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@ArcticChar Thanks, I've seen that the answer was downvoted so I thought it can be motivated by a controversial on the question which I'm not informed about. Have you any other informations about that? Thanks

@ArcticChar Then what is your opinion about an answer with integral test?

@ArcticChar Why was the question edited in this form?

@SineoftheTime I mean that according to the question as it is now, also an answer by integral test or condensation test would be fine, I don't understand why it has been changed in this form.

Yes indeed in this form, also an answer by integral test would be fine. I've also noted there is already an answer by condensation test.

@AlixBlaine When you consider any function you must define it before you can start to make any consideration. Trigonometric functions can be definied in different ways.

@AlixBlaine $\sin^2(25.96°)+\cos^2(25.96°) = 0.191619... + 0.808381... =1.0000...$

@AlixBlaine If you square you get one adding them

@AlixBlaine What is your definition for trigonometric functions?

@AlixBlaine this is wrong $\cos^2(25.96°) = 0.899099868$ we have $\cos(25.96°) = 0.899099868$

refer to en.wikipedia.org/wiki/Trigonometry#Identities

@AlixBlaine There is nothing to prove, simply use definition + Pythagorean theorem

The foundamental identity $\sin^2v + \cos^2v = 1$ is true by definition and Pythagorean theorem.

@JohnBentin Ah ok! Now your point is more clear to me. Indeed I'm assuming it as a given, more geometrically than analytically. From my side I think your answer would be useful to illustrate a different style for the presentation.

Thanks a lot for your suggestions, really appreciated. I'll take a closer look to your version and mine to make thinks more clear and well presented.

For the part "new variable should be defined in terms of the original variables", doesn't it suffice set $a=\rho \cos \theta$ and $b=\rho \sin \theta$?

I'm not native in english, so please forgive me!

That beeing said, if you think there is something wrong in my presentation I'm going to fix it! Thanks

@JohnBentin Nothing personal with your edit, it is just a matter of personal preference I think and personal style. I try to make things simpler as possible, for that reason I don't like to define$f(a,b)$ or set $\theta=\arccos a/\rho$, I like introduce the idea of polar cordinates more directly, and using the fact that the 2 expressions are equivalent. Really nothing personal and your presentation was very fine.

@JohnBentin Sorry, I would be grateful for some more explanation if you can now or later. Bye

@JohnBentin I'm considering a chain $A\!\iff\! B\!\iff\! C \!\iff\! D$ to deduce $A$ form $D$. Sorry I don't get the point, but if you can explain that I can fix it. Thanks

all steps are "$\iff$"

@JohnBentin I'm continuing from the previous expression $(a + \sqrt{a^2 + b^2})^3-6ab^2>0$, isn't it clear in this way?

It’s a little bit heavy but I let it in this way for the moment. Thanks

@ThomasAndrews Exactly, I mean the case $a<0$ with $b=0$ for the original expression! Thanks, I fix that.

@Semiclassical I agree, I would say indeed $\Delta \tau = \kappa \cdot \Delta \theta$. I've also explained that in a comment. Maybe it is not clear in the answer, I try to make it clearer.

for here we can find all the result, also the fact we start form 8 turns (that is (=0.30/0.375)

"should be "the torque at the start of the first turn is 0.3 N-m, while the torque at the end of the last turn is 0.45 N-m""

@SirMrpirateroberts The part not well stated is the one indicated by Semiclassical

@Semiclassical Interesting fact! I wasn't aware of that practical issue! Also springs don't follow Hooke's law for "large" displacement ;)

@SirMrpirateroberts I hope it is clear now! You don't need moment of inertia (which can be anything) nor also frquency nor others things.

@Semiclassical I agree with you!

@Semiclassical Yes we have already understood this point, looking at the result given for "c"

Yes it is equivalent determine the work, maybe point "c" in the book is at the end because they are expecting we calculate energy using work (but it is a completely equivalent approach)

I've updated the answer with all the results

that is 16pi

the initial theta is 0.30/0.375=8 full turns

Of course! They are asking for the energy stored for the 4 turns and this is found subtracting the initial energy to the final total energy, which gives the result.

Maybe I've found the point!

Definitively I think there is a typo for the answer given for "a", it is in contrast with "c"

For point "a" we obtain a similar answer using "18 pi" for the angle, but given the value for K, to have a final torque of 0.45 Nm we need 12 full turns

I'm trying to find out what's going on

Yes I agree it should be given as a first result. Maybe they are using another way to find mechanical energy in "a"

If you add the full question in your post I can reorganize a full answer

The power is easily obtained dividing the consumption of energy in time

But with this value we need to consider the total energy to solve point "a"

That is K=(0.0375 Nm)/(2 pi)=0.00596831...

The value for "c" is obtained taking 0.45 Nm as tha final torque value.

What are proposed answers for "b" and "c"?

Could you provide a screenshot of the exercise form the book?