What is the arclength of #r=-3cos(theta/16+(pi)/16) # on #theta in [(-5pi)/16,(9pi)/16]#?
Arclength is given by:
Rearrange:
From here we can use the small-angle approximation for an easy answer.
Alternatively take the series expansion of the square root:
Apply the trigonometric power reduction formula:
Integrate directly:
Hence:
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The arclength of r = -3cos(theta/16 + pi/16) on theta in [(-5pi)/16, (9pi)/16] is 12 units.
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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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