Explanation Given : Graph of function y 2 + x 2 = 1 in one-quarter area and in interval [ 0 , 1 ] . There is a prebuilt formula to calculate the arc length of a given function in a given interval and for that we need the differentiation of this function with respect to x also, so that its value can be put in said formula of calculating d s and accordingly the required arc length is obtained by integrating this small arc length d s under the said interval [ a , b ] Formula to be used : We have the formula that, d s = 1 + ( d y d x ) 2 d x Further arc length in the given interval [ a , b ] is calculated using the formula, Arc length = ∫ a b d s Calculation : Now on differentiating given function with respect to x , we get 2 y d y d x + 2 x = 0 d y d x = − x y Thus, d s = 1 + ( − x y ) 2 d x = 1 y x 2 + y 2 d x = 1 y d x ( ∵ x 2 + y 2 = 1 ) = 1 1 − x 2 d x So, required arc length for N = 1 ∫ 0 0.25 d s = ∫ 0 0.25 1 1 − x 2 d x = ( sin − 1 x ) 0 0.25 = sin − 1 ( 0.25 ) − sin − 1 ( 0 ) = 0.25268 − 0 = 0.25268 Further the arc length in second segment for N = 2 , ∫ 0.25 0.5 d s = ∫ 0.25 0...

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Chapter 9.2, Problem 35E

To determine

To prove: prove that length of arc of function $y^{2} + x^{2} = 1$ over the interval $[0, 1]$ by computing

the length of the polygonal approximation with $N = 4$ segments

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Chapter 9.2, Problem 34E

Chapter 9.2, Problem 36E

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Chapter 9 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E

Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Prob. 32E Ch. 9.2 - Prob. 1PQ Ch. 9.2 - Prob. 2PQ Ch. 9.2 - Prob. 3PQ Ch. 9.2 - Prob. 4PQ Ch. 9.2 - Prob. 5PQ Ch. 9.2 - Prob. 1E Ch. 9.2 - Prob. 2E Ch. 9.2 - Prob. 3E Ch. 9.2 - Prob. 4E Ch. 9.2 - Prob. 5E Ch. 9.2 - Prob. 6E Ch. 9.2 - Prob. 7E Ch. 9.2 - Prob. 8E Ch. 9.2 - Prob. 9E Ch. 9.2 - Prob. 10E Ch. 9.2 - Prob. 11E Ch. 9.2 - Prob. 12E Ch. 9.2 - Prob. 13E Ch. 9.2 - Prob. 14E Ch. 9.2 - Prob. 15E Ch. 9.2 - Prob. 16E Ch. 9.2 - Prob. 17E Ch. 9.2 - Prob. 18E Ch. 9.2 - Prob. 19E Ch. 9.2 - Prob. 20E Ch. 9.2 - Prob. 21E Ch. 9.2 - Prob. 22E Ch. 9.2 - Prob. 23E Ch. 9.2 - Prob. 24E Ch. 9.2 - Prob. 25E Ch. 9.2 - Prob. 26E Ch. 9.2 - Prob. 27E Ch. 9.2 - Prob. 28E Ch. 9.2 - Prob. 29E Ch. 9.2 - Prob. 30E Ch. 9.2 - Prob. 31E Ch. 9.2 - Prob. 32E Ch. 9.2 - Prob. 33E Ch. 9.2 - Prob. 34E Ch. 9.2 - Prob. 35E Ch. 9.2 - Prob. 36E Ch. 9.2 - Prob. 37E Ch. 9.2 - Prob. 38E Ch. 9.2 - Prob. 39E Ch. 9.2 - Prob. 40E Ch. 9.2 - Prob. 41E Ch. 9.2 - Prob. 42E Ch. 9.2 - Prob. 43E Ch. 9.2 - Prob. 44E Ch. 9.2 - Prob. 45E Ch. 9.2 - Prob. 46E Ch. 9.2 - Prob. 47E Ch. 9.2 - Prob. 48E Ch. 9.2 - Prob. 49E Ch. 9.2 - Prob. 50E Ch. 9.2 - Prob. 51E Ch. 9.2 - Prob. 52E Ch. 9.2 - Prob. 53E Ch. 9.2 - Prob. 54E Ch. 9.2 - Prob. 55E Ch. 9.2 - Prob. 56E Ch. 9.2 - Prob. 57E Ch. 9.2 - Prob. 58E Ch. 9.2 - Prob. 59E Ch. 9.2 - Prob. 60E Ch. 9.2 - Prob. 61E Ch. 9.2 - Prob. 62E Ch. 9.2 - Prob. 63E Ch. 9.2 - Prob. 64E Ch. 9.2 - Prob. 65E Ch. 9.3 - Prob. 1PQ Ch. 9.3 - Prob. 2PQ Ch. 9.3 - Prob. 3PQ Ch. 9.3 - Prob. 4PQ Ch. 9.3 - Prob. 5PQ Ch. 9.3 - Prob. 1E Ch. 9.3 - Prob. 2E Ch. 9.3 - Prob. 3E Ch. 9.3 - Prob. 4E Ch. 9.3 - Prob. 5E Ch. 9.3 - Prob. 6E Ch. 9.3 - Prob. 7E Ch. 9.3 - Prob. 8E Ch. 9.3 - Prob. 9E Ch. 9.3 - Prob. 10E Ch. 9.3 - Prob. 11E Ch. 9.3 - Prob. 12E Ch. 9.3 - Prob. 13E Ch. 9.3 - Prob. 14E Ch. 9.3 - Prob. 15E Ch. 9.3 - Prob. 16E Ch. 9.3 - Prob. 17E Ch. 9.3 - Prob. 18E Ch. 9.3 - Prob. 19E Ch. 9.3 - Prob. 20E Ch. 9.3 - Prob. 21E Ch. 9.3 - Prob. 22E Ch. 9.3 - Prob. 23E Ch. 9.3 - Prob. 24E Ch. 9.3 - Prob. 25E Ch. 9.3 - Prob. 26E Ch. 9.3 - Prob. 27E Ch. 9.3 - Prob. 28E Ch. 9.3 - Prob. 29E Ch. 9.3 - Prob. 30E Ch. 9.4 - Prob. 1PQ Ch. 9.4 - Prob. 2PQ Ch. 9.4 - Prob. 3PQ Ch. 9.4 - Prob. 4PQ Ch. 9.4 - Prob. 5PQ Ch. 9.4 - Prob. 6PQ Ch. 9.4 - Prob. 1E Ch. 9.4 - Prob. 2E Ch. 9.4 - Prob. 3E Ch. 9.4 - Prob. 4E Ch. 9.4 - Prob. 5E Ch. 9.4 - Prob. 6E Ch. 9.4 - Prob. 7E Ch. 9.4 - Prob. 8E Ch. 9.4 - Prob. 9E Ch. 9.4 - Prob. 10E Ch. 9.4 - Prob. 11E Ch. 9.4 - Prob. 12E Ch. 9.4 - Prob. 13E Ch. 9.4 - Prob. 14E Ch. 9.4 - Prob. 15E Ch. 9.4 - Prob. 16E Ch. 9.4 - Prob. 17E Ch. 9.4 - Prob. 18E Ch. 9.4 - Prob. 19E Ch. 9.4 - Prob. 20E Ch. 9.4 - Prob. 21E Ch. 9.4 - Prob. 22E Ch. 9.4 - Prob. 23E Ch. 9.4 - Prob. 24E Ch. 9.4 - Prob. 25E Ch. 9.4 - Prob. 26E Ch. 9.4 - Prob. 27E Ch. 9.4 - Prob. 28E Ch. 9.4 - Prob. 29E Ch. 9.4 - Prob. 30E Ch. 9.4 - Prob. 31E Ch. 9.4 - Prob. 32E Ch. 9.4 - Prob. 33E Ch. 9.4 - Prob. 34E Ch. 9.4 - Prob. 35E Ch. 9.4 - Prob. 36E Ch. 9.4 - Prob. 37E Ch. 9.4 - Prob. 38E Ch. 9.4 - Prob. 39E Ch. 9.4 - Prob. 40E Ch. 9.4 - Prob. 41E Ch. 9.4 - Prob. 42E Ch. 9.4 - Prob. 43E Ch. 9.4 - Prob. 44E Ch. 9.4 - Prob. 45E Ch. 9.4 - Prob. 46E Ch. 9.4 - Prob. 47E Ch. 9.4 - Prob. 48E Ch. 9.4 - Prob. 49E Ch. 9.4 - Prob. 50E Ch. 9.4 - Prob. 51E Ch. 9 - Prob. 1CRE Ch. 9 - Prob. 2CRE Ch. 9 - Prob. 3CRE Ch. 9 - Prob. 4CRE Ch. 9 - Prob. 5CRE Ch. 9 - Prob. 6CRE Ch. 9 - Prob. 7CRE Ch. 9 - Prob. 8CRE Ch. 9 - Prob. 9CRE Ch. 9 - Prob. 10CRE Ch. 9 - Prob. 11CRE Ch. 9 - Prob. 12CRE Ch. 9 - Prob. 13CRE Ch. 9 - Prob. 14CRE Ch. 9 - Prob. 15CRE Ch. 9 - Prob. 16CRE Ch. 9 - Prob. 17CRE Ch. 9 - Prob. 18CRE Ch. 9 - Prob. 19CRE Ch. 9 - Prob. 20CRE Ch. 9 - Prob. 21CRE Ch. 9 - Prob. 22CRE Ch. 9 - Prob. 23CRE Ch. 9 - Prob. 24CRE Ch. 9 - Prob. 25CRE Ch. 9 - Prob. 26CRE Ch. 9 - Prob. 27CRE Ch. 9 - Prob. 28CRE

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To prove: prove that length of arc of function y 2 + x 2 = 1 over the interval [ 0 , 1 ] by computing the length of the polygonal approximation with N = 4 segments

Solution Summary: The author explains that the length of arc of function y 2 + x 2 = 1 over the interval is 1.5718 units up to four decimal places.

Concept explainers

Videos

To prove: prove that length of arc of function $y^{2} + x^{2} = 1$ over the interval $[0, 1]$ by computing

the length of the polygonal approximation with $N = 4$ segments

Want to see the full answer?

Chapter 9 Solutions

To prove: prove that length of arc of function y 2 + x 2 = 1 over the interval [ 0 , 1 ] by computing the length of the polygonal approximation with N = 4 segments

Solution Summary: The author explains that the length of arc of function y 2 + x 2 = 1 over the interval is 1.5718 units up to four decimal places.

Concept explainers

Videos

To prove: prove that length of arc of function y2+x2=1 over the interval [0,1] by computing the length of the polygonal approximation with N=4 segments

Want to see the full answer?

Chapter 9 Solutions

To prove: prove that length of arc of function $y^{2} + x^{2} = 1$ over the interval $[0, 1]$ by computing

the length of the polygonal approximation with $N = 4$ segments