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Lecture XVII

Abstract

We introduce the concepts of vector functions, scalar and vectorelds and stress their relevance in applied sciences. We study curvesin three-dimensional Euclidean space and introduce the concept of arclength, curvature and torsion for a given curve in R 3 .

Vector functions, vector and scalar elds

Denition 1 A vector-valued function is a map associating vectors toreal numbers, that is

r : R R 3 such that r (t) = f (t)e x + g(t)e y + h(t)e z ,where ex , ey , and ez are the unit vectors along the x, y, and z -axes, respec-tively. The domain of r is the intersection of the domains of the functions f , g and h.

Notice that if the parameter t is interpreted as time, then we can think tor (t) as a function describing the trajectory of a particle of mass m in thethree-dimensional Euclidean space. Indeed, we could obtain r (t) by solvingNewton 1 s equation

md2 rdt2

= F ,

where F is the external force acting on the particle.

Example 1 Find the domain of denition of the vector-valued function

r (t) = t3 e x + ln (3 t)e y + te z .Notice that

t3

is a polynomial and its domain of denition is the whole real line.

ln(3 t) will be dened if 3t > 0, that is for t < 3. t is dened if t 0.

Hence, we conclude that the domain of denition of the given vector-valued function is 0 t < 3, that is the interval [0, 3).

1 Isaac Newton (1643-1727) was an English physicist, mathematician, astronomer, nat-ural philosopher, alchemist and theologian.

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In mathematics a vector eld is a construction in vector calculus whichassociates a vector to every point in a subset of the Euclidean space R 3 , thatis a map

V : R 3 R3 such that V (x,y,z ) = F (x,y,z )e x+ G(x,y,z )e y+ H (x,y,z )e z .

Vector elds are often used in physics and engineering to model, for example,the speed and direction of a moving uid throughout space, or the strengthand direction of some force, such as the magnetic, electric or gravitationalforce, as it changes from point to point. In mathematics, physics and en-gineering a scalar eld associates a scalar value to every point in a space,that is a map

S : R 3 R such that ( x,y,z ) S (x,y,z ).Scalar elds are required to be coordinate-independent, meaning that anytwo observers using the same units will agree on the value of the scalareld at the same point in space (or spacetime). Examples used in physicsand engineering include the temperature distribution throughout space, thepressure distribution in a uid, the density of a uid or solid object, theelectrostatic potential, etc. .

Denition 2 We dene the limit of a vector-valued function r (t) as t goes to a

R as

limt a

r (t) = limt a

f (t) ex + limt a

g(t) ey + limt a

h(t) ez .

If the above limit exists we say that r (t) is continuous at t = a.

Example 2 Find the limit for t 1 of r (t) = t + 3 ex + t 1t2 1

ey + tan t

t ez .

According to the denition above we have to compute the corresponding limits for the components of the given vector-valued function. Thus, we nd that

limt 1 t + 3 = 4 = 2,limt 1

t 1t2 1

= limt 1

t 1(t 1)(t + 1)

= limt 1

1t + 1

= 12

,

limt 1

tan tt

= tan(1) 1.557.Hence, we conclude that

limt 1

r (t) = 2 e x + ey2

+ tan(1) ez .

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In mathematics, a parametric equation is a relation dened by the use of parameters. A simple kinematical example is when we use a time parameterto determine the position, velocity, and other information about a body orparticle in motion. In general, a parametric equation is expressed by meansof a set of equations. For example, the simplest equation for a parabola

y = x2

can be parametrized by using a free parameter t, and setting

x = t, y = t2 .

Although the preceding example is a somewhat trivial case, consider forinstance the following parametrization of a circle of radius R

x = R cos t, y = R sin t,

where t is in the range 0 to 2. Converting a set of parametric equations toa single equation involves eliminating the variable t from the simultaneousequations x = x(t) and y = y(t). If one of these equations can be solvedfor t, the expression obtained can be substituted into the other equation toobtain an equation involving x and y only. In the case of the example with

the parametric equations of a circle we can square both x and y, add themtogether and end up with the well-known formula

x2 + y2 = R2 .

Notice that once we have a vector-valued function

r (t) = f (t)e x + g(t)e y + h(t)e z

describing the trajectory of a particle in R 3 the curve on which this particlemoves can also be described in parametric form by setting

x = f (t), y = g(t), z = h(t).

Notice that at each time t the above equation will x a certain point of coordinates ( x,y,z ) in space.

Example 3 Consider the vector-valued function

r (t) = cos t ex + sin t ey + te z .

