lecture 16 : arc lengthapilking/math10560/lectures/lecture 16.pdf · annette pilkington lecture 16...
TRANSCRIPT
Arc Length
Arc Lenth
In this section, we derive a formula for the length of a curve y = f (x) on aninterval [a, b]. We will assume that f is continuous and differentiable on theinterval [a, b] and we will assume that its derivative f ′ is also continuous onthe interval [a, b]. We use Riemann sums to approximate the length of thecurve over the interval and then take the limit to get an integral.
We see from the picture that
L = limn→∞
nXi=1
|Pi−1Pi |
Letting ∆x = b−an
= |xi−1 − xi |, weget
|Pi−1Pi | =q
(xi − xi−1)2 + (f (xi ) − f (xi−1))2 = ∆x
vuut1 +h f (xi ) − f (xi−1)
xi − xi−1
i2
Now by the mean value theorem from last semester, we havef (xi )−f (xi−1)
xi−xi−1= f ′(x∗i ) for some x∗i in the interval [xi−1, xi ]. Therefore
L = limn→∞
nXi=1
|Pi−1Pi | = limn→∞
nXi=1
q1 + [f ′(x∗
i)]2∆x =
Z b
a
q1 + [f ′(x)]2dx
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
If f is continuous and differentiable on the interval [a, b] andf ′ 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].
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = 2x3/2
3from (1, 2
3) to (2, 4
√2
3).
I f (x) = 2x3/2
3, f ′(x) = 3
2· 2
3x1/2 =
√x , [f ′(x)]2 = x , a = 1 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
1
√1 + x dx
I =R 3
2
√u du, where u = 1 + x , u(1) = 2, u(2) = 3.
I = u3/2
3/2
˛̨̨̨˛
3
2
= 23[3√
3− 2√
2].
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
If f is continuous and differentiable on the interval [a, b] andf ′ 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].
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = 2x3/2
3from (1, 2
3) to (2, 4
√2
3).
I f (x) = 2x3/2
3, f ′(x) = 3
2· 2
3x1/2 =
√x , [f ′(x)]2 = x , a = 1 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
1
√1 + x dx
I =R 3
2
√u du, where u = 1 + x , u(1) = 2, u(2) = 3.
I = u3/2
3/2
˛̨̨̨˛
3
2
= 23[3√
3− 2√
2].
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
If f is continuous and differentiable on the interval [a, b] andf ′ 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].
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = 2x3/2
3from (1, 2
3) to (2, 4
√2
3).
I f (x) = 2x3/2
3, f ′(x) = 3
2· 2
3x1/2 =
√x , [f ′(x)]2 = x , a = 1 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
1
√1 + x dx
I =R 3
2
√u du, where u = 1 + x , u(1) = 2, u(2) = 3.
I = u3/2
3/2
˛̨̨̨˛
3
2
= 23[3√
3− 2√
2].
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
If f is continuous and differentiable on the interval [a, b] andf ′ 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].
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = 2x3/2
3from (1, 2
3) to (2, 4
√2
3).
I f (x) = 2x3/2
3, f ′(x) = 3
2· 2
3x1/2 =
√x , [f ′(x)]2 = x , a = 1 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
1
√1 + x dx
I =R 3
2
√u du, where u = 1 + x , u(1) = 2, u(2) = 3.
I = u3/2
3/2
˛̨̨̨˛
3
2
= 23[3√
3− 2√
2].
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
If f is continuous and differentiable on the interval [a, b] andf ′ 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].
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = 2x3/2
3from (1, 2
3) to (2, 4
√2
3).
I f (x) = 2x3/2
3, f ′(x) = 3
2· 2
3x1/2 =
√x , [f ′(x)]2 = x , a = 1 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
1
√1 + x dx
I =R 3
2
√u du, where u = 1 + x , u(1) = 2, u(2) = 3.
I = u3/2
3/2
˛̨̨̨˛
3
2
= 23[3√
3− 2√
2].
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = ex +e−x
2, 0 ≤ x ≤ 2.
I f (x) = ex +e−x
2, f ′(x) = ex−e−x
2,
[f ′(x)]2 = e2x−2ex e−x +e−2x
4= e2x−2+e−2x
4, a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
q1 + e2x−2+e−2x
4dx =
R 2
0
qe2x +2+e−2x
4dx
I =R 2
0
r“ex +e−x
2
”2
dx =R 2
0ex +e−x
2dx .
I = ex−e−x
2
˛̨̨̨˛
2
0
= e2−e−2
2.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = ex +e−x
2, 0 ≤ x ≤ 2.
I f (x) = ex +e−x
2, f ′(x) = ex−e−x
2,
[f ′(x)]2 = e2x−2ex e−x +e−2x
4= e2x−2+e−2x
4, a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
q1 + e2x−2+e−2x
4dx =
R 2
0
qe2x +2+e−2x
4dx
I =R 2
0
r“ex +e−x
2
”2
dx =R 2
0ex +e−x
2dx .
