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What is the arclength of #r=4theta # on #theta in [-pi/4,pi]#?

Answer 1

Luke Alvarez

#approx 27.879#

This is an outline method. The grind of some of the work has been done by computer.

Arc length #s = int dot s \ dt#

and #dot s = sqrt (vec v * vec v)#

Now, for #vec r = 4 theta \ hat r #

#vec v = dot r hat r + r dot theta hat theta#

#= 4 dot theta \ hat r + 4 theta dot theta \ hat theta#

#= 4 dot theta ( hat r + theta \ hat theta )#

So #dot s = 4 dot theta sqrt(1 + theta ^2)#

Arc length #s = 4 int_(t_1)^(t_2) sqrt(1 + theta ^2) \ dot theta \ dt#

# = 4 int_(-pi/4)^(pi) sqrt(1 + theta ^2) \ d theta#

#= 2 [ theta sqrt(theta^2+1) +sinh^(-1) theta ]_(-pi/4)^(pi)# computer solution. See Youtube linked here for the method

#approx 27.879# computer solution

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Answer 2

Isaiah Allen

The arc length of ( r = 4\theta ) on the interval ( \theta ) in ( [-\frac{\pi}{4}, \pi] ) is ( \frac{5\sqrt{2}}{2} + 4\pi ) units.

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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.