You are Here: Home

# Official TSR Mathematical Society

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Find out how cards are replacing warnings on TSR...read more 03-12-2013
1. Re: Official TSR Mathematical Society
(Original post by olipal)
I didn't agree with that euither - so I just decided to copy and paste what he said to me.

This isn't my writing but my friends; please just help him out as I am also intriguured by the answer!!!!!!!!!!
2. Re: Official TSR Mathematical Society
(Original post by GreenLantern1)
I didn't agree with that euither - so I just decided to copy and paste what he said to me.

This isn't my writing but my friends; please just help him out as I am also intriguured by the answer!!!!!!!!!!
Sorry, it was a typo, I meant the gradient of x = 5 is undefined and the y=3 has gradient 0. My mistake
Last edited by usycool1; 02-05-2012 at 22:44.
3. Re: Official TSR Mathematical Society
I wonder if I interrupted you guys.

I've asked this once in the Physics forum but I personally think that this is something about maths as well. I don't know if this is interesting but I guess it will be very challenging:

Pendulum we meet always have a rigid rod, what if it has a deformable string?

My approach is like this: I will set a Lagrangian equations like what people did in solving complex pendulum. The deformable rod will be treated as a (very) complicated pendulum with numerous bob with equal mass on it.

As string can't deform a lot (There is always curvature when you fold a rod), so my solution to the limit of deformation will be modifying the equation so that the bending of that particular point of the rod will have a very high damping, (like , k will be a constant depending on material and is the bending angle to the normal of that particular point of the rod.)

So when I can set up an equation, I can gradually increase the points to a very large number and the motion of a pendulum will be simulated.

Anyway to add the work done against friction into Lagrangian mechanics? And any easier method to do this simulation?

-Agos
4. Re: Official TSR Mathematical Society
(Original post by GreenLantern1)
I only posted this for my friend not myself (so it has just been copied from what he has written to me).

(Original post by usycool1)
Sorry, it was a typo, I meant the gradient of x = 5 is undefined and the y=3 has gradient 0. My mistake
Cool typo.

Take two lines, and , and draw them in a similar way as in the picture below.

Notice that is the gradient of , and is the gradient of . So, the two lines can be expressed in terms of these angles.

These will be perpendicular if , so you'd use .

Spoiler:
Show

Substituting this, you obtain by co-function identities.

Therefore, the product of their gradients is for a rotation of any angle . Translations also do not change anything.

In other words, any two perpendicular lines are rotations and translations of the and axes, and the product of gradients of those two is .
5. Re: Official TSR Mathematical Society

Cool typo.

Take two lines, and , and draw them in a similar way as in the picture below.

Notice that is the gradient of , and is the gradient of . So, the two lines can be expressed in terms of these angles.

These will be perpendicular if , so you'd use .

Spoiler:
Show

Substituting this, you obtain by co-function identities.

Therefore, the product of their gradients is for a rotation of any angle . Translations also do not change anything.

In other words, any two perpendicular lines are rotations and translations of the and axes, and the product of gradients of those two is .

Well done, but it still doesn't answer thw question of those lines in specific that were asked with. So unless youa re related to the french mathematician Jacques Hadamard why are you not answering the qesation.

P.S It isn't a bad excuse I copy and pasted his question on to here. If you don't believe me go have a look on the Current Year 11 thread. You are an odd one aren't ypou.
6. Re: Official TSR Mathematical Society
(Original post by GreenLantern1)
Well done, but it still doesn't answer thw question of those lines in specific that were asked with. You are an odd one aren't ypou.
All you had to see is that you can change coordinates to and , where the expression for the product of gradients of and axes makes sense.
Anyway, I now see that you are not looking for an answer, but for an argument, and I am not interested in this particular one. Sorry!

I have always been an odd one, does that make a difference?
7. Re: Official TSR Mathematical Society

I have always been an odd one, does that make a difference?
I think "odd one" was a euphemism for "****".
8. Re: Official TSR Mathematical Society
(Original post by nohomo)
I think "odd one" was a euphemism for "****".
Then I guess it makes a difference.
9. Re: Official TSR Mathematical Society
(Original post by GreenLantern1)
Posted for Usycool:

When perpendicular lines meet, the product of their gradient is -1, right?

However, say if we have the lines y = 3 and x = 5.

The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
The rule is probably only true for finite gradients.

