How to Apply Derivatives in Physics Problems?

[Heisenberg7]

Homework Statement

Below

Relevant Equations

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I would like to discuss a few ways to apply derivatives in physics (I don't understand it fully). I don't need a full solution, I only need to understand how to successfully apply the derivatives
First example,

Thin insulating ring of mass M, uniformly charged by charge $+q$ has a small cut of length $dl$ ($\Delta l$) (look at the photo). The ring is placed in a horizontal plane and it can rotate about the vertical axis that passes through the point O(its center). In the initial instant, the ring is at rest. Then we "activate" an electric field $E$ which is normal to the straight line $OA$. What maximum velocity will the ring reach?

My attempt: We can look at this ring as a superposition of charge $-dq$ and a full ring of charge $+q+dq$. Obviously the full ring of charge $+q+dq$ can't move so there will only be torque due to the charge $-dq$. Here I ran into some issues. Since we're talking about maximum velocity it's fair to assume that we're going to have derivatives. But, how? The solution in the book states that the velocity will reach its maximum when torque is equal to zero (the point at which the point A will have moved 90 degrees to the left). Then they set up a conservation of energy equation and solved for $v$. Would there be a nice way to apply derivatives, or is it enough to say the velocity will reach its maximum when torque is equal to zero?

Second example,
This one was on my previous post. I would like to know why this works. Why is it possible for us to apply a derivative in this case? Here's the full solution:

The problem statement: A funny car accelerates from rest through a measured track distance in time T with the engine operating at a constant power P. If the track crew can increase the engine power by a differential amount dP, what is the change in the time required for the run?

I'm sorry for including 2 problems in one post. I would like to mention again that I don't need a full solution, just an explanation on why this works(2nd one) and can it work(1st one, also a way to setup that equation)?

 

[Heisenberg7]

(1st example) I guess it's also reasonable to make a comparison with for example a ring with a cut in a gravitational field. Then it makes sense (more intuitive).

 

[kuruman]

Would there be a nice way to apply derivatives,

There is and it's called Newton's second law. Energy conservation is the most expedient way to tackle this problem. It is equivalent to the problem where a pendulum starts with its string in the horizontal position and you are asked to find the angular speed at the lowest point of the motion.

Although the statement of the problem does not explicitly say so, it must be that the center of the ring is constrained from accelerating to the right.

I'm sorry for including 2 problems in one post. I would like to mention again that I don't need a full solution, just an explanation on why this works(2nd one) and can it work(1st one, also a way to setup that equation)?

Please don't do that. Threads should be kept separate to avoid the confusion of who is answering what.

 

[Steve4Physics]

Relevant Equations: Below
Thin insulating ring of mass M, uniformly charged by charge $+q$ has a small cut of length $dl$ ($\Delta l$)

Best to avoid calling the length of the negative charge ’$dl$’ as that refers to an infinitesimal length. Just call the length, say, $L$ (where $L \\\ll$ ring’s radius).

Since we're talking about maximum velocity it's fair to assume that we're going to have derivatives.

No necessarily. Here you can use energy: max. KE gained = max. PE lost.

But, how?

If you are dead-set on using derivatives, you could set up a differential equation for angular displacement as a function of time, $\theta(t)$. Solve for $\theta(t)$; then the angular velocity as a function of time is $\dfrac {d\theta}{dt}$ - so you can see when it is zero.

If you are not yet used to using calculus to solve simple harmonic motion problems, look up some examples.

The solution in the book states that the velocity will reach its maximum when torque is equal to zero (the point at which the point A will have moved 90 degrees to the left). Then they set up a conservation of energy equation and solved for $v$. Would there be a nice way to apply derivatives

If you want to use derivatives, see above. But using onservation of energy simplfies things here.

, or is it enough to say the velocity will reach its maximum when torque is equal to zero?

It's enough if you understand why it is true!

... I would like to mention again that I don't need a full solution, just an explanation on why this works(2nd one) and can it work

The detailed solution is given. If you don't understand it, you need to be specific about which part(s) you don't understand.

(1st one, also a way to setup that equation)?

What do you meant by 'that equation'? The two problem are quite different - you can't use equations relating to the second problem to solve the first problem.

 

[Heisenberg7]

There is and it's called Newton's second law. Energy conservation is the most expedient way to tackle this problem. It is equivalent to the problem where a pendulum starts with its string in the horizontal position and you are asked to find the angular speed at the lowest point of the motion.

Ah, yes. Thought someone would mention this. Well yes, I guess so. But, I just wanted to know if there was a nice way to solve this using derivatives. Energy conservation is the way to go then.

 

[Heisenberg7]

Best to avoid calling the length of the negative charge ’dl’ as that refers to an infinitesimal length. Just call the length, say, L (where L≪ ring’s radius).

