Curvature - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Curvature/main.pdf ·...
Transcript of Curvature - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Curvature/main.pdf ·...
CurvatureMATH 311, Calculus III
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan Curvature
Intuitive Idea
Curvature is a measure of instantaneously how much a curvebends per unit length.
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
J. Robert Buchanan Curvature
Arc Length Revisited
Recall: if a curve is traced out by the vector-valued functionr(t) = 〈f (t),g(t),h(t)〉 for a ≤ t ≤ b, the arc length of the curvefrom u = a to u = t is
s(t) =
∫ t
a‖r′(u)‖du.
In some cases we can solve this equation for t andre-parametrize the curve in terms of arc length.
J. Robert Buchanan Curvature
Arc Length Revisited
Recall: if a curve is traced out by the vector-valued functionr(t) = 〈f (t),g(t),h(t)〉 for a ≤ t ≤ b, the arc length of the curvefrom u = a to u = t is
s(t) =
∫ t
a‖r′(u)‖du.
In some cases we can solve this equation for t andre-parametrize the curve in terms of arc length.
J. Robert Buchanan Curvature
Parametrizing by Arc Length
Example
Suppose r(t) = 〈cos 2t , sin 2t , t〉 (a helix). Find an arc lengthparameterization for this curve.
Since
s(t) =
∫ t
0
√(−2 sin 2u)2 + (2 cos 2u)2 + 1 du =
∫ t
0
√5 du =
√5t
then t = s/√
5 and
x = cos(
2s√5
)y = sin
(2s√
5
)z =
s√5.
J. Robert Buchanan Curvature
Parametrizing by Arc Length
Example
Suppose r(t) = 〈cos 2t , sin 2t , t〉 (a helix). Find an arc lengthparameterization for this curve.
Since
s(t) =
∫ t
0
√(−2 sin 2u)2 + (2 cos 2u)2 + 1 du =
∫ t
0
√5 du =
√5t
then t = s/√
5 and
x = cos(
2s√5
)y = sin
(2s√
5
)z =
s√5.
J. Robert Buchanan Curvature
Unit Tangent Vector
Q: Why is the arc length parameterization important?
A: ‖r′(s)‖ =√
(f ′(s))2 + (g′(s))2 + (h′(s))2 =
√(dsds
)2
= 1.
Recall: If a curve is traced out by the vector-valued functionr(t), then the vector r′(t) is tangent to the curve for each valueof t .
DefinitionIf r′(t) 6= 0 then the vector
T(t) =r′(t)‖r′(t)‖
is called the unit tangent vector to the curve.
J. Robert Buchanan Curvature
Unit Tangent Vector
Q: Why is the arc length parameterization important?
A: ‖r′(s)‖ =√
(f ′(s))2 + (g′(s))2 + (h′(s))2 =
√(dsds
)2
= 1.
Recall: If a curve is traced out by the vector-valued functionr(t), then the vector r′(t) is tangent to the curve for each valueof t .
DefinitionIf r′(t) 6= 0 then the vector
T(t) =r′(t)‖r′(t)‖
is called the unit tangent vector to the curve.
J. Robert Buchanan Curvature
Unit Tangent Vector
Q: Why is the arc length parameterization important?
A: ‖r′(s)‖ =√
(f ′(s))2 + (g′(s))2 + (h′(s))2 =
√(dsds
)2
= 1.
Recall: If a curve is traced out by the vector-valued functionr(t), then the vector r′(t) is tangent to the curve for each valueof t .
DefinitionIf r′(t) 6= 0 then the vector
T(t) =r′(t)‖r′(t)‖
is called the unit tangent vector to the curve.
J. Robert Buchanan Curvature
Unit Tangent Vector
Q: Why is the arc length parameterization important?
A: ‖r′(s)‖ =√
(f ′(s))2 + (g′(s))2 + (h′(s))2 =
√(dsds
)2
= 1.
Recall: If a curve is traced out by the vector-valued functionr(t), then the vector r′(t) is tangent to the curve for each valueof t .
DefinitionIf r′(t) 6= 0 then the vector
T(t) =r′(t)‖r′(t)‖
is called the unit tangent vector to the curve.
