Arc Length Consider a curve y = f ( x ), a x b . Alan H. - PowerPoint PPT Presentation

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem , we’ll obtain a Riemann Sum and conclude the length of the curve is equal to a specific definite integral. Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem , we’ll obtain a Riemann Sum and conclude the length of the curve is equal to a specific definite integral. As usual, create a partition P = { x 0 , x 1 , x 2 , . . . , x n } of the interval [ a , b ]. Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem , we’ll obtain a Riemann Sum and conclude the length of the curve is equal to a specific definite integral. As usual, create a partition P = { x 0 , x 1 , x 2 , . . . , x n } of the interval [ a , b ]. Let ∆ s k be the length of the portion of the curve for which x k − 1 ≤ x ≤ x k . Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem , we’ll obtain a Riemann Sum and conclude the length of the curve is equal to a specific definite integral. As usual, create a partition P = { x 0 , x 1 , x 2 , . . . , x n } of the interval [ a , b ]. Let ∆ s k be the length of the portion of the curve for which x k − 1 ≤ x ≤ x k . Since the length s of the curve is equal to � n k =1 ∆ s k , Alan H. SteinUniversity of Connecticut

Arc Length Consider a curve y = f ( x ), a ≤ x ≤ b . We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem , we’ll obtain a Riemann Sum and conclude the length of the curve is equal to a specific definite integral. As usual, create a partition P = { x 0 , x 1 , x 2 , . . . , x n } of the interval [ a , b ]. Let ∆ s k be the length of the portion of the curve for which x k − 1 ≤ x ≤ x k . Since the length s of the curve is equal to � n k =1 ∆ s k , we may obtain an approximation to s by adding together our approximations to each of the ∆ s k . Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ If f is differentiable on [ a , b ], it will satisfy the hypotheses of the Mean Value Theorem Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ If f is differentiable on [ a , b ], it will satisfy the hypotheses of the Mean Value Theorem and we may infer the existence of some point c k ∈ ( x k − 1 , x k ) such that f ( x k ) − f ( x k − 1 ) = f ′ ( c k )( x k − x k − 1 ). Hence, Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ If f is differentiable on [ a , b ], it will satisfy the hypotheses of the Mean Value Theorem and we may infer the existence of some point c k ∈ ( x k − 1 , x k ) such that f ( x k ) − f ( x k − 1 ) = f ′ ( c k )( x k − x k − 1 ). Hence, ( x k − x k − 1 ) 2 + ( f ′ ( c k )( x k − x k − 1 )) 2 � ∆ s k ≈ Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ If f is differentiable on [ a , b ], it will satisfy the hypotheses of the Mean Value Theorem and we may infer the existence of some point c k ∈ ( x k − 1 , x k ) such that f ( x k ) − f ( x k − 1 ) = f ′ ( c k )( x k − x k − 1 ). Hence, ( x k − x k − 1 ) 2 + ( f ′ ( c k )( x k − x k − 1 )) 2 � ∆ s k ≈ � = (1 + f ′ ( c k ) 2 )( x k − x k − 1 ) 2 Alan H. SteinUniversity of Connecticut

Approximating the Length of a Subarc Since the portion of the curve for which x k − 1 ≤ x ≤ x k has endpoints ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )), we may approximate ∆ s k by the length of the line segment joining ( x k − 1 , f ( x k − 1 )) and ( x k , f ( x k )). Using the distance formula, we obtain ( x k − x k − 1 ) 2 + ( f ( x k ) − f ( x k − 1 )) 2 . � ∆ s k ≈ If f is differentiable on [ a , b ], it will satisfy the hypotheses of the Mean Value Theorem and we may infer the existence of some point c k ∈ ( x k − 1 , x k ) such that f ( x k ) − f ( x k − 1 ) = f ′ ( c k )( x k − x k − 1 ). Hence, ( x k − x k − 1 ) 2 + ( f ′ ( c k )( x k − x k − 1 )) 2 � ∆ s k ≈ � = (1 + f ′ ( c k ) 2 )( x k − x k − 1 ) 2 � � 1 + f ′ ( c k ) 2 ( x k − x k − 1 ) = 1 + f ′ ( c k ) 2 ∆ x k . = Alan H. SteinUniversity of Connecticut

The Riemann Sum and the Arc Length We may then approximate the length of the curve by Alan H. SteinUniversity of Connecticut

The Riemann Sum and the Arc Length We may then approximate the length of the curve by s = � n k =1 ∆ s k ≈ � n � 1 + f ′ ( c k ) 2 ∆ x k . k =1 Alan H. SteinUniversity of Connecticut

The Riemann Sum and the Arc Length We may then approximate the length of the curve by s = � n k =1 ∆ s k ≈ � n � 1 + f ′ ( c k ) 2 ∆ x k . k =1 � This is a Riemann Sum R ( 1 + f ′ ( x ) 2 , P , a , b ) for the function 1 + f ′ ( x ) 2 and we conclude the arc length is equal to the integral � Alan H. SteinUniversity of Connecticut