Will the Bug Ever Reach the End of the Stretching Rubber Band?

[Nathanael]

Homework Statement

"You have a 1 km long rubber band with one end attached to the wall, and the other in your hand. The bug begins to crawl towards you on the rubber band, starting from the wall, at a rate of 1 cm/sec. As he crawls the first centimeter you extend the rubber band 1 km; when he crawls the second centimeter you extend the rubber band another 1 km, and so on, every second. The question is: Does the bug ever reach you, and if so, in how much time?"

Homework Equations

L=1km
v=1km/s
u=1 cm/s
x = distance between the bug and the wall

The Attempt at a Solution

I came up with the following equation:

dx=(u+\frac{x}{L+vt})dt

but I don't know how to solve it. I'm thinking of integrating it from 0 to T and then setting it equal to L+vT and then solving for T but I don't know how to do this (I don't have experience with differential equations)
Is it possible to get a value for T with this equation, and if so, how would I do it?

Thanks.

 

[DrClaude]

I came up with the following equation:

dx=(u+\frac{x}{L+vt})dt

Have you made a dimensional analysis to see if this equation makes sense?

"You have a 1 km long rubber band with one end attached to the wall, and the other in your hand. The bug begins to crawl towards you on the rubber band, starting from the wall, at a rate of 1 cm/sec. As he crawls the first centimeter you extend the rubber band 1 km; when he crawls the second centimeter you extend the rubber band another 1 km, and so on, every second. The question is: Does the bug ever reach you, and if so, in how much time?"

I'm not sure how to interpret the question. The blue highlighted part I interpret to mean that the rubber band is stretched continuously, while the red I interpret to mean that the rubber band is stretched instantaneously every second.

 

[Orodruin]

dx=(u+\frac{x}{L+vt})dt

Watch out here, you are adding a velocity and a dimensionless number.

I would suggest instead working with the fraction of the rubber band that the bug has behind him. When it reaches 1 the bug has arrived. The problem will then require only basic integration.

@DrClaude: In the classical formulation of this problem, the bug crawls 1 cm/s and the band is being extended at a rate of 1 km/s. The only difference is whether to consider the problem a difference equation or a differential equation.

 

[Nathanael]

Watch out here, you are adding a velocity and a dimensionless number.

Have you made a dimensional analysis to see if this equation makes sense?

Yes, it was just a typo. I meant to put dx=(u+v\frac{x}{L+vt})dt (Sorry, I should type more carefully.)
That's the ratio of the distance behind him to the entire length, multiplied by the speed of stretching, so the dimensions are now correct.

I'm not sure how to interpret the question. The blue highlighted part I interpret to mean that the rubber band is stretched continuously, while the red I interpret to mean that the rubber band is stretched instantaneously every second.

It does seem a bit contradictory. It doesn't matter though, (it's not homework), so consider it to be continuously stretched. (That's the case that my equation applies to.)

I would suggest instead working with the fraction of the rubber band that the bug has behind him. When it reaches 1 the bug has arrived. The problem will then require only basic integration.

Let's say r=\frac{x}{L+vt} then I get:

dr=\frac{u}{L+vt}dt+\frac{vr}{L+vt}dt

Then I would set the integral equal to 1.
I don't see how this helps. Maybe that's not what you meant, I'm tired. I think I'll go to bed.

 

[Orodruin]

Let's say r=\frac{x}{L+vt} then I get:

dr=\frac{u}{L+vt}dt+\frac{vr}{L+vt}dt

[STRIKE]$dx/dt$ is not equal to $u$ and the second term should have a minus sign since $L+vt$ was in the denominator. It also has a contribution from the stretch of the rubber band as you described in your first post (in fact you should be able to insert your expression for it here). An alternative is just making the change of variables $r(L+vt) = x$ in your (corrected) original equation.[/STRIKE]

Edit: I realize now that the above might not have been what you did. When you do the differentiation of the right-hand side of your substitution, remember that $L+vt$ is not a constant, but that $d(L+vt) = v\, dt$.

 

[Nathanael]

The problem is I don't know how to calculate the integral. It seems recursive. (I could write it as an infinite chain of fractions.)
x depends on the integral, which depends on x, (which depends on the integral, and so on and so forth indefinitely).

I've come across these infinite fraction chains before but I don't know what to do with them. I'm guessing it doesn't help to treat it as a continued fraction? And there's probably a simpler way to evaluate it? I just don't know how.

 

[D H]

Your first derivative, \frac{dx}{dt} = u + v \frac x{L+vt}, is fairly easy to integrate. Clearing the denominator and rearranging yields (L+vt)\frac {dx}{dt} - u(L+vt) = vx. That's a first order ODE in normal linear form. Use an integrating factor and you're done.