J. Robert Buchanan Curvature
Illustration (1 of 2)
Example
Find the unit tangent vector to r(t) = 〈cos 2t , sin 2t , t〉.
T(t) =r′(t)‖r′(t)‖
=〈−2 sin 2t ,2 cos 2t ,1〉√
(−2 sin 2t)2 + (2 cos 2t)2 + (1)2
=1√5〈−2 sin 2t ,2 cos 2t ,1〉
J. Robert Buchanan Curvature
Illustration (1 of 2)
Example
Find the unit tangent vector to r(t) = 〈cos 2t , sin 2t , t〉.
T(t) =r′(t)‖r′(t)‖
=〈−2 sin 2t ,2 cos 2t ,1〉√
(−2 sin 2t)2 + (2 cos 2t)2 + (1)2
=1√5〈−2 sin 2t ,2 cos 2t ,1〉
J. Robert Buchanan Curvature
Illustration (2 of 2)
-1
0
1
x
-1
0
1y
0
2
4
6
z
J. Robert Buchanan Curvature
Curvature
DefinitionThe curvature, denoted κ, of a curve is the scalar
κ =
∥∥∥∥dTds
∥∥∥∥where T is the unit tangent vector to the curve and s is the arclength parameter.
By the FTC Part II, if s(t) =∫ t
a ‖r′(u)‖du, then ds
dt = ‖r′(t)‖.
By the Chain Rule
κ =
∥∥∥∥dTds
∥∥∥∥ =
∥∥∥∥∥ dTdtdsdt
∥∥∥∥∥ =‖T′(t)‖‖r′(t)‖
.
J. Robert Buchanan Curvature
Curvature
DefinitionThe curvature, denoted κ, of a curve is the scalar
κ =
∥∥∥∥dTds
∥∥∥∥where T is the unit tangent vector to the curve and s is the arclength parameter.
By the FTC Part II, if s(t) =∫ t
a ‖r′(u)‖du, then ds
dt = ‖r′(t)‖.
By the Chain Rule
κ =
∥∥∥∥dTds
∥∥∥∥ =
∥∥∥∥∥ dTdtdsdt
∥∥∥∥∥ =‖T′(t)‖‖r′(t)‖
.
J. Robert Buchanan Curvature
Example (1 of 3)
Find the curvature of the following curve.
r(t) = 〈a cos t ,a sin t〉
Assume a > 0.
r′(t) = 〈−a sin t ,a cos t〉
T(t) =〈−a sin t ,a cos t〉√
(−a sin t)2 + (a cos t)2= 〈− sin t , cos t〉
κ =‖〈− cos t ,− sin t〉‖
a=
1a
J. Robert Buchanan Curvature
Example (1 of 3)
Find the curvature of the following curve.
r(t) = 〈a cos t ,a sin t〉
Assume a > 0.
r′(t) = 〈−a sin t ,a cos t〉
T(t) =〈−a sin t ,a cos t〉√
(−a sin t)2 + (a cos t)2= 〈− sin t , cos t〉
κ =‖〈− cos t ,− sin t〉‖
a=
1a
J. Robert Buchanan Curvature
Example (2 of 3)
Find the curvature of the following curve.
r(t) = 〈at + x0,bt + y0, ct + z0〉
r′(t) = 〈a,b, c〉
T(t) =〈a,b, c〉√
a2 + b2 + c2
κ = 0
J. Robert Buchanan Curvature
Example (2 of 3)
Find the curvature of the following curve.
r(t) = 〈at + x0,bt + y0, ct + z0〉
r′(t) = 〈a,b, c〉
T(t) =〈a,b, c〉√
a2 + b2 + c2
κ = 0
J. Robert Buchanan Curvature
Example (2 of 3)
Find the curvature of the following curve.
r(t) = 〈at + x0,bt + y0, ct + z0〉
r′(t) = 〈a,b, c〉
T(t) =〈a,b, c〉√
a2 + b2 + c2
κ = 0
J. Robert Buchanan Curvature
Half-Angle Formulas
The Half-Angle Formulas are sometimes useful for simplifyingexpressions for curvature.
cos2 θ =12(1 + cos 2θ)
sin2 θ =12(1− cos 2θ)
J. Robert Buchanan Curvature
Example (3 of 3)
Find the curvature of the following curve.
r(t) = 〈cos 2t ,2 sin 2t ,4t〉
r′(t) = 〈−2 sin 2t ,4 cos 2t ,4〉
T(t) =〈−2 sin 2t ,4 cos 2t ,4〉√
(−2 sin 2t)2 + (4 cos 2t)2 + (4)2
=1√
26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉
These derivatives are getting complicated.
