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

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) =

a‖r′(u)‖du.

In some cases we can solve this equation for t andre-parametrize the curve in terms of arc length.

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) =

a‖r′(u)‖du.

In some cases we can solve this equation for t andre-parametrize the curve in terms of arc length.

Parametrizing by Arc Length

Example

Suppose r(t) = 〈cos 2t , sin 2t , t〉 (a helix). Find an arc lengthparameterization for this curve.

s(t) =

√(−2 sin 2u)2 + (2 cos 2u)2 + 1 du =

√5 du =

then t = s/√

x = cos(

2s√5

)y = sin

(2s√

s√5.

Parametrizing by Arc Length

Example

Suppose r(t) = 〈cos 2t , sin 2t , t〉 (a helix). Find an arc lengthparameterization for this curve.

s(t) =

√(−2 sin 2u)2 + (2 cos 2u)2 + 1 du =

√5 du =

then t = s/√

x = cos(

2s√5

)y = sin

(2s√

s√5.

Unit Tangent Vector

Q: Why is the arc length parameterization important?

A: ‖r′(s)‖ =√

(f ′(s))2 + (g′(s))2 + (h′(s))2 =

√(dsds

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.

Unit Tangent Vector

A: ‖r′(s)‖ =√

(f ′(s))2 + (g′(s))2 + (h′(s))2 =

√(dsds

T(t) =r′(t)‖r′(t)‖

Unit Tangent Vector

A: ‖r′(s)‖ =√

(f ′(s))2 + (g′(s))2 + (h′(s))2 =

√(dsds

T(t) =r′(t)‖r′(t)‖

Unit Tangent Vector

A: ‖r′(s)‖ =√

(f ′(s))2 + (g′(s))2 + (h′(s))2 =

√(dsds

T(t) =r′(t)‖r′(t)‖

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〉

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〉

Curvature

DefinitionThe curvature, denoted κ, of a curve is the scalar

∥∥∥∥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

∥∥∥∥dTds

∥∥∥∥ =

∥∥∥∥∥ dTdtdsdt

∥∥∥∥∥ =‖T′(t)‖‖r′(t)‖

Curvature

DefinitionThe curvature, denoted κ, of a curve is the scalar

∥∥∥∥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

∥∥∥∥dTds

∥∥∥∥ =

∥∥∥∥∥ dTdtdsdt

∥∥∥∥∥ =‖T′(t)‖‖r′(t)‖

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〉‖

Example (1 of 3)

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〉‖

Example (2 of 3)

r(t) = 〈at + x0,bt + y0, ct + z0〉

r′(t) = 〈a,b, c〉

T(t) =〈a,b, c〉√

a2 + b2 + c2

κ = 0

Example (2 of 3)

r(t) = 〈at + x0,bt + y0, ct + z0〉

r′(t) = 〈a,b, c〉

T(t) =〈a,b, c〉√

a2 + b2 + c2

κ = 0

Example (2 of 3)

r(t) = 〈at + x0,bt + y0, ct + z0〉

r′(t) = 〈a,b, c〉

T(t) =〈a,b, c〉√

a2 + b2 + c2

κ = 0

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)

Example (3 of 3)

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

26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉

These derivatives are getting complicated.

Example (3 of 3)

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

26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉

Example (3 of 3)

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

26 + 6 cos 4t〈−2 sin 2t ,4 cos 2t ,4〉

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

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〉

√896− 384 cos 4t(26 + 6 cos 4t)3

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〉

√896− 384 cos 4t(26 + 6 cos 4t)3

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 .

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 .

Example

Find the curvature of the parabola y = x2.

κ =|2|

(1 + (2x)2)3/2 =2

(1 + 4x2)3/2

Example

Find the curvature of the parabola y = x2.

κ =|2|

(1 + (2x)2)3/2 =2

(1 + 4x2)3/2

Homework

Read Section 11.4.Exercises: 1–53 odd

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