Your second derivative in post #4 is even easier. You computed the derivative incorrectly. With r = \frac x {L+vt}, the derivative is \frac {dr}{dt} = \frac{\dot x (L+vt) - vx}{(L+vt)^2}. See if you can take it from there.

 

[BvU]

We've had this one several times before; can't find it because I don't know how to operate the PF advanced search (help!).

there's even a Wiki on it here

 

[Nathanael]

We've had this one several times before; can't find it because I don't know how to operate the PF advanced search (help!)

Sorry about that

Your first derivative, \frac{dx}{dt} = u + v \frac x{L+vt}, is fairly easy to integrate. Clearing the denominator and rearranging yields (L+vt)\frac {dx}{dt} - u(L+vt) = vx. That's a first order ODE in normal linear form. Use an integrating factor and you're done.

Thanks for your help but I still haven't made it to the answer. I'm sure it is simple for you guys but I'm still learning. I think I will save this problem for later while I learn more mathematics.

 

[D H]

If you are in the US, you'll typically learn how to solve \frac {dx} {dt} = u + v \frac {x(t)} {L+vt} in your second or third college calculus class.

Solving for r(t) where r(t) = \frac {x(t)} {L+vt} only requires high school AP calculus. You should be able to do that. Your problem is that you made an error in calculating the derivative of r(t). You didn't use the quotient rule.

 

[Satvik Pandey]

Yes, it was just a typo. I meant to put dx=(u+v\frac{x}{L+vt})dt (Sorry, I should type more carefully.)

Could you please tell me how you got this equation?

 

[Nathanael]

Could you please tell me how you got this equation?

I assumed that the rubber band is stretched uniformly. So the "speed of stretching" of a piece of the rubber band is propotional to the length of that piece, and to the speed that the entire rubber band is being stretched.
[[[EDIT:]]]
[[[I should've said it's proportional to the ratio of the piece's length with the entire length (not just to the piece's length)]]]
(I call the speed that the entire rubber band is being stretched "v" and the length of the entire band "L")

I guess the mathematical way to say this idea (of "uniform stretching") would be dv=v\frac{dL}{L}

For example, a piece of the rubber band of length k would be stretched with a speed of v\frac{k}{L}Let's now call the position of the bug x (as measured from the wall)

The length of rubber band as a function of time is L+vt (where L is the initial length)

The speed of the bug (\frac{dx}{dt}) would be the walking speed of the bug (u) plus the speed the rubber band behind the bug being stretched.

So you get:

\frac{dx}{dt}=u+v\frac{x}{L+vt}I hope this was clear (I am sometimes bad at explaining)

 

[Satvik Pandey]

I assumed that the rubber band is stretched uniformly. So the "speed of stretching" of a piece of the rubber band is propotional to the length of that piece, and to the speed that the entire rubber band is being stretched.
[EDIT:]
[I should've said it's proportional to the ratio of the piece's length with the entire length (not just to the piece's length)]
(I call the speed that the entire rubber band is being stretched "v" and the length of the entire band "L")

I guess the mathematical way to say this idea (of "uniform stretching") would be dv=v\frac{dL}{L}

For example, a piece of the rubber band of length k would be stretched with a speed of v\frac{k}{L}


Let's now call the position of the bug x (as measured from the wall)

The length of rubber band as a function of time is L+vt (where L is the initial length)

The speed of the bug (\frac{dx}{dt}) would be the walking speed of the bug (u) plus the speed the rubber band behind the bug being stretched.

So you get:

\frac{dx}{dt}=u+v\frac{x}{L+vt}


I hope this was clear (I am sometimes bad at explaining)

Thank you Nathanael.I got that.
[

 

[Nathanael]

I haven't officially taken calculus, yet, (my high school didn't have it) and I've only read the first 5 chapters of a calculus text (strang's). I've been intending for a little while to finish reading Calculus, but I've made little progress because physics is far more fun. My understanding of calculus (however unthrorough) comes from physics problems.

Solving for r(t) where r(t) = \frac {x(t)} {L+vt} only requires high school AP calculus.

What didn't make sense to me in solving this, is that, I only know x(t) as "the solution to that other equation" so how am I supposed to integrate an ambiguous function?
[[[EDIT:]]]
[[[It makes more sense to me how it's possible, because although x(t) is ambiguous, the integral of x(t) (from 0 to the "T" which satisfies the problem) is known to be L+vT; but I still don't know how to use this knowledge]]]

If you are in the US, you'll typically learn how to solve \frac {dx} {dt} = u + v \frac {x(t)} {L+vt} in your second or third college calculus class.

