What is the arclength of #r=-10sin(theta/4+(9pi)/8) # on #theta in [(pi)/4,(7pi)/4]#?

Question

Isaiah Allen · Accepted Answer

#L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(3pi)/4)cos((n-k)(11pi)/2))/(n-k)}# units.

#r=-10sin(theta/4+(9pi)/8)# #r^2=100sin^2(theta/4+(9pi)/8)# #r'=-10/4cos(theta/4+(9pi)/8)# #(r')^2=100/16cos^2(theta/4+(9pi)/8)# Arclength is given by: #L=int_(pi/4)^((7pi)/4)sqrt(100sin^2(theta/4+(9pi)/8)+100/16cos^2(theta/4+(9pi)/8))d theta# Apply the substitution #theta/4+(9pi)/8=phi# #L=40int_((19pi)/16)^((25pi)/16)sqrt(sin^2phi+1/16cos^2phi)dphi# Apply the Trigonometric identity #sin^2x+cos^2x=1#: #L=40int_((19pi)/16)^((25pi)/16)sqrt(1-15/16cos^2phi)dphi# Since #15/16cos^2phi<1#, take the series expansion of the square root: #L=40int_((19pi)/16)^((25pi)/16)sum_(n=0)^oo((1/2),(n))(-15/16cos^2phi)^ndphi# Isolate the #n=0# term and simplify: #L=40int_((19pi)/16)^((25pi)/16)dphi+40sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((19pi)/16)^((25pi)/16)cos^(2n)phidphi# Apply the Trigonometric power-reduction formula: #L=40((25pi)/16-(19pi)/16)+40sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((19pi)/16)^((25pi)/16){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)phi)}dphi# Integrate directly: #L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n[((2n),(n))phi+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)phi)/(n-k)]_((19pi)/16)^((25pi)/16)# Insert the limits of integration: #L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(25pi)/8)-sin((n-k)(19pi)/8))/(n-k)}# Apply the Trigonometric sum-to-product identity: #L=15pi+40sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(3pi)/4)cos((n-k)(11pi)/2))/(n-k)}#

Ezekiel Alexander · Answer

undefined To find the arc length of the curve defined by $ r = -10 \sin\left(\frac{\theta}{4} + \frac{9\pi}{8}\right) $ on the interval $ \left[\frac{\pi}{4}, \frac{7\pi}{4}\right] $, you can use the formula for polar arc length:

$$ L = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta $$

Where $ \frac{dr}{d\theta} $ is the derivative of $ r $ with respect to $ \theta $. In this case, $ r = -10 \sin\left(\frac{\theta}{4} + \frac{9\pi}{8}\right) $, so you'll need to find $ \frac{dr}{d\theta} $ first by taking the derivative of $ r $ with respect to $ \theta $.

Once you have $ \frac{dr}{d\theta} $, you can plug it into the formula along with $ r $, and then integrate over the given interval. After evaluating the integral, you'll get the arc length $ L $.