Pleasse Help Me to Find Parameter of CIrcle by integration.

[urduworld]

please help me to findout this problem
i am putting this question second time bcoz i was unable to findout its category i am new here
answer should be 2(pi)a

 

[Mark44]

What you want is the perimeter of your circle, not the parameter.
In your integral for arc length, you are apparently going to calculate the arc length of the first quadrant of the circle, and then multiply by 4.

You have x^2 + y^2 = a^2. Either solve for y as a function of x, and then differentiate to get dy/dx, OR differentiate the given equation implicitly to get dy/dx. Put that expression in for dy/dx in your formula for arc length.

Finally, you need to put in limits of integration, and then calculate the integral.

 

[urduworld]

yes i got your point, but i don't know what limit i have to apply. i was applying limit
x=0 and x=2(pi)a/4 am i right here, i have solve it approximately at end but i am unable to solve last four steps

 

[Dick]

It looks like you are trying to integrate the arclength over the first quadrant and multiply by 4. That's fine. That means your x limits are 0 to a. But you aren't really showing what you are doing or following Mark44's advice. What is y'?

 

[urduworld]

we will find value of y from eq 1 + y^2 = a^2 - x^2
i will put it (dy/dx)^2

 

[Dick]

we will find value of y from eq 1 + y^2 = a^2 - x^2
i will put it (dy/dx)^2

Ok, I think. Do so.

 

[urduworld]

Please Help me to solve this Parameter of the circle

1. i have try my best and i and reach at a point and unable to solve it complete please help me to find the parameter of the circle

.2 also what limit i should apply ( i am thinking to apply 2(pi)a/4) am i right

3.What i further do

If you are unable to see http://i481.photobucket.com/albums/rr178/urduworld/aaa.gif"

Thanks in advance

 

[tiny-tim]

1. i have try my best and i and reach at a point and unable to solve it complete please help me to find the parameter of the circle

Hi urduworld!

(have an integral: ∫ and a square-root: √ and a pi: π and a theta: θ )

ok, you're trying to find the circumference of a circle of radius a by integrating arc-length round the circle,

and you've arrived at 4∫ dx/√(a² - x²)

Now do a trig substitution.

.2 also what limit i should apply ( i am thinking to apply 2(pi)a/4) am i right

Since you've decided to find the length of just the first quadrant, your x is going from 0 to a, isn't it? So those are your limits: ∫₀^a … dx.

(Incidentally, if you'd started with θ as your variable, and dθ as your arc-length, then you could have done the whole circle, with limits of 0 ≤ θ ≤ 2π, or one quadrant with limits of 0 ≤ θ ≤ π/2, the result multiplied by 4, which would give the same answer).

 

[Mark44]

You posted the same problem in another thread: https://www.physicsforums.com/showthread.php?t=354071.

You shouldn't start a new thread for the same problem.

Also, the distance around a circle is its circumference, or perimeter. Parameter means something different.

 

[urduworld]

You posted the same problem in another thread: https://www.physicsforums.com/showthread.php?t=354071.

You shouldn't start a new thread for the same problem.

Also, the distance around a circle is its circumference, or perimeter. Parameter means something different.

yes i have posted this second time because that time i can't got answer :(

 

[urduworld]

ohh i really got point i have got that i have have to substitute a^2 - x^2 with (u or any alphabet) and then i have to apply limit am i right Thanks tiny :)

 

[Mark44]

If you don't get an answer in a thread, don't post a new thread, "bump" the thread. That will bring it back up to the top of the displayed posts, and someone will usually respond to it.

In any case, Dick responded to your other thread with what you needed to do. Did you follow his suggestion?

 

[berkeman]

yes i have posted this second time because that time i can't got answer :(

Threads merged. Do NOT do this again.

 

[Mark44]

ohh i really got point i have got that i have have to substitute a^2 - x^2 with (u or any alphabet) and then i have to apply limit am i right Thanks tiny :)

No, not an ordinary substitution - a trig substitution. Do you know how to do one of these substitutions?

 

[tiny-tim]

ohh i really got point i have got that i have have to substitute a^2 - x^2 with (u or any alphabet) and then i have to apply limit am i right Thanks tiny :)

No.

That would give you ∫ 2√(a² - u)du/√u , which is even worse.

 

[urduworld]

then what i have to do i can't understand trig substitution :(

 

[urduworld]

i don't know trig substitution. please just hint me what i have to do

 

[berkeman]

then what i have to do i can't understand trig substitution :(

Trigonometric substitution means using trig identities to come up with a form of the equation that is easier to integrate. See this page for example, for some useful trig identities. I'm betting that your textbook also has a section like this...

http://en.wikipedia.org/wiki/Trig_identities

.

 

[tiny-tim]

Or http://en.wikipedia.org/wiki/Trigonometric_substitution"

 

[urduworld]

Thanks Tiny-Tim, Mark 44 and Berkman :) You really help me and i have solve this question, PF really helps me :) Again Thanks in Last