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We want to understand the form of the trajectory whose points are described by the above vector-valued function as the parameter t varies. If we denote by C such a trajectory, the parametric equations for the curve C are

x = cos t, y = sin t, z = t.

Since x2 + y2 = 1 whatever is the value of z we conclude that the points of the curve C must lay on the lateral surface of a cylinder of radius one.Furthermore, if we consider for instance different values of t, for instance we start by taking t = 0, t = / 2, t = 3/ 2 and so on we discover that the trajectory of the particle is a helix . For a picture download the le helix.gif on the web page of the course. The plot has been generated with the command

spacecurve in Maple and by taking t in the interval [6, 6].Denition 3 Let r (t) be the vector-valued function r (t) = f (t)e x + g(t)e y + h(t)e z .

We dene the derivative r (t) of r (t) as

r (t) = lim t 0

r (t + t) r (t) t

.

If the above limit exists, then we have

r (t) = df

dt e

x +

dg

dt e

y +

dh

dt e

z.

Concerning the geometrical interpretation of r (t), if r (t) describes a curveC in R 3 the derivative r (t) is the tangent vector to the curve C at the pointP = ( x(t), y(t), z (t)). In what follows we shall denote the unit tangent vectorby

T (t) = r (t)

|r (t)|.

From the physical point of view r (t) can be interpreted as the velocity of a particle having trajectory C described by the vector-valued function r (t).The interpretation of the second derivative of r (t) is that of an acceleration. If

and denote the dot and cross product, respectively, we have the followingdifferentiation rulesddt

(u v ) = u v + u v ,ddt

(u v ) = u v + u v ,ddt

[u (f (t))] = f (t)u (f (t)) , (chain rule )

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where u (t) and v (t) are two arbitrary vector-valued functions.

Denition 4 Let

r (t) = f (t)e x + g(t)e y + h(t)e z .

be a continuous vector-valued function. We dene the integral of r (t) as follows

b

adt r (t) =

b

adt f (t) e x +

b

adt g(t) e y +

b

adt h(t) e z .

A curve in the plane can be approximated by connecting a nite number of points on the curve using line segments to create a polygonal path. Since itis straightforward to calculate the length of each linear segment (using thePythagorean theorem in Euclidean space, for example), the total length of theapproximation can be found by summing the lengths of each linear segment.If the curve is not already a polygonal path, better approximations to thecurve can be obtained by following the shape of the curve increasingly moreclosely. The approach is to use an increasingly larger number of segments of smaller lengths. The lengths of the successive approximations do not decreaseand will eventually keep increasingpossibly indenitely, but for smooth curvesthis will tend to a limit as the lengths of the segments get arbitrarily small.

For some curves there is a smallest number L that is an upper bound on thelength of any polygonal approximation. If such a number exists, then thecurve is said to be rectiable and the curve is dened to have arc length L.

Denition 5 Let C be a curve in R 3 described by the vector-valued function

r (t) = x(t)e x + y(t)e y + z (t)e z

where t [a, b]. If C is described only once as t increases from a to b, then its length or arc length is given by the formula

L = b

adt |r (t)| =

b

adt

dxdt

2

+dydt

2

+dz dt

2

.

Example 4 Find the length of the helix

r (t) = cos t ex + sin t ey + t ez

from the point (1, 0, 0) to (1, 0, 2). First of all, we have to nd the values of t corresponding to the two given points. For the rst point we obtain

1 e x + 0 e y + 0 e z = cos t ex + sin t ey + t ez .

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The above vectorial equation gives rise to the following three equations

cos t = 1, sin t = 0, t = 0.

which are clearly satised for t = 0. In an analogous way we nd that r (t)will identify the point (1, 0, 2) whenever t = 2. Furthermore,

r (t) = sin t ex + cos t ey + e z ,|r (t)| = sin2 t + cos 2 t + 1 = 2.Hence, the length will be given by

L =

2

0

dt

|r (t)

|=

2

0

dt 2 = 2 2.It is often convenient to parametrize a curve with respect to arc length be-cause arc length arises from the shape of a curve and does not depend ona particular coordinate system. Moreover, with this reparameterization wecan tell where we are on the curve after we have travelled a distance s alongthe curve.Denition 6 Let C be a curve in R 3 described by the vector-valued function

r (t) = x(t)e x + y(t)e y + z (t)e zwhere t[a, b]. Further suppose that C is described only once as t increases

from a to b. We dene the arc length function s of the curve C as

s(t) = t

ad |r ( )| =

t

ad dxd

2

+dyd

2

+dz d

2

. (1)

Notice that from the Fundamental Theorem of Calculus we also have dsdt

= |r (t)|. (2)Example 5 Reparameterize the helix

r (t) = cos t ex + sin t ey + t ez

with respect to arc length from (1, 0, 0) in the direction of increasing t. We already know that the initial point is associated to t = 0. Computing the integral in (1) we obtain

s(t) = t

0d |r ( )| = 2

t

0d = 2t = t =

s 2.