Last edited by raheem94; 03-05-2012 at 19:46.
10. Re: Official TSR Mathematical Society
What was Lebesgue's favourite play?
Spoiler:
Show
Measure for Measure
11. Re: Official TSR Mathematical Society
(Original post by ben-smith)
What was Lebesgue's favourite play?
Good one. Have you come across this before?

"If you've got a hairy line integral around a simple curve, magic things happen." — APM
Spoiler:
Show

"I could have made it even hairier, but after a while I decided enough was enough." — APM
Spoiler:
Show

"An integral like nothing I've ever seen before, but... courage!" — APM

12. Re: Official TSR Mathematical Society
13. Re: Official TSR Mathematical Society
(Original post by GreenLantern1)
Posted for Usycool:

When perpendicular lines meet, the product of their gradient is -1, right?

However, say if we have the lines y = 3 and x = 5.

The gradient of the y=3 in undefined, and the gradient of the x = 5 is 1.

So is this not an exception for the rule that perpendicular lines's gradients have a product of -1? Or is their another explanation for this?
Bit late to the party but, why do you think this poses a problem for the rule?

Gradient of x = 5 is 0.

Let's assume for a second we don't know the gradient of y = 3, but we do know our lines are perpendicular:

So 0 times what gives -1? There's no value that satisfies so m2 is undefined, as we already know it is.
Last edited by mimx; 11-05-2012 at 16:42.
14. Re: Official TSR Mathematical Society
(Original post by mimx)
Bit late to the party but, why do you think this poses a problem for the rule?

Gradient of x = 5 is 0.

Let's assume for a second we don't know the gradient of y = 3, but we do know our lines are perpendicular:

So 0 times what gives -1? There's no value that satisfies so m2 is undefined, as we already know it is.
Very interestiing! If you can ask your maths teacher. I am on study leave so won;t see them till next year
15. Re: Official TSR Mathematical Society
quick question

by using the substitution t=tanx evaluate the definite integral 1/(9cos^2x - sin^2x) dx with the limits being pi/4 and 0

will pos rep whoever gives me the answer. thanks
16. Re: Official TSR Mathematical Society
(Original post by Entrepreneur123)
quick question

by using the substitution t=tanx evaluate the definite integral 1/(9cos^2x - sin^2x) dx with the limits being pi/4 and 0

will pos rep whoever gives me the answer. thanks
Before making the substitution, divide through by . Then the substitution and the rest is very easy.
17. Re: Official TSR Mathematical Society
i don't know if this is the right place to ask this

anyway, on my calculator, how to i convert decimals to pi?? i need this for an upcoming maths exam

i have a casio fx-115ms calculator, i've searched online and found no answers to this. my friend's calculator is able to toggle between decimals and pi using the same button to convert decimals to fractions (which on her calculator is the s<=>d button), yet the equivalent button on my calculator does not do this??

my brother's calculator (casio fx-83es) is able to toggle between decimals and pi/fractions using this button while on the mathio mode, i'm don't see this mode in my calculator (unless it's under a different name and i'm missing it)

thanks, sorry again if this isn't the right place to ask
18. Re: Official TSR Mathematical Society
guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?
19. Re: Official TSR Mathematical Society
(Original post by poyyo)
i don't know if this is the right place to ask this

anyway, on my calculator, how to i convert decimals to pi?? i need this for an upcoming maths exam

i have a casio fx-115ms calculator, i've searched online and found no answers to this. my friend's calculator is able to toggle between decimals and pi using the same button to convert decimals to fractions (which on her calculator is the s<=>d button), yet the equivalent button on my calculator does not do this??

my brother's calculator (casio fx-83es) is able to toggle between decimals and pi/fractions using this button while on the mathio mode, i'm don't see this mode in my calculator (unless it's under a different name and i'm missing it)

thanks, sorry again if this isn't the right place to ask
not all the calculators have that ability. but out of the interest remember to use 3.1416 as pi. it helps accuracy.
20. Re: Official TSR Mathematical Society
guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?
What's the game?

## Step 2: Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank

this is what you'll be called on TSR

2. this can't be left blank

never shared and never spammed

3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. By completing the slider below you agree to The Student Room's terms & conditions and site rules

2. Slide the button to the right to create your account

You don't slide that way? No problem.

Last updated: November 30, 2013
Study resources