88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a

2.4 Curvature

2.4.1 De…nitions and Examples

The notion of curvature measures how sharply a curve bends. We would expect the curvature to be 0 for a straight line, to be very small for curves which bend very little and to be large for curves which bend sharply. If we move along a curve, we see that the direction of the tangent vector will not change as long as the curve is ‡at. Its direction will change if the curve bends. The more the curve bends, the more the direction of the tangent vector will change. So, it makes sense to study how the tangent vector changes as we move along a curve.

But because we are only interested in the direction of the tangent vector, not its magnitude, we will consider the unit tangent vector. Curvature is de…ned as follows:

De…nition 150 (Curvature) Let C be a smooth curve with position vector

!r (s) where s is the arc length parameter. The curvature of C is de…ned to be:

= d!T

ds (2.11)

where!T is the unit tangent vector. Note the letter used to denote the curvature is the greek letter kappa denoted .

Remark 151 The above formula implies that !T be expressed in terms of s, arc length. And therefore, we must have the curve parametrized in terms of arc length.

Example 152 Find the curvature of a circle of radius a.

We saw earlier that the parametrization of a circle of radius a with respect to arc length length was !r (s) =D

a coss a; a sins

a E

. First, we need to compute !T (s).

By de…nition,

!T (s) = !r⁰(s) k!r⁰(s)k So, we must …rst compute !r⁰(s).

!r⁰(s) = a 1 asins

a ; a 1 acoss

a

= D

sins a; coss

a E

We can see that k!r⁰(s)k = 1 (but we already knew that from theorem 148. Thus

!T (s) =D sins

a; coss a E

It follows that

d!T

ds = 1

acoss a; 1

asins a

and therefore

= d!

T ds

= 1

a

In other words, the curvature of a circle is the inverse of its radius. This agrees with our intuition of curvature. Curvature is supposed to measure how sharply a curve bends. The larger the radius of a circle, the less it will bend, that is the less its curvature should be. This is indeed the case. The larger the radius, the smaller its inverse.

Example 153 Find the curvature of the circular helix.

Earlier, we found that the parametrization of the circular helix with respect to arc length was !r (s) = cos s

p2; sin s p2; s

p2 . As before, we need to compute

!T (s) which can be obtained from !r⁰(s).

!r⁰(s) = 1 p2sin s

p2; 1 p2cos s

p2; 1 p2 Thus

!T (s) = 1 p2sin s

p2; 1 p2cos s

p2; 1 p2

= 1

p2 sin s

p2; cos s p2; 1 Therefore

d!T ds = 1

p2 p1

2cos s p2; 1

p2sin s p2; 0 It follows that

= d!

T ds

= 1

2

The de…nition for the curvature works well when the curve is parametrized with respect to arc length, or when this can be done easily. However, since s can be expressed as a function as t, there should also be formulas for the curvature in term of t. The next theorems give us various formulas for the curvature.

Theorem 154 Let C be a smooth curve with position vector !r (t) where t is any parameter. Then the following formulas can be used to compute .

=

!T⁰(t)

k!r⁰(t)k (2.12)

=k!r⁰(t) !r⁰⁰(t)k

k!r⁰(t)k³ (2.13)

Proof. We prove each formula separately.

1. Proof of =

!T⁰(t) k!r⁰(t)k

Using the chain rule, we have d!T dt = d!T

ds ds dt =d!T

ds k!r⁰(t)k from formula 2.10. Thus,

d!T ds =

d!T dt k!r⁰(t)k

=

!T⁰(t) k!r⁰(t)k Hence

= d!T ds

=

!T⁰(t) k!r⁰(t)k

2. Proof of =k!r⁰(t) !r⁰⁰(t)k k!r⁰(t)k³

We express !r⁰(t) and !r⁰⁰(t) in terms of T and T⁰:then compute their cross product.

Computation of !r⁰. Since!T = !r⁰

k!r⁰k and ds

dt = k!r⁰k, we get that

!r⁰= ds dt

!T

Computation of !r⁰⁰ Taking the derivative with respect to t of the previous formula gives us

!r⁰⁰= d²s dt²

!T +ds dt

!T⁰

Computation of !r⁰(t) !r⁰⁰(t). From the two previous formulas and using the properties of cross products, we see that

!r⁰ !r⁰⁰= ds dt

d²s dt²

!T !T + ds dt

2 !T !T⁰

Since the cross product of a vector by itself is always the zero vector, we see that

k!r⁰ !r⁰⁰k = ds dt

2 !T !T⁰

= ds

dt

2 !T !T⁰ sin

where is the angle between!T and!T⁰. Since !T = 1, by proposi- tion 125, we know that!