We conclude that

r (s) = cos s 2 ex + sin

s 2 ey +

s 2 e z .

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The curvature of a curve C can be thought as a measure of how quickly thecurve changes direction at a given point. Recalling that the direction at agiven point on a curve is given by the unit tangent vector T at that point itresults that the curvature will be connected to the rate of change of T alongC . This intuitive idea is made rigorous in the following denition.

Denition 7 Let C be a curve in R 3 described by a vector-valued function r (t). The curvature of C is

=dTds

, T = r (t)

|r (t)|.

Even though the above formula reects our heuristic idea of curvature of acurve it is not well suited for practical purposes. In fact, it results to be moreconvenient to derive a formula for the curvature involving only derivativeswith respect to t. This can be achieved by observing that

dTds

= dT

dtdtds

=

dTdtdsdt

= T

|r |,

where we used (2) in the last step. Hence, we found an equivalent formula

for the curvature given by = |T |

|r |. (3)

Example 6 Show that the curvature of a circle of radius R is = R 1 . First of all, points lying on a circle of radius R can be described by a vector-valued function

r (t) = R cos t ex + R sin t ey.

In order to apply formula (3) we need to compute the following quantities

r (t) =

R sin t ex + R cos t ey ,

|r (t)| = R2 sin2 t + R2 cos2 t = R,T =

r (t)

|r (t)| = sin t ex + cos t ey ,

|T | = 1.Hence, we conclude that

= |T ||r |

= 1R

.

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Finally, a very useful formula for the computation of curvature involving onlyr (t) and its rst and second derivatives is

= |r (t) r (t)||r (t)|3

. (4)

Example 7 Find the curvature of the curve described by

r (t) = t ex + t2 e y + t3 e z .

In order to apply (4) we have to compute

r (t) = ex + 2 t ey + 3t2

e z = |r (t)|3

= (1 + 4 t2

+ 9 t4

)3 / 2

,r (t) = 2 ey + 6t ez .

On the other side

r (t) r (t) = dete x ey ez1 2t 3t2

0 2 6t= 6 t2 e x 6t ey + 2 ez

and

|r (t) r (t)| = 2 9t4 + 9 t2 + 1 .Thus, we conclude that

(t) = 2 9t4 + 9 t2 + 1(1 + 4t2 + 9 t4 )3 / 2

= 2 9t4 + 9 t2 + 1(1 + 4 t2 + 9 t4 )3 .Example 8 We want to rewrite the formula (4) for the special case of a plane curve y = y(x). Clearly, points on this curve have coordinates (x, y(x))and are identied by a position vector

r (x) = x ex + y(x) ey.

Moreover,

r (x) = ex + y (x) ey = |r (x)|3 = [1 + ( y )2 ]3 / 2 ,

r (x) = y e y .

On the other side

r (x) r (x) = dete x ey ez1 y 00 y 0

= y (x) ez

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and

|r (x) r (x)| = |y |.Thus, we conclude that

(x) = |y |[1 + (y )2 ]3 / 2

.

The torsion of a curve measures how sharply it is twisting and if r (t) is avector-valued function describing a curve in space the torsion of the curvecan be computed with the formula

= [r (t) r (t)] r (t)|r (t) r (t)|2 .

It is not difficult to check that the torsion of a helix is constant.

Practice problems

1. Find the domain of denition of the vector functions

(a) r (t) = t2 e x + t 1 ey + 5 t ez ;(b) r (t) = ln t ex +

t

t 1 ey + e

t e z .

2. Find

(a) the limit for t 0+ of r (t) = cos t ex + sin t ey + t ln t ez ;

(b) the limit for t 0 of

r (t) = et 1

t

ex + 1 + t 1

t

ey + 3

1 + t

ez .

3. The equation of a cone in R 3 is given by z 2 = x2 + y2 . Show that thecurve with parametric equations x = t cos t, y = t sin t and z = t lieson that cone and use this fact to help sketch the curve.

4. Show that the curve with parametric equations x = t2 , y = 1 3t, andz = 1 + t3 passes through the points (1 , 4, 0) and (9, 8, 28) but notthrough the point 4 , 7, 6).

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5. Find a vector valued-function representing the curve of intersection of the circular cylinder {(x,y,z )R 3 | x2 + y2 = 1 , z R }and the planey + z = 2.

