What is the arclength of #r=-3cos(theta/16+(pi)/16) # on #theta in [(-5pi)/16,(9pi)/16]#?

Answer 1

#L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n{((2n),(n))(7pi)/128+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))(sin((n-k)(25pi)/128)-sin((n-k)(11pi)/128))/(n-k)}# units.

#r=-3cos(theta/16+pi/16)# #r^2=9cos^2(theta/16+pi/16)#
#r'=3/16sin(theta/16+pi/16)# #(r')^2=9/256sin^2(theta/16+pi/16)#

Arclength is given by:

#L=int_((-5pi)/16)^((9pi)/16)sqrt(9cos^2(theta/16+pi/16)+9/256sin^2(theta/16+pi/16))d theta#
Apply the substitution #theta/16+pi/16=phi#:
#L=3int_((11pi)/256)^((25pi)/256)sqrt(cos^2phi+1/256sin^2phi)dphi#

Rearrange:

#L=3int_((11pi)/256)^((25pi)/256)sqrt(1-255/256sin^2phi)dphi#

From here we can use the small-angle approximation for an easy answer.

Alternatively take the series expansion of the square root:

#L=3int_((11pi)/256)^((25pi)/256)sum_(n=0)^oo((1/2),(n))(-255/256sin^2phi)^ndphi#
Isolate the #n=0# term and simplify:
#L=3int_((11pi)/256)^((25pi)/256)dphi+3sum_(n=1)^oo((1/2),(n))(-255/256)^nint_((11pi)/256)^((25pi)/256)sin^(2n)phidphi#

Apply the trigonometric power reduction formula:

#L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/256)^nint_((11pi)/256)^((25pi)/256){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))cos((2n-2k)phi)}dphi#

Integrate directly:

#L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n[((2n),(n))phi+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))sin((2n-2k)phi)/(n-k)]_((11pi)/256)^((25pi)/256)#

Hence:

#L=(32pi)/128+3sum_(n=1)^oo((1/2),(n))(-255/1024)^n{((2n),(n))(7pi)/128+sum_(k=0)^(n-1)(-1)^(n-k)((2n),(k))(sin((n-k)(25pi)/128)-sin((n-k)(11pi)/128))/(n-k)}#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The arclength of r = -3cos(theta/16 + pi/16) on theta in [(-5pi)/16, (9pi)/16] is 12 units.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7