Arc Length 11/18/2011 Suppose you want to know what the length of a - PowerPoint PPT Presentation

Arc Length 11/18/2011

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) a b

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) ∆ x x 2 . . . x 0 x 1 x n Slice! n X ` = lim (little length) i n !1 i =1

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) one piece − → ∆ x x 2 . . . x 0 x 1 x n x i x i+1 Slice! n X ` = lim (little length) i n !1 i =1

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) ∆ l one piece ∆ y − → ∆ x x 2 . . . ∆ x x 0 x 1 x n x i x i+1 Slice! n X ` = lim ( ∆ ` ) i n !1 i =1

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) ∆ l one piece ∆ y − → ∆ x x 2 . . . ∆ x x 0 x 1 x n x i x i+1 Let n go to 1 Slice! � Z x = b n X ` = lim ( ∆ ` ) i = d ` d l n !1 x = a i =1 dy dx

Suppose you want to know what the length of a curve y = f ( x ) is from the point ( a , f ( a )) to the point ( b , f ( b )): y=f(x) ∆ l one piece ∆ y − → ∆ x x 2 . . . ∆ x x 0 x 1 x n x i x i+1 Let n go to 1 Slice! � Z x = b n X ` = lim ( ∆ ` ) i = d ` d l n !1 x = a i =1 dy dx 2 + dy 2 p d ` = dx

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 p d ` =

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx r dx 2 + dy 2 = dx dx 2

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx r dx 2 + dy 2 r dx 2 dx 2 + dy 2 = dx = dx 2 dx dx 2

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx r dx 2 + dy 2 r dx 2 dx 2 + dy 2 = dx = dx 2 dx dx 2 s✓ dx ◆ 2 ◆ 2 ✓ dy = + dx dx dx

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx r dx 2 + dy 2 r dx 2 dx 2 + dy 2 = dx = dx 2 dx dx 2 s✓ dx ◆ 2 ◆ 2 ✓ dy = + dx dx dx q 1 + ( f 0 ( x )) 2 dx =

Manipulating into something we can actually calculate... d l Remember, y = f ( x ) . dy dx dx 2 + dy 2 dx dx 2 + dy 2 = p p d ` = dx r dx 2 + dy 2 r dx 2 + dy 2 dx 2 = dx = dx 2 dx dx 2 s✓ dx ◆ 2 ◆ 2 ✓ dy = + dx dx dx q 1 + ( f 0 ( x )) 2 dx = Z b q 1 + ( f 0 ( x )) 2 dx So ` = x = a

Example Find the length of the arc y = x 3 / 2 , from x = 0 to x = 1. 3 2 1 1 2

Find the length of the curve y = x 4 + 1 32 x 2 from x = 1 to x = 2. 15 10 5 1 2

Find the length of the curve y = x 4 + 1 32 x 2 from x = 1 to x = 2. 15 10 5 1 2 16 x � 3 = 64 x 6 − 1 f ( x ) = x 4 + 1 ⇒ f 0 ( x ) = 4 x 3 − 1 32 x � 2 = 16 x 3

Find the length of the curve y = x 4 + 1 32 x 2 from x = 1 to x = 2. 15 10 5 1 2 16 x � 3 = 64 x 6 − 1 f ( x ) = x 4 + 1 ⇒ f 0 ( x ) = 4 x 3 − 1 32 x � 2 = 16 x 3 Keeping the algebra tame: Let A = (2 x ) 3 = 8 x 3 and so A 2 = 64 x 6 , and f 0 ( x ) = A 2 � 1 2 A .

Find the length of the curve y = x 4 + 1 32 x 2 from x = 1 to x = 2. 15 10 5 1 2 16 x � 3 = 64 x 6 − 1 f ( x ) = x 4 + 1 ⇒ f 0 ( x ) = 4 x 3 − 1 32 x � 2 = 16 x 3 Keeping the algebra tame: Let A = (2 x ) 3 = 8 x 3 and so A 2 = 64 x 6 , and f 0 ( x ) = A 2 � 1 2 A . So ◆ 2 ✓ A 2 − 1 1+( f 0 ( x )) 2 = 1+ 2 A

Most of the time, the resulting integral is “hard” (not elementary) Set up (but do not integrate) the integrals which compute the length of the following functions: 1. f ( x ) = x 2 from x = − 3 to 2 2. f ( x ) = x 2 + 5 from x = − 3 to 2 3. f ( x ) = − x 2 + ⇡ from x = − 3 to 2 4. f ( x ) = sin( x ) from x = 0 to π 2 5. f ( x ) = e x from x = 0 to 1 √ 1 − x 2 from x = − 1 to 1 6. f ( x ) =