T ?!T⁰ thus k!r⁰ !r⁰⁰k = ds

dt

2 !T⁰

= k!r⁰(t)k² !T⁰ from formula 2:10:

Therefore

!T⁰ = k!r⁰ !r⁰⁰k k!r⁰(t)k² So

=

!T⁰(t) k!r⁰(t)k

= k!r⁰ !r⁰⁰k k!r⁰(t)k³

Remark 155 Formula 2.12 is consistent with the de…nition of curvature. It says that if t is any parameter used for a curve C, then the curvature of C is

=

!T⁰(t)

k!r⁰(t)k. In the case the parameter is s, then the formula and using the fact that k!r⁰(s)k = 1, the formula gives us the de…nition of curvature.

Theorem 156 If C is a curve with equation y = f (x) where f is twice di¤ er- entiable then

= jf⁰⁰(x)j 1 + (f⁰(x))²

3 2

(2.14)

Proof. The proof follows easily from formula 2.13. First, let us remark that it is easy to parametrize the curve given by y = f (x) as a 3-D parametric curve.

We can simply use 8

<

:

x = x y = f (x)

z = 0

Using x as the name of the parameter. Thus, the position vector of our curve is !r (x) = hx; f (x) ; 0i. It follows that

!r⁰(x) = h1; f⁰(x) ; 0i and !r⁰⁰(x) = h0; f⁰⁰(x) ; 0i

Thus !r⁰ !r⁰⁰= h0; 0; f⁰⁰(x)i

Hence

k!r⁰ !r⁰⁰k = jf⁰⁰(x)j and

k!r⁰k = q

1 + (f⁰(x))² Therefore

= k!r⁰ !r⁰⁰k k!r⁰(t)k³

= jf⁰⁰(x)j q

1 + (f⁰(x))²

3

= jf⁰⁰(x)j 1 + (f⁰(x))²

3 2

We illustrate these formulas with some examples. Of course, the formula we use depends on the information given.

Example 157 Find the curvature of the curve given by !r (t) = 2t; t²; 1 3t³ as a function of t

Since we have the position vector describing the curve but it is not given with respect to arc length, we can …nd by using either formula 2.12 or formula 2.13.

Since this is an example, we show both methods. Since both methods require !r⁰ and k!r⁰k, we compute both here.

!r⁰(t) = 2; 2t; t² Therefore

k!r⁰(t)k = p

4 + 2t²+ t⁴

= q

(2 + t²)²

= 2 + t² since 2 + t²> 0

Method 1 Here, we use formula 2.12. Using this formula, we need to …nd

!T⁰(t)

!T (t) = !r⁰(t) k!r⁰(t)k

= 2; 2t; t² 2 + t²

= 2

2 + t²; 2t

2 + t²; t² 2 + t² So,

!T⁰(t) = 1

(2 + t²)² 4t; 4 2t²; 4t therefore

!T⁰(t) =

p16t²+ 16 16t²+ 4t⁴+ 16t² (2 + t²)²

=

p4 (t⁴+ 4t²+ 4) (2 + t²)²

= 2

2 + t² It follows that

=

!T⁰(t) k!r⁰(t)k

= 2

(2 + t²)²

Method 2 Here, we use formula 2.13.Using this formula, we need to …nd

!r⁰⁰(t).

!r⁰⁰(t) = h0; 2; 2ti Next, we …nd k!r⁰(t) !r⁰⁰(t)k.

!r⁰(t) !r⁰⁰(t) = 2; 2t; t² h0; 2; 2ti

= 2t²; 4t; 4 Therefore

k!r⁰(t) !r⁰⁰(t)k = 2t²; 4t; 4

= p

4t⁴+ 16t²+ 16

= 2 t²+ 2

Finally,

= k!r⁰(t) !r⁰⁰(t)k k!r⁰(t)k³

= 2 t²+ 2 (t²+ 2)³

= 2

(2 + t²)² The same answer as above.

Example 158 Find the curvature of y = x² as a function of x. Then, …nd the curvature of the same curve when x = 0, x = 1.

Here, we use formula 2.14. Let f (x) = x². First, we need to …nd f⁰(x) and f⁰⁰(x).

f⁰(x) = 2x f⁰⁰(x) = 2 So

= jf⁰⁰(x)j 1 + (f⁰(x))²

3 2

= 2

(1 + 4x²)³² When x = 0, we get

= 2 When x = 1, we get

= 2 5³²

Make sure you have read, studied and understood what was done above before attempting the problems.

2.4.2 Problems

Do odd # 1, 3 (only …nd!T and ), 5, 9 - 15 (only …nd!T and ), 19, 23, 25 at the end of 10.4 in your book.