I've written it as the limit of a series, but there are two problems for figuring it out. The series is:

lim_{\Delta t\rightarrow 0} [^{(k/\Delta t)}_{(n=0)}\Sigma(x_n)]=vk\Delta t+L
Where x_n=nu\Delta t+\frac{x_{n-1}}{L+nv\Delta t}

The first problem is the way I've defined k (the upper limit). It is not explicitly stated (I couldn't think of a way to write it explicitly, so I just wrote the relationship). This makes it quite confusing to deal with.

The second problem is, (as I touched on in an earlier post,) the sequence is recursive. The definition of x_n depends on x_{n-1}. This isn't theoretically a problem (because we know x_0=0) but it makes it too complicated for me. The only "solvable" (non recursive) way I can think of to write it is, as long chain of fractions (which becomes infinite in the limit).

That is beyond my capabilities, so I will just save this problem for a future day.
EDIT:
I probably should've written " k " as " T " because that's what it represents in the equation (it is the solution of the problem)

 

[Nathanael]

Thank you Nathanael.

You're welcome



P.S. (to D H)

If you are in the US, you'll typically learn how to solve \frac {dx} {dt} = u + v \frac {x(t)} {L+vt} in your second or third college calculus class.

If you don't mind, can you show me how you would solve this?
Even if I don't entirely understand, it could possibly give me something to think about.
(I like learning math from physical ideas like this, where you can actually picture what's happening)

 

[Satvik Pandey]

Diffrential of form
\frac { dy }{ dx } +Py=Q
(where P is constant and Q can be constant as well as function of y)
can be solved as

y{ e }^{ \int { Pdx } }=\int { \{ Q.{ e }^{ Pdx } } \} dx\quad +C

\frac{dx}{dt} -v\frac{x}{L+vt} =u

Can it be solved like this?

y{ e }^{ \int { \frac { v dt }{ L+vt } } }=\int { u.{ e }^{ \int { \frac { v dt }{ L+vt } } } } dt\quad +C

 

[ehild]

\frac{dx}{dt}=u+v\frac{x}{L+vt}

Nathanael, I understand that you do not want to apply Maths methods you do not know and understand yet, but you might be curious about the solution, are you not? You can cheat a bit then and look at wolframalpha.com, but first bring the equation to a simpler form, by introducing dimensionless quantities where possible.

So choose y=x/L and denote a=u/L and b=v/L. You get

\frac{dy}{dt}=a+\frac{by}{1+bt} with a= u/L and b=v/L and the initial condition y=0 at t=0.

I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0. That is a separable equation, it can be written as

\frac{df}{dt}=-P(t)f and the remaining part is fg'+Q(t)=0

I rearrange it (it does not seem legal, but it will be)

\frac{df}{f}=-P(t) dt

and integrate

\int {\frac{df}{f}}=\int{-P(t) dt}

Do not bother with the integration constant. Substitute f into the remaining part of the equation:
fg'+Q(t)=0: g'=-Q(t)/f=0
Integrate, but add the integration constant now. Substitute back into y=fg.

The method is equivalent with the integrating factor method in principle, but I never can remember that

Try, just for fun. ehild

 

[Satvik Pandey]

Was my method correct,ehild?

 

[ehild]

Yes, it can be solved by the integrating factor method, but what is∫(vdt/(L+vt))?


ehild

 

[Satvik Pandey]

Yes, it can be solved by the integrating factor method, but what is∫(vdt/(L+vt))?


ehild

Substituting (L+vt)=a.So da=vdt.

=\int { \frac { da }{ a } } =\log { a } =\log { (L+vt) }

 

[ehild]

Correct, go ahead.

 

[Satvik Pandey]

Correct, go ahead.

Let R=\log { (L+vt) }
At t=0 R= log(L)
Is it right?

 

[ehild]

Well, go on...

 

[ehild]

Take care with the minus: P=-v/(L+vt)

 

[Satvik Pandey]

Well, go on...

y{ e }^{ -log { (L+vt) } }=u\int { { e }^{ -log(L+vt) } } dt\quad +C

How to solve u\int { { e }^{- log(L+vt) } } dt\quad +C?

Should I substitute log(L+vt)?

I am not very comfortable in integrating logarithmic and exponential function.

Why we have not considered constant of ∫(vdt/(L+vt))?

 

[ehild]

What is $e^{-\log(x)}$

 

[Nathanael]

but first bring the equation to a simpler form, by introducing dimensionless quantities where possible.

