Prove that the arc length of the polar curve #r = a(1-cos theta)# is #8a#?

Answer 1

The polar arc length of a curve is given by:

# L = int_alpha^beta \ sqrt(r^2 + ((dr)/(d theta))^2) \ d theta #

We have:

# r = a(1-cos theta) # # \ \ = a-acos theta #

Thus:

# (dr)/(d theta) = asin theta #

So, the arc length is:

# L = int_0^(2pi) \ sqrt( (a-acos theta)^2 + (asin theta)^2 ) \ d theta # # \ \ \ = int_0^(2pi) \ sqrt( (a^2-2a^2cos theta + a^2 cos^2 theta) + (asin theta)^2 ) \ d theta # # \ \ \ = int_0^(2pi) \ asqrt( 1-2cos theta + cos^2 theta + sin^2 theta ) \ d theta # # \ \ \ = a \ int_0^(2pi) sqrt( 2-2cos theta ) \ d theta #
# \ \ \ = a \ int_0^(2pi) sqrt( 2)sqrt(1-cos theta ) \ d theta #

Using the trig identity:

# cos2x -= 1-2sin^2x => 2sin^2x-=1-cos2x#

We then have:

# L = a \ int_0^(2pi) sqrt( 2)sqrt(2sin^2 (theta/2) ) \ d theta # # \ \ \ = a \ int_0^(2pi) 2sin (theta/2) \ d theta # # \ \ \ = 2a \ [ -2cos (theta/2) ]_0^(2pi)# # \ \ \ = -4a \ (cos pi - cos 0)# # \ \ \ = -4a \ (-1-1)# # \ \ \ = (-4a)(-2) # # \ \ \ = 8a # QED
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Answer 2

To find the arc length of the polar curve ( r = a(1 - \cos(\theta)) ), you can use the formula for arc length in polar coordinates:

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

First, find ( \frac{dr}{d\theta} ):

[ \frac{dr}{d\theta} = \frac{d}{d\theta}(a(1 - \cos(\theta))) ]

[ = a \cdot \frac{d}{d\theta}(1 - \cos(\theta)) ]

[ = a \cdot \sin(\theta) ]

Now, substitute ( r = a(1 - \cos(\theta)) ) and ( \frac{dr}{d\theta} = a\sin(\theta) ) into the formula for arc length:

[ L = \int_{0}^{2\pi} \sqrt{(a(1 - \cos(\theta)))^2 + (a\sin(\theta))^2} , d\theta ]

[ = \int_{0}^{2\pi} \sqrt{a^2(1 - 2\cos(\theta) + \cos^2(\theta)) + a^2\sin^2(\theta)} , d\theta ]

[ = \int_{0}^{2\pi} \sqrt{a^2 - 2a^2\cos(\theta) + a^2\cos^2(\theta) + a^2\sin^2(\theta)} , d\theta ]

[ = \int_{0}^{2\pi} \sqrt{a^2 - 2a^2\cos(\theta) + a^2} , d\theta ]

[ = \int_{0}^{2\pi} \sqrt{2a^2(1 - \cos(\theta))} , d\theta ]

[ = \int_{0}^{2\pi} \sqrt{2}a\sqrt{1 - \cos(\theta)} , d\theta ]

[ = \sqrt{2}a \int_{0}^{2\pi} \sqrt{1 - \cos(\theta)} , d\theta ]

Since ( \sqrt{1 - \cos(\theta)} ) is an even function, the integral over one period is twice the integral from 0 to ( \pi ).

[ = 2\sqrt{2}a \int_{0}^{\pi} \sqrt{1 - \cos(\theta)} , d\theta ]

[ = 2\sqrt{2}a \cdot \frac{\pi}{2} ]

[ = \sqrt{2}\pi a ]

Given that ( a ) is a positive constant, ( \sqrt{2}\pi a = 8a ). Therefore, the arc length of the polar curve ( r = a(1 - \cos(\theta)) ) is indeed ( 8a ).

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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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