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    (Original post by olipal)
    :lolwut:
    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!!!!!!!!!!
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    (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 :facepalm:
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    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 \theta, k will be a constant depending on material and \theta 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
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    (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).
    That's a bad excuse.

    (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 :facepalm:
    Cool typo.


    Take two lines, y_1 and y_2, and draw them in a similar way as in the picture below.

    Click image for larger version. 

Name:	perpendicular-lines.png 
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    Notice that \tan\theta is the gradient of y_1, and -\tan\varphi is the gradient of y_2. So, the two lines can be expressed in terms of these angles.

    y_1 = \tan(\theta)x, \ \ y_2 = -\tan(\varphi)x

    These will be perpendicular if \displaystyle \theta + \varphi = \frac{\pi}{2}, so you'd use \displaystyle \varphi = \frac{\pi}{2} - \theta.

    Spoiler:
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    Substituting this, you obtain \displaystyle y_2 = -\tan(\varphi)x = -\tan(\pi/2 - \theta)x = -\cot(\theta)x by co-function identities.

    Therefore, the product of their gradients is -\tan\theta\cot\theta = -1 for a rotation of any angle \theta. Translations also do not change anything.


    In other words, any two perpendicular lines are rotations and translations of the x and y axes, and the product of gradients of those two is -1.
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    (Original post by jack.hadamard)
    That's a bad excuse.



    Cool typo.


    Take two lines, y_1 and y_2, and draw them in a similar way as in the picture below.

    Click image for larger version. 

Name:	perpendicular-lines.png 
Views:	29 
Size:	3.9 KB 
ID:	145598

    Notice that \tan\theta is the gradient of y_1, and -\tan\varphi is the gradient of y_2. So, the two lines can be expressed in terms of these angles.

    y_1 = \tan(\theta)x, \ \ y_2 = -\tan(\varphi)x

    These will be perpendicular if \displaystyle \theta + \varphi = \frac{\pi}{2}, so you'd use \displaystyle \varphi = \frac{\pi}{2} - \theta.

    Spoiler:
    Show

    Substituting this, you obtain \displaystyle y_2 = -\tan(\varphi)x = -\tan(\pi/2 - \theta)x = -\cot(\theta)x by co-function identities.

    Therefore, the product of their gradients is -\tan\theta\cot\theta = -1 for a rotation of any angle \theta. Translations also do not change anything.




    In other words, any two perpendicular lines are rotations and translations of the x and y axes, and the product of gradients of those two is -1.

    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.
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    (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 y_1 and y_2, where the expression for the product of gradients of x and y 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?
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    (Original post by jack.hadamard)

    I have always been an odd one, does that make a difference?
    I think "odd one" was a euphemism for "****".
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    (Original post by nohomo)
    I think "odd one" was a euphemism for "****".
    Then I guess it makes a difference.
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    (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.

    I once made a thread about it, http://www.thestudentroom.co.uk/show....php?t=1937908
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    What was Lebesgue's favourite play?
    Spoiler:
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    Measure for Measure
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    (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:
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    "I could have made it even hairier, but after a while I decided enough was enough." — APM
    Spoiler:
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    "An integral like nothing I've ever seen before, but... courage!" — APM

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    http://www.thestudentroom.co.uk/show...4#post37484624

    help
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    (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:

    m_1 \times m_2 = -1.

    0 \times m_2 = -1.

    So 0 times what gives -1? There's no value that satisfies so m2 is undefined, as we already know it is.
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    (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:

    m_1 \times m_2 = -1.

    0 \times m_2 = -1.

    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
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    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
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    (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 \dfrac{cos^2x}{cos^2x}. Then the substitution and the rest is very easy.
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    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
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    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)?
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    (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.
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    (Original post by mashmammad)
    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?

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