What is the arclength of #r=-2sin(theta/4+(7pi)/8) # on #theta in [(pi)/4,(7pi)/4]#?

Answer 1

#L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)sum_(m=0)^n((n),(k))(-1)^m/(n-k)delta_((n-k)(4m+2))}#

#r=-2sin(theta/4+(7pi)/8)# #r^2=4sin^2(theta/4+(7pi)/8)#
#r'=-1/2costheta(theta/4+(7pi)/8)# #(r')^2=1/4cos^2(theta/4+(7pi)/8)#

Arclength is given by:

#L=int_(pi/4)^((7pi)/4)sqrt(4sin^2(theta/4+(7pi)/8)+1/4cos^2(theta/4+(7pi)/8))d theta#
Apply the substitution #theta/4+(7pi)/8=phi#:
#L=2int_((15pi)/16)^((21pi)/16)sqrt(sin^2phi+1/16cos^2phi)dphi#
Apply the identity #sin^2phi=1-cos^2phi#:
#L=2int_((15pi)/16)^((21pi)/16)sqrt(1-15/16cos^2phi)dphi#
Since #15/16cos^2phi<1#, apply a series expansion for the square root:
#L=2int_((15pi)/16)^((21pi)/16)sum_(n=0)^oo((1/2),(n))(-15/16cos^2phi)^ndphi#
Isolate the #n=0# term and simplify:
#L=2int_((15pi)/16)^((21pi)/16)dphi+2sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((15pi)/16)^((21pi)/16)cos^(2n)phidphi#

Apply the Trigonometric power-reduction formula:

#L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/16)^nint_((15pi)/16)^((21pi)/16){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((n),(k))cos((2n-2k)phi)}dphi#

Integrate directly:

#L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/64)^n[((2n),(n))phi+sum_(k=0)^(n-1)((n),(k))sin((2n-2k)phi)/(n-k)]int_((15pi)/16)^((21pi)/16)#

Insert limits:

#L=(3pi)/4+2sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/8+sum_(k=0)^(n-1)((n),(k))(sin((n-k)(21pi)/8)-sin((n-k)(15pi)/8))/(n-k)}#

Apply the Trigonometric sum-to-product formula:

#L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)((n),(k))(sin((n-k)(3pi)/4)cos((n-k)(9pi)/2))/(n-k)}#
Since #sin((n-k)(3pi)/4)cos((n-k)(9pi)/2)# gives either #0# or #+-1#:
#L=(3pi)/4+sum_(n=1)^oo((1/2),(n))(-15/64)^n{((2n),(n))(3pi)/4+4sum_(k=0)^(n-1)sum_(m=0)^n((n),(k))(-1)^m/(n-k)delta_((n-k)(4m+2))}#

This can probably be further simplified.

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Answer 2

To find the arc length of the curve defined by ( r = -2\sin\left(\frac{\theta}{4} + \frac{7\pi}{8}\right) ) on the interval ( \left[\frac{\pi}{4}, \frac{7\pi}{4}\right] ), we can use the formula for arc length in polar coordinates:

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

where ( r ) is the polar function and ( \frac{{dr}}{{d\theta}} ) represents its derivative with respect to ( \theta ).

By calculating the derivative of ( r ), we find ( \frac{{dr}}{{d\theta}} = -\frac{1}{2}\cos\left(\frac{\theta}{4} + \frac{7\pi}{8}\right) ).

Then, we substitute ( r ) and ( \frac{{dr}}{{d\theta}} ) into the arc length formula and integrate with respect to ( \theta ) over the given interval:

[ L = \int_{\frac{\pi}{4}}^{\frac{7\pi}{4}} \sqrt{\left(-2\sin\left(\frac{\theta}{4} + \frac{7\pi}{8}\right)\right)^2 + \left(-\frac{1}{2}\cos\left(\frac{\theta}{4} + \frac{7\pi}{8}\right)\right)^2} , d\theta ]

After simplifying, the integral becomes:

[ L = \int_{\frac{\pi}{4}}^{\frac{7\pi}{4}} \sqrt{4\sin^2\left(\frac{\theta}{4} + \frac{7\pi}{8}\right) + \frac{1}{4}\cos^2\left(\frac{\theta}{4} + \frac{7\pi}{8}\right)} , d\theta ]

This integral can be challenging to solve directly, so numerical methods or specialized software may be needed to approximate the value of 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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