnitude of acceleration of the particle when the velocity of the particle is zero.
Relevant Equations: Differential Calculus

My approach is as follows:
a = dv/dt
= (dv/dx) * (dx/dt)
= (dv/dx) * v
Putting v = 0:
a = (dv/dx) * 0 = 0 m s^(-2)

But, what I don't understand is this:
If v=0, then dx/dt must also be 0. Consequently, dx must also be 0 at that particular instant. But, we are writing acceleration as (dv/dx) * v. Won't this expression become undefined when dx=0?

Here's a thing. Assume that $v$ can be written as a function of $x$. Then:
$$\frac{dv}{dt} =\frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v$$Therefore: acceleration must be zero whenever velocity is zero, for any motion!? What's going on?

 

[Grelbr42]

You need to be careful with the generic case of considering any-old velocity as a function of position. In some cases $\frac{dV}{dx}$ has a $\frac{1}{V}$ in it. Consider simple harmonic motion with $x(t) = A \sin (w t)$. You get $V(t) = A w \cos (w t)$ and you can easily work out that $V = w \sqrt{A^2 - x^2}$. So then $\frac{d V} {dx} = -\frac{w^2 x}{V}$. And finally, $a = \frac{d V} {dx} V = -w^2 x$, the familiar answer.

To see what happens in the OP, you need to think about what the potential energy might be that produces the given velocity as a function of $x$. And where it has its maximum. And what happens if you put an object at rest at the maximum of a potential. Easy example might be, a ball balanced at the top of a hill.

 

[PeroK]

Note that velocity and acceleration are the first and second derivatives of position wrt time. If $v =0$ and $a \ne 0$ then we have a turning point, where velocity changes sign. In which case we have two velocities for the same position and velocity is not a function of position. If velocity remains a function of position through a point where velocity is zero, then we must have a point of infection, in which case $a =0$ at that point.

 

[malawi_glenn]

As another concrete example, consider a stone being thrown upwards vertically with initial speed $v_0$. Let positive direction be up, the acceleration is constant if we ignore air-resistance and motion close to the surface of the Earth $a = \dfrac{\text d v}{\text d t} = -g$.

We use the SUVAT-equation $v(x(t))^2 - v_0^2 = -2gx$ where $x$ is the distance from the position where the stone was thrown, $x(0) = 0$. We have $v = 0$ when $x = \dfrac{v_0^2}{2g}$.

For motion upwards, positive $v$, we obtain $v(x) = \sqrt{v_0^2 - 2gx}$.

Using the relation from the chain rule $\dfrac{\text d v}{\text d t} = \dfrac{\text d v}{\text d x} v$ and differentiating $v(x) = \sqrt{v_0^2 - 2gx}$ w.r.t. $x$:
$\dfrac{\text d v}{\text d x} = \dfrac{1}{2}\cdot \dfrac{-2g}{\sqrt{v_0^2 - 2gx}} = -\, \dfrac{g}{v}$

we obtain $\dfrac{\text d v}{\text d t} = \dfrac{\text d v}{\text d x} v =-\, \dfrac{g}{v} \cdot v = - g$ as expected.

 

[kuruman]

It seems to me that the intent of the problem's author was to use a straightforward approach, namely solve the separable differential equation $$\frac{dx}{dt}=(x-1)(x-4)$$ to find $x(t)$ and hence $v(t)$ and $a(t)$. Then one can easily perhaps determine when (not where) the velocity is zero and evaluate the acceleration at such time(s).

 

[SammyS]

It seems to me that the intent of the problem's author was to use a straightforward approach, namely solve the separable differential equation $$\frac{dx}{dt}=(x-1)(x-4)$$ to find $x(t)$ and hence $v(t)$ and $a(t)$. Then one can easily determine when (not where) the velocity is zero and evaluate the acceleration at such time(s).

Yes, that differential equation is consistent with the problem in the OP.

However, there is a difficulty in determining when (at what time) the velocity is zero.

In general, you are free to choose an arbitrary value, $x=x_0$ at $t=0$ or in general pick $x=x_T$ at $t=T$. But as @PeroK suggests: (Bold added for emphasis.)

Here's a thing. Assume that $v$ can be written as a function of $x$. Then:
$$\frac{dv}{dt} =\frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v$$Therefore: acceleration must be zero whenever velocity is zero, for any motion!? What's going on?

If you pick a time for the velocity to be zero, the acceleration also being zero implies that the particle will be stationary at either $x=1$ or at $x=4$.

Also, note:
Setting some value for $x_0$ such as $x_0=0$ does seem to imply that the particle will come to rest at $x=1$. However, the particle will never get to $x=1$, i.e. it takes an infinite amount of time to get there.

 

[kuruman]

Yes, that differential equation is consistent with the problem in the OP.

However, there is a difficulty in determining when (at what time) the velocity is zero.

After solving the differential equation, I agree that there is great difficulty determining when the velocity is zero (I edited my previous about that). I see now that using $a=v\dfrac{dv}{dx}~$ for this particular problem to derive a potential function $U(x)$ and then plotting it, illustrates @PeroK's query "what's going on?" quite nicely.

 

[PeroK]

From a mathematical point of view: if $v(x)$ is well defined and differentiable at some point $x_0$, then $v(x_0) =0$ implies $a(x_0)=0$.

My answer to "what's going on" is that in general (and for most motions that are studied in physics), these hypotheses fail.

This, IMO, is another example of some mathematics being specified that has little or no relevance to actual physics. Or, at least, the connection to a real physical system is obscure. This seems to have been quite common in the homework forum recently.

 

[kuruman]

This, IMO, is another example of some mathematics being specified that has little or no relevance to actual physics. Or, at least, the connection to a real physical system is obscure.

I am not 100% convinced about that and this why. I took your "what's going on" as an invitation to seek physical meaning to the given expression for the velocity. If we assume that there is a single force acting on the particle, then we can derive a potential function for that force. $$\begin{align} & F=ma=mv\frac{dv}{dx}=m(x-1)(x-4)(2x-5)=-\frac{dU}{dx} \nonumber \\ & \implies U(x)=- \frac{1}{2}x^4+5x^3-\frac{33}{2}x^2+20x. \nonumber \end{align}$$A plot of this potential is shown below.

At $x=1~$m and $x=4~$m we have points of unstable equilibrium where the potential energy is $8 ~$units. If the particle has energy ($8-\epsilon$) units, in the limit $\epsilon\rightarrow 0$, we have turning points at $x=1~$m and $x=4~$m. Meanwhile the period of the oscillatory motion increases and becomes infinite when $\epsilon=0.$ Such a potential could be used for example to model how a diatomic molecule vibrates at low energies but falls apart when too much energy is pumped into it. So I think that there could be physical meaning to this and that is why I am not 100% convinced.

BTW, we went through the $v({dv}/{dx})=0$ business 3 years ago here and I think it would be wise to let that sleeping dog lie.