What is the arc length of #f(x) = sinx # on #x in [pi/12,(5pi)/12] #?

Question

What is the arc length of #f(x) = sinx  # on #x in [pi/12,(5pi)/12]  #?

Aurora Alvarado · Accepted Answer

#L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}#

#f(x)=sinx#

#f'(x)=cosx#

Arc length is given by:

#L=int_(pi/12)^((5pi)/12)sqrt(1+cos^2x)dx#

For #x in [pi/12,(5pi)/12]#, #cos^2x<1#. Take the series expansion of the square root:

#L=int_(pi/12)^((5pi)/12)sum_(n=0)^oo((1/2),(n))cos^(2n)xdx#

Isolate the #n=0# case:

#L=int_(pi/12)^((5pi)/12)dx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12)cos^(2n)xdx#

Apply the Trigonometric power-reduction formula:

#L=[x]_ (pi/12)^((5pi)/12)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx#

Simplify:

#L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^nint_(pi/12)^((5pi)/12){((2n),(n))+2sum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx#

Integrate directly:

#L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n[((2n),(n))x+sum_(k=0)^(n-1)((n),(k))sin(2(n-k)x)/(n-k)]_(pi/12)^((5pi)/12)#

Simplify:

#L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+sum_(k=0)^(n-1)((n),(k))(sin((n-k)(5pi)/6)-sin((n-k)pi/6))/(n-k)}#

Apply the Trigonometric sum-to-product identity #sina-sinb=2sin((a-b)/2)cos((a+b)/2)#:

#L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}#

Isolate the #n=1# case:

#L=pi/3+pi/12+sum_(n=2)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}#

Paisley Johnson · Answer

undefined To find the arc length of $ f(x) = \sin(x) $ on the interval $ x \in \left[\frac{\pi}{12}, \frac{5\pi}{12}\right] $, you can use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx $$

Where:
- $ f'(x) $ is the derivative of the function.
- $ a $ and $ b $ are the limits of integration.

Given $ f(x) = \sin(x) $, the derivative $ f'(x) $ is $ \cos(x) $. Thus, the formula for arc length becomes:

$$ L = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \sqrt{1 + \cos^2(x)} \, dx $$

Now, you can integrate $ \sqrt{1 + \cos^2(x)} $ over the given interval $\left[\frac{\pi}{12}, \frac{5\pi}{12}\right]$ to find the arc length $ L $.