entertainmyfaith
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a pendulum of length 6cm follows the arc of a circle.
its straight line displacement as a vector is given by d = 6sinΘ i + 6(1-cosΘ) j
show that the magnitude of the displacement is 6√(2(1-cosΘ))

so i've converted 6sinΘ and 6(1-1-(1/2)Θ2), which altogether became 6Θ - (1/2)Θ2 ; to then find the magnitude i did √(6Θ)2 - √(1/2Θ2)2 but then i just end up w the equation i first started with so i feel like i'm going in circles- any advice on what i should be doing?
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RDKGames
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(Original post by entertainmyfaith)
a pendulum of length 6cm follows the arc of a circle.
its straight line displacement as a vector is given by d = 6sinΘ i + 6(1-cosΘ) j
show that the magnitude of the displacement is 6√(2(1-cosΘ))

so i've converted 6sinΘ and 6(1-1-(1/2)Θ2), which altogether became 6Θ - (1/2)Θ2 ; to then find the magnitude i did √(6Θ)2 - √(1/2Θ2)2 but then i just end up w the equation i first started with so i feel like i'm going in circles- any advice on what i should be doing?
I'm not sure what you are trying to do here, it seems unjustified to even start using small angle approximations when you don't even know that \theta is small, but the question simply asks for the magnitude of the displacement vector \mathbf{d}.

This is the same as simply showing |\mathbf{d}| = 6\sqrt{2(1-\cos \theta)} with no small angle approximations required.
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Gent2324
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is this a level maths? if so what topic
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entertainmyfaith
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(Original post by RDKGames)
I'm not sure what you are trying to do here, it seems unjustified to even start using small angle approximations when you don't even know that \theta is small, but the question simply asks for the magnitude of the displacement vector \mathbf{d}.

This is the same as simply showing |\mathbf{d}| = 6\sqrt{2(1-\cos \theta)} with no small angle approximations required.
ah, i see; completely forgot about sin2x +cos2x =1
ok so i got √(36sin2Θ + 36 -36cos2Θ)
then √(36sin2Θ + 36 - 36(1-sin2Θ))
√(36sin2Θ + 36 - 36 + 36sin2Θ)
= √(72sin2Θ)
6√(2sin2Θ)
6√(2(1-cos2Θ))
though i know 2 cannot be rooted any further, why is there still a root over cosΘ in the equation that i'm meant to reach as my final answer as i thought you could cos2Θ down to cosΘ?
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entertainmyfaith
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(Original post by Gent2324)
is this a level maths? if so what topic
yeah a level- part of a question on topic small degree approximations.
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RDKGames
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(Original post by entertainmyfaith)
ah, i see; completely forgot about sin2x +cos2x =1
ok so i got √(36sin2Θ + 36 -36cos2Θ)
I understand that 36\sin^2 \theta comes from squaring the i component, which means

36-36\cos^2 \theta comes from squaring 6(1-\cos \theta) ... which is wrong, and you made a rookie error by assuming that (a-b)^2 = a^2 - b^2 I'm afraid

Can you amend it?
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entertainmyfaith
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(Original post by RDKGames)
I understand that 36\sin^2 \theta comes from squaring the i component, which means

36-36\cos^2 \theta comes from squaring 6(1-\cos \theta) ... which is wrong, and you made a rookie error by assuming that (a-b)^2 = a^2 - b^2 I'm afraid

Can you amend it?
actually tragic how i'm still making stupid errors:lol:
expanded the bracket, managed to get some long ass equation where the 36cos2Θs cancelled out, so i was left with √72-72cosΘ which i then cancelled down to 6√2(1-cosΘ) which is what i'm meant to reach
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RDKGames
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(Original post by entertainmyfaith)
actually tragic how i'm still making stupid errors:lol:
expanded the bracket, managed to get some long ass equation where the 36cos2Θs cancelled out, so i was left with √72-72cosΘ which i then cancelled down to 6√2(1-cosΘ) which is what i'm meant to reach
Yep, and you could've also kept a factor of 6 out of your working out entirely.

Since \mathbf{d} = 6\sin \theta \mathbf{i} + 6(1-\cos \theta) \mathbf{j} you can factor 6 out to get

\mathbf{d} = 6[\sin \theta \mathbf{i} + (1-\cos \theta) \mathbf{j}] = 6\mathbf{a}

where \mathbf{a} is just the vector \sin \theta \mathbf{i} + (1-\cos \theta) \mathbf{j}.

So then |\mathbf{d}| is simply |6\mathbf{a}|, which simplifies to |6||\mathbf{a}| = 6|\mathbf{a}|,

I.e. |\mathbf{d}| = 6 \sqrt{\sin^2 \theta + (1-\cos\theta)^2} and simplify the inside of the square root down with simpler coefficients.
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entertainmyfaith
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(Original post by RDKGames)
Yep, and you could've also kept a factor of 6 out of your working out entirely.

Since \mathbf{d} = 6\sin \theta \mathbf{i} + 6(1-\cos \theta) \mathbf{j} you can factor 6 out to get

\mathbf{d} = 6[\sin \theta \mathbf{i} + (1-\cos \theta) \mathbf{j}] = 6\mathbf{a}

where \mathbf{a} is just the vector \sin \theta \mathbf{i} + (1-\cos \theta) \mathbf{j}.

So then |\mathbf{d}| is simply |6\mathbf{a}|, which simplifies to |6||\mathbf{a}| = 6|\mathbf{a}|,

I.e. |\mathbf{d}| = 6 \sqrt{\sin^2 \theta + (1-\cos\theta)^2} and simplify the inside of the square root down with simpler coefficients.
oh i didn't consider that! thank you
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DFranklin
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(Original post by entertainmyfaith)
oh i didn't consider that! thank you
Also, (and I suspect this might relate to the next part of the question), knowing that 2 sin^2(x/2) = 1 - cos x means you can get often get rid of the square root (although strictly speaking, \sqrt{\sin^2(x/2)} = |\sin(x/2)|, so some care is needed regarding the modulus signs).
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