What is the arclength of #r=-10sin(theta/4+(5pi)/16) # on #theta in [(-5pi)/16,(9pi)/16]#?

Answer 1

#L=(133pi)/20-48/5sin(7/16pi)cos(11/8pi)+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}# units.

#r=−10sin(θ/4+(5pi)/16)# #r^2=100sin^2(θ/4+(5pi)/16)#
#r'=−10/4cos(θ/4+(5pi)/16)# #(r')^2=100/25cos^2(θ/4+(5pi)/16)#

Arclength is given by:

#L=int_((-5pi)/16)^((9pi)/16)sqrt(100sin^2(θ/4+(5pi)/16)+100/25cos^2(θ/4+(5pi)/16))d theta#
Apply the substitution #θ/4+(5pi)/16=phi# and rearrange:
#L=40int_((15pi)/64)^((29pi)/64)sqrt(1-24/25cos^2phi)dphi#
For #phi in [(15pi)/64,(29pi)/64]#, #24/25cos^2phi<1#. Take the series expansion of the square root:
#L=40int_((15pi)/64)^((29pi)/64)sum_(n=0)^oo((1/2),(n))(-24/25cos^2phi)^ndphi#
Isolate the #n=0# term and rearrange:
#L=40int_((15pi)/64)^((29pi)/64)dphi+40sum_(n=1)^oo((1/2),(n))(-24/25)^nint_((15pi)/64)^((29pi)/64)cos^(2n)phidphi#

Apply the trigonometric power-reduction formula:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-24/25)^nint_((15pi)/64)^((29pi)/64){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)phi)}dphi#

Integrate term by term:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n[((2n),(n))phi+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)phi)/(n-k)]_((15pi)/64)^((29pi)/64)#

Insert the limits of integration:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(29pi)/32)-sin((n-k)(15pi)/32))/(n-k)}#

Apply the trigonometric sum-to-product formula:

#L=(35pi)/4+40sum_(n=1)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#
Isolate the #n=1# terms:
#L=(35pi)/4-24/5((7pi)/16+2sin(7/16pi)cos(11/8pi))+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#

Simplify:

#L=(133pi)/20-48/5sin(7/16pi)cos(11/8pi)+40sum_(n=2)^oo((1/2),(n))(-6/25)^n{((2n),(n))(7pi)/32+2sum_(k=0)^(n-1)((2n),(k))(sin((n-k)(7pi)/16)cos((n-k)(11pi)/8))/(n-k)}#
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Answer 2

To find the arc length, you would use the formula:

[L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{{dr}}{{d\theta}}\right)^2} d\theta]

Given (r = -10\sin\left(\frac{\theta}{4} + \frac{5\pi}{16}\right)), you would need to find (\frac{{dr}}{{d\theta}}) and then evaluate the integral from (-\frac{5\pi}{16}) to (\frac{9\pi}{16}).

First, find (\frac{{dr}}{{d\theta}}): [\frac{{dr}}{{d\theta}} = -10 \cdot \frac{d}{d\theta}\left(\sin\left(\frac{\theta}{4} + \frac{5\pi}{16}\right)\right)]

Apply the chain rule: [\frac{{dr}}{{d\theta}} = -10 \cdot \frac{1}{4} \cdot \cos\left(\frac{\theta}{4} + \frac{5\pi}{16}\right)]

Now integrate: [L = \int_{-\frac{5\pi}{16}}^{\frac{9\pi}{16}} \sqrt{(-10\sin(\theta/4 + 5\pi/16))^2 + (-10/4\cos(\theta/4 + 5\pi/16))^2} d\theta]

Compute the integral to find the arc length.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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