Polar Arc Length?
Does anyone know how to find out the length of an arc/ curve on polar co-ordinates?
Or even a link to a place that explains to a really basic level would be nice
I'm only interested due to doing basic Polar Co-ordinates in A2 Further Maths
Or even a link to a place that explains to a really basic level would be nice
I'm only interested due to doing basic Polar Co-ordinates in A2 Further Maths
Original post by ComputerMaths97
Does anyone know how to find out the length of an arc/ curve on polar co-ordinates?
Or even a link to a place that explains to a really basic level would be nice
I'm only interested due to doing basic Polar Co-ordinates in A2 Further Maths
Or even a link to a place that explains to a really basic level would be nice
I'm only interested due to doing basic Polar Co-ordinates in A2 Further Maths
First hit in Google search
(edited 8 years ago)
Original post by ghostwalker
Another useless comment, excellent.
It hardly explains it, if you knew anything about Maths I'm sure you'd know an A2 Maths student wouldn't know what half of it even means let alone how to calculate it.
"
"Original post by ComputerMaths97
Another useless comment, excellent.
It hardly explains it, if you knew anything about Maths I'm sure you'd know an A2 Maths student wouldn't know what half of it even means let alone how to calculate it.
""
It hardly explains it, if you knew anything about Maths I'm sure you'd know an A2 Maths student wouldn't know what half of it even means let alone how to calculate it.
"
"Since you're doing A2 Further Maths, and are asking about arc length in polar co-ordinates, I would presume you've at least covered it in cartesian coordinates. There's nothing in the derivation in that link that should be beyond you.
Khan academy have a video if you prefer.
(edited 8 years ago)
Original post by ghostwalker
Since you're doing A2 Further Maths, and are asking about arc length in polar co-ordinates, I would presume you've at least covered it in cartesian coordinates. There's nothing in the derivation in that link that should be beyond you.
Khan academy have a video if you prefer.
Khan academy have a video if you prefer.
"I would presume" and "should" is the reason I ask online and don't search google. I don't want to need assumed understanding, I want it to teach in the most basic way.
Thank you!
Original post by ComputerMaths97
"I would presume" and "should" is the reason I ask online and don't search google. I don't want to need assumed understanding, I want it to teach in the most basic way.
Thank you!
Thank you!
This sounds to me like tailor made teaching which normally you have to pay for ...
Original post by TeeEm
This sounds to me like tailor made teaching which normally you have to pay for ...
Your point is?
Was I not allowed to ask a question on TSR, and ask for that much detail?
Oh I'm sorry, shoot me now.
Original post by ComputerMaths97
Your point is?
Was I not allowed to ask a question on TSR, and ask for that much detail?
Oh I'm sorry, shoot me now.
Was I not allowed to ask a question on TSR, and ask for that much detail?
Oh I'm sorry, shoot me now.
Judging from your posts 3 and 5 you must have had a bad day .... It happens sometimes.
Luckily Ghostwalker was not having a bad day too, and bothered to reply.
I am awfully sorry I cannot help you.
All the best.
Original post by ComputerMaths97
Your point is?
Was I not allowed to ask a question on TSR, and ask for that much detail?
Oh I'm sorry, shoot me now.
Was I not allowed to ask a question on TSR, and ask for that much detail?
Oh I'm sorry, shoot me now.
For a tiny bit of arc of length ds (approx a straight line), using Pythagoras gives
(ds)^2 is approx (dx)^2 + (dy)^2
Now go and read the link you were given.
This is what I do not understand, never worked with sec before. I understand the pi/4 bit, but why does it become sec^3x?
Original post by ComputerMaths97
This is what I do not understand, never worked with sec before. I understand the pi/4 bit, but why does it become sec^3x?
This is what I do not understand, never worked with sec before. I understand the pi/4 bit, but why does it become sec^3x?
What substitution was made to turn the '1' into a 'pi/4'? That should explain where the sec^3 has come from (with a bit of help from a well-known trig identity)
So I can see that we're using the idea that the range is from 0 to 1, and by using tanx = 0 and tanx = 1, we get x= pi/4 and x=0. Where theta = tan(x), we found that sqrt(theta^2 + 1) = sec(x) from the trig identity tan^2(x) + 1 = sec^2(x)
OOOOH using:
So the sqrt(theta^2 + 1) = sec(x) and the d(theta) becomes sec^2(x) with respect to x, so it becomes sec^3(x) but the range is now in terms of x not theta!!!
So my only question is, why does d(theta) = sec^2(x) dx ??? Is this just common knowledge?
Original post by ComputerMaths97
So my only question is, why does d(theta) = sec^2(x) dx ??? Is this just common knowledge?
So my only question is, why does d(theta) = sec^2(x) dx ??? Is this just common knowledge?
In a word, yes.
If y = tan x then dy/dx = sec^2(x) so in an integral you would replace dy with sec^2(x)dx.
You can work out dy/dx by writing tan x = sin x / cos x and then applying the quotient rule.
Original post by davros
In a word, yes.
If y = tan x then dy/dx = sec^2(x) so in an integral you would replace dy with sec^2(x)dx.
You can work out dy/dx by writing tan x = sin x / cos x and then applying the quotient rule.
If y = tan x then dy/dx = sec^2(x) so in an integral you would replace dy with sec^2(x)dx.
You can work out dy/dx by writing tan x = sin x / cos x and then applying the quotient rule.
Thank you so much, can't explain how useful you've been!
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