How Does the Chain Rule Apply to Newton's Second Law in Calculus?

[Vanrichten]

Hello,

I'm new to the language of calculus. I am learning about Newtons second law and I'm trying to understand it's forms in calculus. Excuse my notation, just trying to keep it as simple as possible.

F=m * dV/dt
I know V= dx/dt

My textbook says you can 'apply chain rule' to obtain the following

dV/dt = dV/dx * dx/dt

and this becomes V * dV/dx

I'm really having trouble with grasping how the chain rule is applied here, to me when I read this it sounds like a magic wand is being waved and somehow the result above comes about. So my questions are

1) How exactly is the chain rule applied here showing the work done in the intermediate steps to get dV/dt = dV/dx * dx/dt

2) In answering 1, I hope it will answer my question of how you end up with the V at the end.

 

[verty]

Here is my way to understand this. Think of each $dV$ or $dx$ as having a hidden $dt$ below it. So ${dV \over dx} = {dV/dt \over dx/dt}$. If $dt$ is below, it's not hidden. Does it make more sense now?

 

[tnich]

Hello,

I'm new to the language of calculus. I am learning about Newtons second law and I'm trying to understand it's forms in calculus. Excuse my notation, just trying to keep it as simple as possible.

F=m * dV/dt
I know V= dx/dt

My textbook says you can 'apply chain rule' to obtain the following

dV/dt = dV/dx * dx/dt

and this becomes V * dV/dx

I'm really having trouble with grasping how the chain rule is applied here, to me when I read this it sounds like a magic wand is being waved and somehow the result above comes about. So my questions are

1) How exactly is the chain rule applied here showing the work done in the intermediate steps to get dV/dt = dV/dx * dx/dt

2) In answering 1, I hope it will answer my question of how you end up with the V at the end.

The chain rule is by definition
$\frac {dx}{dz} = \frac {dx}{dy} \frac {dy}{dz}$
or alternatively
$\frac {df(g(z))}{dz} = \frac {df(g(z))}{dg(t)} \frac {dg(z)}{dz}$
So there are no intermediate steps in $\frac {dV}{dt} = \frac {dV}{dx} \frac {dx}{dt}$. It is just using the definition of the chain rule. Perhaps what is confusing you is the choice of $x$ as the intermediate variable between $V$ and $t$. There is no mystery there. The textbook author observed that he could express acceleration as a function of time $a(t)$ and that he could also express it as a function of distance $a(x)$. To relate those two functions, he used the chain rule
$a(t) =\frac {dV(t)}{dt} = \frac {dV(x(t))}{dx(t)} \frac {dx(t)}{dt}$

or $a= \frac {dV}{dt} = \frac {dV}{dx} \frac {dx}{dt}$

 

[rcgldr]

Note that this specific usage of chain rule is only useful when acceleration is a function of position, A = F(x). Since A = dV/dt, using chain rule results in
$$A = \frac{dV}{dx} \ \frac{dx}{dt} = \frac {V \ dv}{dx} = F(x)$$
$$V \ dv = F(x)dx $$
after which both sides can be integrated and a constant added for the initial state to get velocity as a function of position.

 

[Vanrichten]

The chain rule is by definition
$\frac {dx}{dz} = \frac {dx}{dy} \frac {dy}{dz}$
or alternatively
$\frac {df(g(z))}{dz} = \frac {df(g(z))}{dg(t)} \frac {dg(z)}{dz}$
So there are no intermediate steps in $\frac {dV}{dt} = \frac {dV}{dx} \frac {dx}{dt}$. It is just using the definition of the chain rule. Perhaps what is confusing you is the choice of $x$ as the intermediate variable between $V$ and $t$. There is no mystery there. The textbook author observed that he could express acceleration as a function of time $a(t)$ and that he could also express it as a function of distance $a(x)$. To relate those two functions, he used the chain rule
$a(t) =\frac {dV(t)}{dt} = \frac {dV(x(t))}{dx(t)} \frac {dx(t)}{dt}$

or $a= \frac {dV}{dt} = \frac {dV}{dx} \frac {dx}{dt}$

Thanks, this made things a bit more clear.

I've done more research on the chain rule and I think I'm starting to understand. The correct term for this is In Leibniz notation, if y=f(u) and u=g(x) and are both differentiable functions, then dy/dx = dy /du du/ dx

For my particular case, you could say the 'y=f(u)' term was my velocity function written as

v=f(x)

And my 'u=g(x)' becomes x my position is also a function of time, t, so this becomes a composite function where

x=g(t) t is time

Finally,

dv/dt = dv/dx * dx/dt

And since v=dx/dt I can say that

dv/dt=dv/dx *v

 

[Vanrichten]

Here is my way to understand this. Think of each $dV$ or $dx$ as having a hidden $dt$ below it. So ${dV \over dx} = {dV/dt \over dx/dt}$. If $dt$ is below, it's not hidden. Does it make more sense now?

Hmmm,that makes some sense but I'm not sure I have the intuition to know WHY it's hidden. Is this always the case?

 

[Vanrichten]

Note that this specific usage of chain rule is only useful when acceleration is a function of position, A = F(x). Since A = dV/dt, using chain rule results in
$$A = \frac{dV}{dx} \ \frac{dx}{dt} = \frac {V \ dv}{dx} = F(x)$$
$$V \ dv = F(x)dx $$
after which both sides can be integrated and a constant added for the initial state to get velocity as a function of position.

Ahh thanks, now I fully understand this...Integrating the left hand side gives 1/2mV^2 which can be evaluated from the initial to final position right?

And then the kinetic energy is equal to the work done

 

[rcgldr]

Integrating the left hand side gives 1/2mV^2 which can be evaluated from the initial to final position right?

Depends if you want to stop there. If you're trying to determine time versus position, then you have to integrate again, but with square root of the integral of F(x)dx, which often ends up with a difficult integral. Here is a link to an answer with 3 links to examples of how long it takes for two objects to collide due to gravity, with an initial state of zero velocity and some distance apart.

https://www.physicsforums.com/threads/solution-to-Newtons-equation.944403/#post-5980309

 

[verty]

Hmmm,that makes some sense but I'm not sure I have the intuition to know WHY it's hidden. Is this always the case?

Because $dx$, $dv$, etc can be thought of as rates of change. Then things like this make sense:
$dx = x'(t) dt$ (another way of writing ${dx \over dt} = x'(t)$)
$\int f(x) dx = \int f(x(t)) x'(t) dt$ (another way of writing get the sum of f(x) given rate dx, this is the sum of the product of f(x) with x's related rate).

Anyway, it makes sense to me. I should learn the official answer but I haven't had any problems with this thinking. If it's more confusing or you want to learn the official answer, ignore it.

 

[TIBIN DANIEL BIJU]

Dear friend:
Chain rule is the rule for differentiation. we use it when we want to find the differential of a function w.r.t. another variable where the given function can be expressed as a function of the given variable.
Here acceleration is a function of velocity which is again the function of time. so we apply chain rule for differentiation according to the formula for chain rule.
hope you got the point...

 

[rcgldr]

Here acceleration is a function of velocity

For this particular example, acceleration is a function of position, which is why chain rule is being used to convert dv/dt into (dv/dx) (dx/dt).