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Arc Example x = g(y) Simpson’s rule Arc Length Function Arc Lenth

In this section, we derive a formula for the length of a y = f (x) on an interval [a, b]. We will assume that f is continuous and differentiable on the interval [a, b] and we will assume that its f 0 is also continuous on the interval [a, b]. We use Riemann sums to approximate the length of the curve over the interval and then take the to get an .

We see from the picture that

n X L = lim |P P | n→∞ i−1 i i=1

b−a Letting ∆x = n = |xi−1 − xi |, we get

v u f (x ) − f (x ) q 2 2 u h i i−1 i2 |−1Pi | = (xi − xi−1) + (f (xi ) − f (xi−1)) = ∆xt1 + xi − xi−1 Now by the from last semester, we have f (xi )−f (xi−1) 0 ∗ ∗ = f (x ) for some x in the interval [xi−1, xi ]. Therefore xi −xi−1 i i n n X X q Z b q L = lim |P P | = lim 1 + [f 0(x∗)]2∆x = 1 + [f 0(x)]2dx n→∞ i−1 i n→∞ i i=1 i=1 a

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length

If f is continuous and differentiable on the interval [a, b] and f 0 is also continuous on the interval [a, b]. We have a formula for the length of a curve y = f (x) on an interval [a, b].

r Z b Z b h dy i2 L = p1 + [f 0(x)]2dx or L = 1 + dx a a dx

√ 2x3/2 2 4 2 Example Find the arc length of the curve y = 3 from (1, 3 ) to (2, 3 ). 2x3/2 0 3 2 1/2 √ 0 2 I f (x) = 3 , f (x) = 2 · 3 x = x,[f (x)] = x, a = 1 and b = 2. √ R b p 0 2 R 2 I L = a 1 + [f (x)] dx = 1 1 + x dx R 3 √ I = 2 u du, where u = 1 + x, u(1) = 2, u(2) = 3. ˛3 3/2 ˛ √ √ = u ˛ = 2 [3 3 − 2 2]. I 3/2 ˛ 3 ˛2

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length

r Z b Z b h dy i2 L = p1 + [f 0(x)]2dx or L = 1 + dx a a dx

ex +e−x Example Find the arc length of the curve y = 2 , 0 ≤ x ≤ 2. ex +e−x 0 ex −e−x I f (x) = 2 , f (x) = 2 , 0 2 e2x −2ex e−x +e−2x e2x −2+e−2x [f (x)] = 4 = 4 , a = 0 and b = 2. q q R b p 0 2 R 2 e2x −2+e−2x R 2 e2x +2+e−2x I L = a 1 + [f (x)] dx = 0 1 + 4 dx = 0 4 dx

I (By an application of Pilkington’s Law :) which says that if AB = 1/4, then 1 + (A − B)2 = (A + B)2.)

r 2 R 2 “ ex +e−x ” R 2 ex +e−x I = 0 2 dx = 0 2 dx. ˛2 x −x ˛ 2 −2 = e −e ˛ = e −e . I 2 ˛ 2 ˛0

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length

r Z b Z b h dy i2 L = p1 + [f 0(x)]2dx or L = 1 + dx a a dx

Example Set up the integral which gives the arc length of the curve y = ex , 0 ≤ x ≤ 2. x 0 x 0 2 2x I f (x) = e , f (x) = e ,[f (x)] = e , a = 0 and b = 2. √ R b p 0 2 R 2 2x I L = a 1 + [f (x)] dx = 0 1 + e dx.

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length, of form x = g(y).

For a curve with equation x = g(y), where g(y) is continuous and has a continuous derivative on the interval c ≤ y ≤ d, we can derive a similar formula for the arc length of the curve between y = c and y = d.

v u Z d q Z d u h dx i2 L = 1 + [g0(y)]2dy or L = t1 + dy c c dy

4 4 Example Find the length of the curve 24xy = y + 48 from the ( 3 , 2) to 11 ( 4 , 4). y 4+48 y 3 2 I Solving for x in terms of y, we get x = 24y = 24 + y = g(y).

0 3y 2 2 y 2 2 0 2 y 4 4 y 2 4 y 4 1 4 I g (y) = 24 − y 2 = 8 − y 2 .[g (y)] = 64 − y 2 · 8 + y 4 = 64 − 2 + y 4 q R d p 0 2 R 4 y 4 1 4 I L = c 1 + [f (y)] dy = 2 1 + 64 − 2 + y 4 dy.

r 2 R 4 q y 4 1 4 R 4 “ y 2 2 ” I = 2 64 + 2 + y 4 dy = 2 8 + y 2 dy. ˛4 2 3 ˛ = R 4 y + 2 dy = y − 2 ˛ = 17 . I 2 8 y 2 24 y ˛ 6 ˛2

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length, Simpson’s rule

We cannot always find an for the integrand to evaluate the arc length. However, we can use Simpson’s rule to estimate the arc length.

Example Use Simpson’s rule with n = 10 to estimate the length of the curve √ x = y + y, 2 ≤ y ≤ 4

√1 I dx/dy = 1 + 2 y ,

r h i2 r h i2 r L = R 4 1 + dx dy = R 4 1 + 1 + √1 dy = R 4 2 + √1 + 1 dy I 2 dy 2 2 y 2 y 4y

I With n = 10, Simpson’s rule gives us

∆y ˆ ˜ L ≈ S10 = g(2) + 4g(2.2) + 2g(2.4) + 4g(2.6) + 2g(2.8) + 4g(3) + 2g(3.2) + 4g(3.4) + 2g(3.6) + 4g(3.8) + g(4) 3 q √1 1 4−2 where g(y) = 2 + y + 4y and ∆y = 10 . (see notes for details).

I We get S10 ≈ 3.269185.

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length Function

The along a curve with equation y = f (x) from a fixed point (a, f (a)) is a function of x. It is called the arc length function and is given by Z x s(x) = p1 + [f 0(t)]2dt. a From the fundamental theorem of , we see that s0(x) = p1 + [f 0(x)]2. In the language of differentials, this translates to r “ dy ”2 ds = 1 + dx or (ds)2 = (dx)2 + (dy)2 dx

Annette Pilkington Lecture 16 : Arc Length Arc Length Example x = g(y) Simpson’s rule Arc Length Function Arc Length Function

The distance along a curve with equation y = f (x) from a fixed point (a, f (a)) is a function of x. It is called the arc length function and is given by Z x s(x) = p1 + [f 0(t)]2dt. a

2x3/2 Example Find the arc length function for the curve y = 3 taking P0(1, 3/2) as the starting point. 2x3/2 √ d 3 I We have dx = x I Z x q √ Z x √ Z 1+x √ s(x) = 1 + ( t)2dt = 1 + tdt = udu 1 1 2 where u = 1 + t

I ˛x+1 u3/2 ˛ √ √ = ˛ = 2(x + 1) x + 1/3 − 4 2/3 3/2 ˛ ˛2

Annette Pilkington Lecture 16 : Arc Length