Oh, okay. It's what they used in the book, so I simply went with it.

The detailed solution is given. If you don't understand it, you need to be specific about which part(s) you don't understand.

I'm not sure about the last part where it takes the derivative of the equation, $$PT^3 = \frac{9}{8} m L^2$$ Why would one take a derivative in this case? I mean sure, it could be that they just want to get that infinitesimal time $dt$, but is there anything more behind the reasoning?

What do you meant by 'that equation'? The two problem are quite different - you can't use equations relating to the second problem to solve the first problem.

I'm sorry I didn't make it clear. I simply wanted to know if it's possible to set up a nice derivative and solve it using derivatives.

 

[Steve4Physics]

I'm not sure about the last part where it takes the derivative of the equation, $$PT^3 = \frac{9}{8} m L^2$$ Why would one take a derivative in this case? I mean sure, it could be that they just want to get that infinitesimal time $dt$, but is there anything more behind the reasoning?

I’m not entirely sure what the problem is but this might help and hopefully is not overkill...

First, note that $dP$ and $dT$ is infinitesimals. $\delta P$ and $\delta T$ represent small changes. So the question should really ask:

“If the track crew can increase the engine power by a small amount $\delta P$, what is the corresponding change in the time, $\delta T$, required for the run?”

A small change in power, $\delta P$, will result in a small change in run-time, $\delta T$. We need an equation which relates $\delta T$ to $\delta P$.

Note that $\frac {\delta T}{\delta P} \approx \frac {dT}{dP}$.

It turns out that for small enough changes, we can replace $dT$ with $\delta T$ and replace $dP$ with $\delta P$.

Since the result of differentiating here is
$dP T^3 + 3PT^2 dT = 0$
it follows that
$\delta P T^3 + 3PT^2 \delta T \approx 0$

So we now have the (approximate) relationship between $\delta T$ and $\delta P$ that the question required. To get this we used the results of differentiation.

I'm sorry I didn't make it clear. I simply wanted to know if it's possible to set up a nice derivative and solve it using derivatives.

You have to treat problems on a case-by-case basis. Sometimes it maybe appropriate to use derivatves - sometimes not. And the way derivatives are used will depend on the problem. Often a differential equation and its solution are required. It's a sets of skills you build up by doing lots of practice problems and studying worked examples.

 

[Heisenberg7]

I’m not entirely sure what the problem is but this might help and hopefully is not overkill...

First, note that $dP$ and $dT$ is infinitesimals. $\delta P$ and $\delta T$ represent small changes. So the question should really ask:

“If the track crew can increase the engine power by a small amount $\delta P$, what is the corresponding change in the time, $\delta T$, required for the run?”

A small change in power, $\delta P$, will result in a small change in run-time, $\delta T$. We need an equation which relates $\delta T$ to $\delta P$.

Note that $\frac {\delta T}{\delta P} \approx \frac {dT}{dP}$.

It turns out that for small enough changes, we can replace $dT$ with $\delta T$ and replace $dP$ with $\delta P$.

Since the result of differentiating here is
$dP T^3 + 3PT^2 dT = 0$
it follows that
$\delta P T^3 + 3PT^2 \delta T \approx 0$

So we now have the (approximate) relationship between $\delta T$ and $\delta P$ that the question required. To get this we used the results of differentiation.


You have to treat problems on a case-by-case basis. Sometimes it maybe appropriate to use derivatves - sometimes not. And the way derivatives are used will depend on the problem. Often a differential equation and its solution are required. It's a sets of skills you build up by doing lots of practice problems and studying worked examples.

Alright, thanks for the feedback!

 

[berkeman]

Second example,
This one was on my previous post. I would like to know why this works.

Please don't do that. Threads should be kept separate to avoid the confusion of who is answering what.

Yes, please do not do this. Questions should not be cross-posted across multiple threads, for the obvious reasons.

 

[haruspex]

Would there be a nice way to apply derivatives, or is it enough to say the velocity will reach its maximum when torque is equal to zero?

Usually it’s the same thing:
$I\dot\omega=\tau$
For a local maximum of $\omega$ we set $\dot\omega=0$, which will be if and only if $\tau=0$.
But there are exceptions.
When the derivative is zero it could be a minimum or an inflexion point, e.g. $\tau$ has been increasing, is momentarily 0, then increases again, as in $\tau=kt^3$.
Or the maximum occurs beyond the range of times considered.
Or there is a discontinuity in the derivative, whereby it switches from positive to negative without ever being zero.

For the second problem:

I'm not sure about the last part where it takes the derivative of the equation,

The task is to find how a small change in one variable affects a dependent variable. That is what derivatives do.