So choose y=x/L and denote a=u/L and b=v/L. You get

\frac{dy}{dt}=a+\frac{by}{1+bt} with a= u/L and b=v/L and the initial condition y=0 at t=0.

I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)

I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0. That is a separable equation, it can be written as

\frac{df}{dt}=-P(t)f and the remaining part is fg'+Q(t)=0

Ok that is pretty cool, I just don't quite understand how you did f'g+P(t)fg=0? That is just for convenience? Why are you at liberty to set up that equation arbitrarily?

I rearrange it (it does not seem legal, but it will be)

\frac{df}{f}=-P(t) dt

and integrate

\int {\frac{df}{f}}=\int{-P(t) dt}

Do not bother with the integration constant.

Why do we not bother with the integration constant? Is it because it will "mix" with the integration constant in the next part?

Integrating both sides I get ln(f)=ln(bt+1) and thus (taking e^ to both sides) f=bt+1? (This doesn't seem quite right)



Thank you very much for your reply ehild! It was very interesting to see how it's done I have no idea how you looked at the equation and this is what came to mind!
You are brilliant (if not for your method, then for your helpfulness)!

(I went through the whole solution but I made a mistake somewhere, because my answer was wrong.)
(I am too tired to do it again tonight, (it's 1:30am) but I will try again, more carefully, tomorrow.)

 

[ehild]

I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)Ok that is pretty cool, I just don't quite understand how you did f'g+P(t)fg=0? That is just for convenience? Why are you at liberty to set up that equation arbitrarily?

Yes y=x/L is the same Orodruin suggested.
fg=y, but you can multiply one of them with something and divide the other with the same something. You choose f so that f'g+P(t)fg=0.

Why do we not bother with the integration constant? Is it because it will "mix" with the integration constant in the next part?

Yes, one integration constant is enough, and you include it in the next part. But you may add an integration constant in the first part. It might be useful if you get some logarithmic function.

Integrating both sides I get ln(f)=ln(bt+1) and thus (taking e^ to both sides) f=bt+1? (This doesn't seem quite right)

Yes, f=1+bt. Why do you think it is not right?

(I went through the whole solution but I made a mistake somewhere, because my answer was wrong.)
(I am too tired to do it again tonight, (it's 1:30am) but I will try again, more carefully, tomorrow.)

OK, have a good night. ehild

 

[Satvik Pandey]

What is $e^{-\log(x)}$

${ e }^{ -log(x) }=\frac { 1 }{ x }$

Using this

$\int { { e }^{ -log(L+vt) } } dt=\int { \frac { dt }{ L+vt } }$

Let $L+vt=a$.
So $dt=\frac { da }{ v }$

$\frac { 1 }{ v } \int { \frac { da }{ a } } =\frac { \log { (L+vt) } }{ v }$

$\frac { x }{ (L+vt) } =\frac { u\log { (L+vt) } }{ v } +C$

 

[Satvik Pandey]

$x=(L+vt)\frac { u\log { (L+vt) } }{ v } +C$

At $t=0 x=0$

$0=\frac { Lu log(L) }{ v } +C$

As $L=1$

So $log(1)=0$

So $C=0$

Is it right?

 

[ehild]

$x=(L+vt)\frac { u\log { (L+vt) } }{ v } +C$

don't you miss parentheses?

ehild

 

[ehild]

I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

But it is not bad that you have understood how to solve a linear first-order differential equation.

ehild

 

[Nathanael]

Substitute f into the remaining part of the equation:
fg'+Q(t)=0: g'=-Q(t)/f=0
Integrate, but add the integration constant now. Substitute back into y=fg.

So:

g=\frac{a}{b}ln(bt+1)+C

And:

y=(bt+1)(\frac{a}{b}ln(bt+1)+C)

From the initial condition we see that C=0

Replacing a, b, and y, with their respective definitions (and multiplying by L) I get:

x=(vt+L)(\frac{u}{v}ln(\frac{vt}{L}+1))=L+vt

Which yields:

t=e^{100000}-1

Which is the correct answer[Edit: much easier when I'm not tired hehe ]

 

[Nathanael]

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

It's amazing how you guys have such an intuitive grasp of differential equations. It is very impressive to me, I don't understand it at all.

But it is not bad that you have understood how to solve a linear first-order differential equation.

I would hardly say I've understood it! I would have never done that in a million years! I only "understood" because you were nice enough to walk me through it like a baby



I have one more question(s) about this method:

I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0.

Does this always work? (For first order linear DE's)
If not then how often does it work?

Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?
(Or do you have to be clever about it?)

