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I asked the following question and was given the attached solution: Idealize the Earth as a perfect sphere of radius R = 6371 km. If you could measure distances with an error of ±1 meter, how large a circle would you have to draw on the Earth’s surface to convince yourself that the Earth is spherical rather than flat?  Note that (or prove yourself) on the surface of a sphere, a circle of radius r will have a circumference of C = 2πR sin(r/R).  Round your answer to the nearest whole number. Hint: recall that you can approximate sin(x) with a Taylor series expansion: sin(x) \approx≈ x - x3/3!   He has me lost at from the taylor expansion, so basically everything. I don't understand what 3! in the numerator of the first step, nor do I understand what the numerator in the second step is. Please help

I asked the following question and was given the attached solution: Idealize the Earth as a perfect sphere of radius R = 6371 km. If you could measure distances with an error of ±1 meter, how large a circle would you have to draw on the Earth’s surface to convince yourself that the Earth is spherical rather than flat?  Note that (or prove yourself) on the surface of a sphere, a circle of radius r will have a circumference of C = 2πR sin(r/R).  Round your answer to the nearest whole number. Hint: recall that you can approximate sin(x) with a Taylor series expansion: sin(x) \approx≈ x - x3/3!   He has me lost at from the taylor expansion, so basically everything. I don't understand what 3! in the numerator of the first step, nor do I understand what the numerator in the second step is. Please help

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I asked the following question and was given the attached solution:

Idealize the Earth as a perfect sphere of radius R = 6371 km. If you could measure distances with an error of ±1 meter, how large a circle would you have to draw on the Earth’s surface to convince yourself that the Earth is spherical rather than flat?  Note that (or prove yourself) on the surface of a sphere, a circle of radius r will have a circumference of C = 2πR sin(r/R). 

Round your answer to the nearest whole number.

Hint: recall that you can approximate sin(x) with a Taylor series expansion: sin(x) \approx≈ x - x3/3!

 

He has me lost at from the taylor expansion, so basically everything. I don't understand what 3! in the numerator of the first step, nor do I understand what the numerator in the second step is. Please help

 

 

 

 

Given Datu: R= 6731 km ennon= ±l M C= 27RSM () 2> From Tayler Expemercn, 31 \13 27R 113 2x3.14X637l X103 a=(1.50 x107) X=5.313 X10 → druw a ciscle (I1) rã x.R aã 5.313X 1ox6371 re 33. 85 N 34 SO, rE 34 |cM So, dnaw ca Cincle with addition of 34 Ikm nadius.
Transcribed Image Text:Given Datu: R= 6731 km ennon= ±l M C= 27RSM () 2> From Tayler Expemercn, 31 \13 27R 113 2x3.14X637l X103 a=(1.50 x107) X=5.313 X10 → druw a ciscle (I1) rã x.R aã 5.313X 1ox6371 re 33. 85 N 34 SO, rE 34 |cM So, dnaw ca Cincle with addition of 34 Ikm nadius.
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