I = ex−e−x
2
˛̨̨̨˛
2
0
= e2−e−2
2.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = ex +e−x
2, 0 ≤ x ≤ 2.
I f (x) = ex +e−x
2, f ′(x) = ex−e−x
2,
[f ′(x)]2 = e2x−2ex e−x +e−2x
4= e2x−2+e−2x
4, a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
q1 + e2x−2+e−2x
4dx =
R 2
0
qe2x +2+e−2x
4dx
I =R 2
0
r“ex +e−x
2
”2
dx =R 2
0ex +e−x
2dx .
I = ex−e−x
2
˛̨̨̨˛
2
0
= e2−e−2
2.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = ex +e−x
2, 0 ≤ x ≤ 2.
I f (x) = ex +e−x
2, f ′(x) = ex−e−x
2,
[f ′(x)]2 = e2x−2ex e−x +e−2x
4= e2x−2+e−2x
4, a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
q1 + e2x−2+e−2x
4dx =
R 2
0
qe2x +2+e−2x
4dx
I =R 2
0
r“ex +e−x
2
”2
dx =R 2
0ex +e−x
2dx .
I = ex−e−x
2
˛̨̨̨˛
2
0
= e2−e−2
2.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Find the arc length of the curve y = ex +e−x
2, 0 ≤ x ≤ 2.
I f (x) = ex +e−x
2, f ′(x) = ex−e−x
2,
[f ′(x)]2 = e2x−2ex e−x +e−2x
4= e2x−2+e−2x
4, a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
q1 + e2x−2+e−2x
4dx =
R 2
0
qe2x +2+e−2x
4dx
I =R 2
0
r“ex +e−x
2
”2
dx =R 2
0ex +e−x
2dx .
I = ex−e−x
2
˛̨̨̨˛
2
0
= e2−e−2
2.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Set up the integral which gives the arc length of the curvey = ex , 0 ≤ x ≤ 2.
I f (x) = ex , f ′(x) = ex , [f ′(x)]2 = e2x , a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
√1 + e2x dx .
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Set up the integral which gives the arc length of the curvey = ex , 0 ≤ x ≤ 2.
I f (x) = ex , f ′(x) = ex , [f ′(x)]2 = e2x , a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
√1 + e2x dx .
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length
L =
Z b
a
p1 + [f ′(x)]2dx or L =
Z b
a
r1 +
hdy
dx
i2
dx
Example Set up the integral which gives the arc length of the curvey = ex , 0 ≤ x ≤ 2.
I f (x) = ex , f ′(x) = ex , [f ′(x)]2 = e2x , a = 0 and b = 2.
I L =R b
a
p1 + [f ′(x)]2dx =
R 2
0
√1 + e2x dx .
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, curves of form x = g(y).
For a curve with equation x = g(y), where g(y) is continuous and has acontinuous derivative on the interval c ≤ y ≤ d , we can derive a similarformula for the arc length of the curve between y = c and y = d .
L =
Z d
c
q1 + [g′(y)]2dy or L =
Z d
c
vuut1 +h dx
dy
i2dy
Example Find the length of the curve 24xy = y 4 + 48 from the point ( 43, 2) to
( 114, 4).
I Solving for x in terms of y , we get x = y4+4824y
= y3
24+ 2
y= g(y).
I g ′(y) = 3y2
24− 2
y2 = y2
8− 2
y2 . [g ′(y)]2 = y4
64− 4
y2 · y2
8+ 4
y4 = y4
64− 1
2+ 4
y4
I L =R d
c
p1 + [f ′(y)]2dy =
R 4
2
q1 + y4
64− 1
2+ 4
y4 dy .
I =R 4
2
qy4
64+ 1
2+ 4
y4 dy =R 4
2
r“y2
8+ 2
y2
”2
dy .
I =R 4
2y2
8+ 2
y2 dy = y3
24− 2
y
˛̨̨̨˛
4
2
= 176.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, Simpson’s rule
We cannot always find an antiderivative for the integrand to evaluate the arclength. 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
I dx/dy = 1 + 12√
y,
I L =R 4
2
r1 +
hdxdy
i2dy =
R 42
r1 +
h1 + 1
2√
y
i2dy =
R 42
r2 + 1√
y+ 1
4ydy
I With n = 10, Simpson’s rule gives us
L ≈ S10 =∆y
3
ˆg(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)
˜
where g(y) =q
2 + 1√y
+ 14y
and ∆y = 4−210
. (see notes for details).
I We get S10 ≈ 3.269185.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, Simpson’s rule
We cannot always find an antiderivative for the integrand to evaluate the arclength. 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
I dx/dy = 1 + 12√
y,
I L =R 4
2
r1 +
hdxdy
i2dy =
R 42
r1 +
h1 + 1
2√
y
i2dy =
R 42
r2 + 1√
y+ 1
4ydy
I With n = 10, Simpson’s rule gives us
L ≈ S10 =∆y
3
ˆg(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)
˜
where g(y) =q
2 + 1√y
+ 14y
and ∆y = 4−210
. (see notes for details).