J. Robert Buchanan Curvature
Example (3 of 3)
Find the curvature of the following curve.
r(t) = 〈cos 2t ,2 sin 2t ,4t〉
r′(t) = 〈−2 sin 2t ,4 cos 2t ,4〉
T(t) =〈−2 sin 2t ,4 cos 2t ,4〉√
(−2 sin 2t)2 + (4 cos 2t)2 + (4)2
=1√
26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉
These derivatives are getting complicated.
J. Robert Buchanan Curvature
Example (3 of 3)
Find the curvature of the following curve.
r(t) = 〈cos 2t ,2 sin 2t ,4t〉
r′(t) = 〈−2 sin 2t ,4 cos 2t ,4〉
T(t) =〈−2 sin 2t ,4 cos 2t ,4〉√
(−2 sin 2t)2 + (4 cos 2t)2 + (4)2
=1√
26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉
These derivatives are getting complicated.
J. Robert Buchanan Curvature
Another Method for Finding κ
TheoremThe curvature of the smooth curve traced out by thevector-valued function r(t) is given by
κ =‖r′(t)× r′′(t)‖‖r′(t)‖3
.
J. Robert Buchanan Curvature
Example
ExampleUse this formula to find the curvature ofr(t) = 〈cos 2t ,2 sin 2t ,4t〉.
r′(t) = 〈−2 sin 2t ,4 cos 2t ,4〉r′′(t) = 〈−4 cos 2t ,−8 sin 2t ,0〉
r′(t)× r′′(t) = 〈32 sin 2t ,−16 cos 2t ,16〉
κ =
√896− 384 cos 4t(26 + 6 cos 4t)3
J. Robert Buchanan Curvature
Example
ExampleUse this formula to find the curvature ofr(t) = 〈cos 2t ,2 sin 2t ,4t〉.
r′(t) = 〈−2 sin 2t ,4 cos 2t ,4〉r′′(t) = 〈−4 cos 2t ,−8 sin 2t ,0〉
r′(t)× r′′(t) = 〈32 sin 2t ,−16 cos 2t ,16〉
κ =
√896− 384 cos 4t(26 + 6 cos 4t)3
J. Robert Buchanan Curvature
Finding κ for Curves in the xy -plane
If a curve in the xy -plane is described by the function y = f (x),then we can parametrize this curve as r(t) = 〈t , f (t),0〉.
Then we calculate the curvature as
κ =‖r′(t)× r′′(t)‖‖r′(t)‖3
=‖〈1, f ′(t),0〉 × 〈0, f ′′(t),0〉‖
‖〈1, f ′(t),0〉‖3
=|f ′′(t)|[
1 + (f ′(t))2]3/2 .
J. Robert Buchanan Curvature
Finding κ for Curves in the xy -plane
If a curve in the xy -plane is described by the function y = f (x),then we can parametrize this curve as r(t) = 〈t , f (t),0〉.
Then we calculate the curvature as
κ =‖r′(t)× r′′(t)‖‖r′(t)‖3
=‖〈1, f ′(t),0〉 × 〈0, f ′′(t),0〉‖
‖〈1, f ′(t),0〉‖3
=|f ′′(t)|[
1 + (f ′(t))2]3/2 .
J. Robert Buchanan Curvature
Example
Find the curvature of the parabola y = x2.
κ =|2|
(1 + (2x)2)3/2 =2
(1 + 4x2)3/2
J. Robert Buchanan Curvature
Example
Find the curvature of the parabola y = x2.
κ =|2|
(1 + (2x)2)3/2 =2
(1 + 4x2)3/2
J. Robert Buchanan Curvature
Homework
Read Section 11.4.Exercises: 1–53 odd
J. Robert Buchanan Curvature