6. Find the derivative of the following vector-valued functions

(a) r (t) = e x e y + e4 t e z ,

(b) r (t) = e t cos t ex + e t sin t ey + ln |t| ez ,

(c) r (t) = a + t b + t2 c ,

(d) r (t) = ta (b + t c),where a , b , and c are arbitrary constant vectors in R 3

7. Find the unit tangent vector T at the point with the given value of theparameter t

(a) r (t) = t ex + ( t t2 ) ey + arctan t ez , t = 1,

(b) r (t) = e2 t cos t ex + e2 t sin t ey + e

2 t e z , t = 2 .

8. Evaluate the integrals

(a)

4

1

dt t ex + te t e y + ezt

2 ,

(b) / 4

0dt [cos (2t) ex + sin (2 t) ey + t sin t ez].

9. If

u (t) = e x 2t2 e y + 3t3 e z , v (t) = t ex + cos t ey + sin t ez ,nd

ddt

[u (t) v (t)] and d

dt[u (t) v (t)].

10. If r (t) = 0 , show that

ddt |r (t)| =

r (t) r (t)|r (t)|

.

11. Reparametrize the curve with respect to arc length measured from thepoint where t = 0 in the direction of increasing t

(a) r (t) = et sin t ex + et cos t ey ,

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(b) r (t) = (1 + 2 t) ex + (3 + t) ey

5t ez ,

(c) r (t) = 3sin t ex + 4 t ey + 3 cos t ez .

12. Reparametrize the curve

r (t) = 2

t2 + 1 1 ex + 2tt2 + 1

ey

with respect to arc length measured from the point (1 , 0) in the direc-tion of increasing t. Express the reparametrization in its simplest form.What can you conclude about the curve?

13. Find the curvature of the curves described by the vector-valued func-tions

(a) r (t) = t2 e x + (sin t t cos t) ey + (cos t + t sin t) ez , t > 0,(b) r (t) = t2 e x + 2 t ey + ln t ez , t > 0,

14. Find the curvature of r (t) = 2t e x + et e y + e t e z at the point (0 , 1, 1).15. Graph the curve with parametric equations

x = t, y = 4t3 / 2 , z =

t2

and nd the curvature at the point (1 , 4, 1).16. Find the curvature of

y = x3 , y = x, y = sin x.17. At what point does the curve have maximum curvature? What happens

to the curvature as x ?y = ln x, y = ex .

18. Find an equation of a parabola that has curvature 4 at the origin.

19. Show that the circular helix r (t) = a cos t ex + a sin t ey + bt ez wherea and b are positive constants, has constant curvature and constanttorsion.

20. Find the curvature and torsion of the curve x = sinh t, y = cosh t, z = tat the point (0 , 1, 0).

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21. The DNA molecule has the shape of a double helix. The radius of eachhelix is about 10 angstroms (1 angstrom = 10 8 cm). Each helix risesabout 34 angstroms during each complete turn, and there are about2.9 108 complete turns. Estimate the length of each helix.

22. We consider the problem of designing a railroad track to make a smoothtransition between sections of straight track. Existing track along thenegative x-axis is to be joined smoothly to a track along the line y = 1for x 1. Find a polynomial P = P (x) of degree 5 such that thefunction dened by

F (x) =

0 if x

0,

P (x) if 0 < x < 1,1 if x 1,

is continuous, has continuous slope and continuous curvature.

23. What force is required so that a particle of mass m has the positionfunction

r (t) = t3 e x + t2 e y + t3 e z ?

24. A force of magnitude 20 N acts directly upward from the xy-plane onan object of mass 4 Kg. The object starts at the origin with initialvelocity v (0) = ex

e y . Find its position function and its speed at

time t.

25. A projectile is red with an initial speed of 500 m/s and angle of eleva-tion 30o. Find the range of the projectile, the maximum height reachedand the speed at impact.

26. The position function of a spaceship is

r (t) = (3 + t) ex + (2 + ln t) ey + 7 4t2 + 1

ez

and the coordinates of a space station are (6 , 4, 9). The captain wantsthe spaceship to coast into the space station. When should the enginesbe turned off?

27. A rocket burning its onboard fuel while moving through space has ve-locity v (t) and mass m(t) at time t. If the exhaust gases escape withvelocity ve relative to the rocket, it can be deduced from NewtonsSecond Law of Motion that

mdvdt

= dm

dt ve .

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(a) Show that

v (t) = v (0) ln m(0)m(t) v e.(b) For the rocket to accelerate in a straight line from rest to twice the

speed of its own exhaust gases, what fraction of its initial masswould the rocket have to burn as fuel?

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