2:56 AM
Sir @JohnRennie , for small angles, both F=-kx and Torque=-kx(theta) should be applicable for a pendulum right?

1 hour later…
4:19 AM
@Swan Yes.

1 hour later…
5:34 AM
@JohnRennie Thanks sir! I'm assuming both the equations would be applicable for all SHMs but the Torque=K(Theta) would just give 0=0 for most cases.
5:49 AM
You can use whichever is most convenient for your problem.
I find the angle version τ = kθ is usually simpler.

1 hour later…
6:56 AM
@JohnRennie Thanks again:-)
OK :-)

3 hours later…
9:30 AM
Hi sir @JohnRennie
@sanya Hi :-)
@JohnRennie I got the same.
OK :-)
Sir how to find the range of a 6 deg polynomial function?
What's the range? The difference between the minimum and maximum values in some interval?
9:37 AM
I guess so. Its like the function's outputs on the y axis for all the x that is in its domain
In general that's a hard problem because when you differentiate to find the minima and maxima you'll get a 5th order polynomial and there is no simple way to find the roots of 5th order polynomials.
If you get asked this in a question there must be a trick for doing it simply.
Can you post the question this came from?
I think there might be some simplification
Heres the que
Yes
That's a cubic in disguise :-)
I tried taking x^2 as t but I dont even know how to find the range of a cubic..
Set y = x² and it turns into a cubic.
9:41 AM
Yes so it becomes y^3 + 2y² +3y+1
which could be simplified to (y+1)^3 - y² but how to approach it further?
Hmm, it would be easy if the cubic was in the numerator. Differentiating gives a quadratic and that's easy to solve.
What is the range of 𝑥?
We cant use differentiation since its not been taught in maths yet
Ah, OK. What is the range of 𝑥?
R I believe
-∞ < 𝑥 < ∞?
9:45 AM
Yes
All real numbers
Well at |x| ⟶ ∞ the function goes to zero. Yes?
Because 1/∞ = 0
Yeah
And if the polynomial has any real roots the the function will go to ∞ at those roots.
Yes
in fact it will go to ±∞ either side of the root.
So if there are any roots the range of 𝑦 will be ±∞.
If there aren't any roots that's a lot harder ...
In the polynomial we get only even powers of 𝑥, and there are no negative terms, so the polynomial cannot ever be negative.
9:50 AM
But we restrict our domain from the values of x where the function becomes undefined I.e the denominagor becomes 0. So I think the domain of x being all R is also wrong if it has roots
@JohnRennie Oh so it wont be 0 ever
@sanya Well it could be zero. But it can't ever go less than zero.
The trouble is that to establish if it could be zero you need to find its minimum value and I don't know how to do that without differentiation.
@JohnRennie How would it be zero?...
As u say it has even powers so wouldnt the addition always be one or greater?
Aha, yes, true. So the minimum value is when x = 0 i.e. y(x) = 1
Is this argument valid? Since the function can be one or greater the range would be 7/3 or less But greater than 0 ? Or would I have to prove it more..formally?
It seems fine to me. We have found that the max and min of the polynomial are ∞ and 1, so the max and min of the reciprocal are 1 and 0.
So the range is +⁷⁄₃ to 0.
10:00 AM
OK :) Thankyou. dont know why I didnt notice the function closely. I just assumed itd be difficult because its 6th deg
If a question looks impossible you can be sure there's a cunning trick for solving it :-)

5 hours later…
3:04 PM
@BoxBoxBoxBox Gonna attend your CP lecture Sir orz