@JohnRennie how do I find the largest $x$ so that the approximation $sin(x) \approx x - \frac{1}{6}x^3$ is accurate to about 1%?

Well $\sin(x) = x -x^3/3! + x^5/5! - x^7/7! ...$

So $\sin(x) - (x - x^3/6) = x^5/5! + O(x^7)$

The question asks for 1% accuracy, not an absolute accuracy of 0.01

$$ \frac{x^5/5!}{\sin(x)} \le 0.01 $$

so $\frac{x^5/120}{sin(x)}\leq 0.01$?

ah

then just play around with a calculator?

I can't think of a way to solve that apart from using a calculator

Or I'd use an Excel spreadsheet.

yeah I don't think I'd want to even if there was a way to solve it

ah yeah

I'll use a sheet

@JohnRennie I don't really understand what this Q is asking. What does it mean by being "included" in the polynomial?

The plots aren't showing up for me unfortunately, so I'm not sure if that would make things clearer.

I'd guess it means the polynomial is $\sum_{i=0}^N x^{2i+1}/(2i+1)!$ and you need to find $N$.

e.g. if the polynomial approximation was $f(x) = x - x^3/3!$ then $N=1$

@JohnRennie We'd need to find $N$ so that...?

I mean, what's the condition here?

I guess I need to see the plot?

If you looked at a graph of $\sin(x)$ and $f(x) = x - x^3/3!$ then the two lines would start to diverge where $f(x)$ ceased to be a good approximation. Suppose you didn't know $f(x)$, then from the value of $x$ where the divergence became significant you could work out how many terms were in $f(x)$.

That's what the question is asking you.

The trouble is that if you can't see the graphs you can't tell what value of $x$ they diverge.

yeah...

I'll have to ask for the plot

that makes sense though

@JohnRennie can we get in trouble for being cavalier with differentials? like doing $dU = TdS - PdV \to \Delta U = T \Delta S - P \Delta V$?

Yes, but that usually doesn't stop us :-)

haha

@JohnRennie Hello sir :-)

@Jasmine hi :-)

Is it true that work energy theorem isnt applicable for system of particles

I think that's too general to answer. As far as I know the work energy theorem always applys, though if you have friction present that causes complications.

C is incorrect and thats the answer but I think 1 is also incorrect as I have seen examples where they applied work energy theorem on system of particles as well

That's a weird one. I would say both A and C are incorrect, but the question implies there is only one incorrect statement.

@JohnRennie yes..

It kind of depends on your definitions. D could be regarded as untrue.

It depends on whether you count fictitious forces.

@JohnRennie why so

In a non-inertial frame the velocities of particles change even though no force is acting on them.

@JohnRennie to apply WET on non-inertial frames we have to apply pseodo force

Yes, I agree, and that's why I think the question is ambiguous.

@JohnRennie oh got your point

Basically it's a rubbish question and I wouldn't worry too much about it.

@JohnRennie Ok :-)

Did you get anywhere with that atom magnetic field question from yetsreday?

I have been thinking about it but I can't see any way the answer could be 32, unless I've completely misunderstood the question.

@JohnRennie yes, what I was doing was completely wrong as an electron cant have force acting on itself due to its field

I calculated by other method and got 4:1

I asked my friends to solve it and they also got 4:1

It's not just me then :-)

This was the formula I used and got 4:1

@JohnRennie Ohh then thats for sure that there is a key mistake :-)

@JohnRennie need help with Saurav's statement

@JohnRennie Sorry , I just noticed you are busy in another room I will better catch you in night or tomorrow with few more mechanics questions

:-)

Hey @JohnRennie I have got a question - not from a book, but my own. Imagine a straight rod being pulled by some external force resulting in a constant velocity $v$. This rod is present in a space where a magnetic field $B$ exists. Certain emf will generate in the rod, since there is change in flux because of the moving rod.

Now lets suppose that $B$ is not constant, but rather a function of both $x$ and $y$ where $x$ represents the distance from the origin on the x axis, and $y$ represents distance from origin along y axis

How would you try to find the emf generated at the ends of rod who is moving in such an environment?

$B(x,y)$ is a known function.

I have to go now I'm afraid. I might be back later.

OK!