 

[ehild]

I have one more question(s) about this method:


Does this always work? (For first order linear DE's)
If not then how often does it work?

This method always works for linear first order differential equations, just like the integrating factor method.

Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?

That belongs to the method. As you wrote, '"it is the trick" of the method.

Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

 

[Nathanael]

Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

\frac{dy}{dt}=\frac{\dot x}{L+vt}-\frac{xv}{(L+vt)^2}

(With initial value y=0 when t=0)
Which could be written as:

\frac{dy}{dt}=\frac{\dot x-yv}{L+vt}

(I got stuck here for a few minutes but then thought of this )
Maybe I should plug in \dot x=u+vy and then I get:

\frac{dy}{dt}=\frac{u}{L+vt}

\int dy=\int \frac{u}{L+vt}dt=\frac{u}{v}ln(L+vt)+C=y

From our initial condition we see that C=0

Now we just set y=1 and solve for t!

ln(1+t)=100000

t=e^{100000}-1

 

[ehild]

Is not Maths fun?

I would have started it with x=y(L+vt)
x'=y'(L+vt)+yv. Substituting back into the original equation:
y'(L+vt)+yv=u+yv--->y'(L+vt)=u---> y'=1/(L+vt)

 

[Nathanael]

Is not Maths fun?

Very fun! But not as fun as physics! I have no idea why I suddenly understood how to do it that way. All you did was say "now what about orodruin's hint"

It feels much better when someone doesn't have to walk you through every single step
(even though I would've never done it without orodruin's hint!)

 

[Orodruin]

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

It's amazing how you guys have such an intuitive grasp of differential equations. It is very impressive to me, I don't understand it at all.

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

 

[ehild]

Very fun! But not as fun as physics!


I have no idea why I suddenly understood how to do it that way. All you did was say "now what about orodruin's hint"

It feels much better when someone doesn't have to walk you through every single step
(even though I would've never done it without orodruin's hint!)

Look at the previous post, I edited it.

Sometimes you need to be guided, but next time you will figure it out by yourself. If you practice enough, your brain will find connections among things you have learnt, and suddenly something new occurs to you that you never heard or read before : You discover something.

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

You set up a model, do the Maths and predict the outcome of an experiment. It was always a great surprise and also a great pleasure for me when my measurement results agreed with theory.

ehild

 

[ehild]

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

Bug in an expanding Universe ... Amazing! Never would have occurred to me, but I never taught Relativity Theory.

ehild

 

[Nathanael]

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

Thanks orodruin! Thank you ehild too!

I wish to give you both more thanks, but, apparently, I've given too many in the past 24 hours (I've only given like 5 )

Look at the previous post, I edited it.

That works too. Both ways are good

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

I've always liked math, long before I knew what physics was, but the main thing I enjoyed was word problems! I think that is why I enjoy physics; it's essentially mathematical word problems
To me, the fun part of math is figuring out how to figure it out, and that's why I find physics much more fun. Physics, to me, is all the fun parts of math without the boring parts! (Maybe one day I'll develop more of a liking for pure math.)

It was always a great surprise and also a great pleasure for me when my measurement results agreed with theory.

I bet!


Thanks for helping me out ehild, you're awesome




Edit:

I have been known to give students similar problems on relativity exams ...

but I never taught Relativity Theory.

So you both are or have been teachers? (Or professors, sorry if it's rude to say teacher )
I think I might want to be a teacher when I'm older (if I have what it takes) that would be a fun job!

 

[Orodruin]

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

You know, Feynman put it the other way around in slightly other words ...

Also, while on the subject:

 

[ehild]

You know, Feynman put it the other way around in slightly other words ...

May I ask, what he said? Somehow I am not fond of Feynman, so I am not familiar with his sayings.

ehild

 

[ehild]

I think I might want to be a teacher when I'm older (if I have what it takes) that would be a fun job!

I am sure you will become a very good teacher.

ehild

 

[Orodruin]

May I ask, what he said? Somehow I am not fond of Feynman, so I am not familiar with his sayings.

ehild

Something to the effect of "Physics is to mathematics what sex is to masturbation" if I don't misremember.

But it is good that everyone has different preferences or physics jobs would be even scarcer than they are now ...

 

[ehild]

I think I start to like Feynman

ehild

 

[Nathanael]

Something to the effect of "Physics is to mathematics what sex is to masturbation"

I think I start to like Feynman

 

[ehild]

Well, we are a bit offtopic. Better to stop it before we get infractions.

ehild

 

[Satvik Pandey]

Thank you ehild,Orodruin,Nathanael I got the answers too.

Thank you very much Nathanael for posting this question.