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Math Help - Arc Length

  1. #1
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    Arc Length

    Hi, this is not necessarily a homework question, but someone else has called me to ask this question. I havent had calcululs in awhile so its kind of vague.

    How do you find the arc length of a circle with radius 5 and a 40 degree angle. That is all he gave me. Im feeling like there is something missing.

    Is this enough info to find the arc length?

    Thank you so much to whoever can help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by msbrown View Post
    Hi, this is not necessarily a homework question, but someone else has called me to ask this question. I havent had calcululs in awhile so its kind of vague.

    How do you find the arc length of a circle with radius 5 and a 40 degree angle. That is all he gave me. Im feeling like there is something missing.

    Is this enough info to find the arc length?

    Thank you so much to whoever can help.
    The angle subtended by an arc of length s from the centre of the circle
    of radius r is:

    theta = s/r,

    where theta is in radians. So:

    s = r.theta.

    Now you have an angle of 40 degrees, which needs to be converted to
    radians. This is 40*pi/180 ~= 0.698 radian. Hence:

    s = 5x0.698 = 3.98

    RonL
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  3. #3
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    Hello, msbrown!

    I bet you could have baby-talked your way through this one . . .


    How do you find the arc length of a circle with radius 5 and a 40 angle?

    The circumference of a circle is: C\:=\:2\pi r

    Your circle has a circumference of: 2\pi(5) = 10\pi units.


    Your angle is: \frac{40^o}{360^o} = \frac{1}{9} of a "full circle".

    Hence, its arc length is \frac{1}{9} of the circumference.


    \text{Arc Length } = \;\frac{1}{9} \times 10\pi \:=\:\frac{10\pi}{9} units.

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  4. #4
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    I don't see how that has anything to do with Calculus. Therefore, the fact that you haven't had Calculus in a while should not make a difference for this problem. Unless, that is, you're talking about the arc length of a function, which is obtained given,

    an x = f(y), where f(y) is a function such that its derivative f'(x) is continuous on the closed interval [a, b]. The arc length, then, of f from y = a to y = b is the integral:

    integral(sqrt(1 + [f'(x)]^2))dy, from a to b).
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by AfterShock View Post
    I don't see how that has anything to do with Calculus. Therefore, the fact that you haven't had Calculus in a while should not make a difference for this problem. Unless, that is, you're talking about the arc length of a function, which is obtained given,

    an x = f(y), where f(y) is a function such that its derivative f'(x) is continuous on the closed interval [a, b]. The arc length, then, of f from y = a to y = b is the integral:

    integral(sqrt(1 + [f'(x)]^2))dy, from a to b).
    Just giving the formula in "Math-speak":

    \int_a^b \sqrt{1 + \left ( f'(x) \right )^2} \, dy

    -Dan
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  6. #6
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    Thank you to all. Yes at first I was thinking of arc length of a function, which is in deed calculus and not the simple explaination that was given, but once I read the first reply, it all came back to me! Thank you and sorry for my total brain freeze.
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  7. #7
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    Thanks Topsquark, I'll learn Latex one of these days. And it should have been:

    f'(y) instead of x (my fault).
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