I We get S10 ≈ 3.269185.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, Simpson’s rule
We cannot always find an antiderivative for the integrand to evaluate the arclength. 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
I dx/dy = 1 + 12√
y,
I L =R 4
2
r1 +
hdxdy
i2dy =
R 42
r1 +
h1 + 1
2√
y
i2dy =
R 42
r2 + 1√
y+ 1
4ydy
I With n = 10, Simpson’s rule gives us
L ≈ S10 =∆y
3
ˆg(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)
˜
where g(y) =q
2 + 1√y
+ 14y
and ∆y = 4−210
. (see notes for details).
I We get S10 ≈ 3.269185.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, Simpson’s rule
We cannot always find an antiderivative for the integrand to evaluate the arclength. 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
I dx/dy = 1 + 12√
y,
I L =R 4
2
r1 +
hdxdy
i2dy =
R 42
r1 +
h1 + 1
2√
y
i2dy =
R 42
r2 + 1√
y+ 1
4ydy
I With n = 10, Simpson’s rule gives us
L ≈ S10 =∆y
3
ˆg(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)
˜
where g(y) =q
2 + 1√y
+ 14y
and ∆y = 4−210
. (see notes for details).
I We get S10 ≈ 3.269185.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
Arc Length, Simpson’s rule
We cannot always find an antiderivative for the integrand to evaluate the arclength. 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
I dx/dy = 1 + 12√
y,
I L =R 4
2
r1 +
hdxdy
i2dy =
R 42
r1 +
h1 + 1
2√
y
i2dy =
R 42
r2 + 1√
y+ 1
4ydy
I With n = 10, Simpson’s rule gives us
L ≈ S10 =∆y
3
ˆg(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)
˜
where g(y) =q
2 + 1√y
+ 14y
and ∆y = 4−210
. (see notes for details).
I We get S10 ≈ 3.269185.
Annette Pilkington Lecture 16 : Arc Length
Arc Length
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
s(x) =
Z x
a
p1 + [f ′(t)]2dt.
From the fundamental theorem of calculus, we see that s ′(x) =p
1 + [f ′(x)]2.In the language of differentials, this translates to
ds =
r1 +
“dy
dx
”2
dx or (ds)2 = (dx)2 + (dy)2
Annette Pilkington Lecture 16 : Arc Length
Arc Length
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
s(x) =
Z x
a
p1 + [f ′(t)]2dt.
Example Find the arc length function for the curve y = 2x3/2
3taking P0(1, 3/2)
as the starting point.
I We haved 2x3/2
3dx
=√
x
I
s(x) =
Z x
1
q1 + (
√t)2dt =
Z x
1
√1 + tdt =
Z 1+x
2
√udu
where u = 1 + t
I
=u3/2
3/2
˛̨̨̨˛x+1
2
= 2(x + 1)√
x + 1/3− 4√
2/3
Annette Pilkington Lecture 16 : Arc Length
Arc Length
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
s(x) =
Z x
a
p1 + [f ′(t)]2dt.
Example Find the arc length function for the curve y = 2x3/2
3taking P0(1, 3/2)
as the starting point.
I We haved 2x3/2
3dx
=√
x
I
s(x) =
Z x
1
q1 + (
√t)2dt =
Z x
1
√1 + tdt =
Z 1+x
2
√udu
where u = 1 + t
I
=u3/2
3/2
˛̨̨̨˛x+1
2
= 2(x + 1)√
x + 1/3− 4√
2/3
Annette Pilkington Lecture 16 : Arc Length
Arc Length
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
s(x) =
Z x
a
p1 + [f ′(t)]2dt.
Example Find the arc length function for the curve y = 2x3/2
3taking P0(1, 3/2)
as the starting point.
I We haved 2x3/2
3dx
=√
x
I
s(x) =
Z x
1
q1 + (
√t)2dt =
Z x
1
√1 + tdt =
Z 1+x
2
√udu
where u = 1 + t
I
=u3/2
3/2
˛̨̨̨˛x+1
2
= 2(x + 1)√
x + 1/3− 4√
2/3
Annette Pilkington Lecture 16 : Arc Length
Arc Length
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
s(x) =
Z x
a
p1 + [f ′(t)]2dt.
Example Find the arc length function for the curve y = 2x3/2
3taking P0(1, 3/2)
as the starting point.
I We haved 2x3/2
3dx
=√
x
I
s(x) =
Z x
1
q1 + (
√t)2dt =
Z x
1
√1 + tdt =
Z 1+x
2
√udu
where u = 1 + t
I
=u3/2
3/2
˛̨̨̨˛x+1
2
= 2(x + 1)√
x + 1/3− 4√
2/3
Annette Pilkington Lecture 16 : Arc Length