ENGI 4430 Parametric Vector Functions Page 2-01

2. Parametric Vector Functions (continued)

Any non-zero vector r can be decomposed into its magnitude r and its direction:

ˆ , where 0r r r r r

Tangent Vector: T

dx dy dz d

dt dt dt dt

rT

If the parameter t is time, then the tangent vector is also

the velocity vector,

d

dt

rv , whose magnitude is the speed

dv

dt

r.

The unit tangent is

d d

dt dt

r rT

Arc Length

In 2 :

2 2 2

s x y

In 3 :

2 2 2 2

s x y z

2 2 2 2

2 2 2 2

s x y z

t t t t

2 2 2ds dx dy dz d

dt dt dt dt dt

r

The vector d

dt

r points in the direction of the tangent T to the curve defined parametrically by

tr r .

d d d ds d dt

dt dt dt dt dt ds

r r r rT

d

ds

rT

[unit tangent = rate of change of r with respect to distance travelled along the curve.]

[However, arc length s is rarely a convenient parameter.]

^

^ ^

ENGI 4430 Parametric Vector Functions Page 2-02

Example 2.01

(a) Find the arc length along the curve defined by

T

1 14 4

2 sin 2 cos2 sint t t t t r , from the point where t = 0 to the point where t = 4π.

(b) Find the unit tangent T .

(a) 2 21 2 12 2cos 2 1 cos 2 2sin sin

4 4 2

dxt t t t

dt

1 2

2sin 2 2sin cos sin cos4 4

dyt t t t t

dt

cosdz

tdt

T2sin sin cos cos

dt t t t

dt

r

4 2 2 2 2 2 2 2

2 2

sin sin cos cos sin sin cos cos

1sin cos 1

ds dt t t t t t t t

dt dt

t t

r

4 4 4

00 01 1

ds dsL dt dt t

dt dt

Therefore

4L

(b)

d d

dt dt

r rT

But 1 "for all "d

t tdt

r

. Therefore

2sin

sin cos

cos

t

t t

t

T

[Note that, in this example, t itself is the arc length from the point 14

0, , 0 , (where t = 0).

There are very few curves for which the parameterization is this convenient!]

^

^

^

ENGI 4430 Parametric Vector Functions Page 2-03

Arc Length for a Polar Curve

For a polar curve defined by , ,r f the parameter is and

cos sinx f y f .

Using the abbreviations , , cos , sin ,r f r f c s

rx

dc s

dr

and r

y

ds c

dr

2 2

2 2

2r c rr csdx dy

d d

22 2 2 2r s r s rr sc 2 2r c

2

2 2 2 2 2 2 2 drr c s r s c r

d

Therefore the arc length L along the polar curve r f from to is

2

2 drL r d

d

Example 2.02

Find the length L of the perimeter of the cardioid 1 cosr

0, 2

1 cos 1r c

sindr

sd

2

2 drr

d

2 21 2 2 2c c s c

2 1 c

ENGI 4430 Parametric Vector Functions Page 2-04

Example 2.02 (continued)

But 21 cos2 2cosx x . Set 22

x x

2

2 22 2 cos2

drr

d

2 2

2 2

0 0

2 cos 2 cos2 2

L d d

Using symmetry in the horizontal axis,

2

0 0 0

2 cos 4 cos 4 2 sin 8 1 02 2 2

L d d

Therefore the perimeter of the cardioid curve is L = 8.

Note:

For 22 , cos cos , cos2 2 2

not

and

2

0

cos 0!2

d

Example 2.03

Find the arc length along the spiral curve 0r ae a , from to .

drr ae ae

d

2 2

2ds

ae ae aed

2 2ds

L d a e d a ed

2L a e e

ENGI 4430 Parametric Vector Functions Page 2-05

Overview of Normal and Binormal:

For a curve in 2 , there is a unique normal line at every point

where the unit tangent vector is defined:

There are only two choices for the unit normal vector.

N is chosen to point in the direction in which the curve is

turning. This definition fails wherever the unit tangent does

not exist. This definition also fails at all points of inflexion and

wherever the curve is a straight line.

However, for a curve in 3 , there is a normal plane at each

point, not a normal line. From the infinite number of

different directions for a normal vector, we need to identify

a unique direction for a principal normal. Again, we shall

choose the direction in which the unit tangent vector is

turning at each point.

1 1 T T T

Using the product rule of differentiation,

0 2 0d d d

ds ds ds

T T TT T T

ord d

ds ds

T T0 T (Note: a unit vector can never be zero).

Select the principal normal N to be the direction in which the unit tangent is, instantaneously,

changing:

d d ds

ds dt dt

T TN N and N are therefore parallel to

d

dt

T.

The unit principal normal then follows:

d d

dt dt

T TN

^

^ ^ ^

^ ^^ ^ ^ ^

^ ^ ^

^^^ ^

^ ^ ^

ENGI 4430 Parametric Vector Functions Page 2-06

The unit binormal is at right angles to both tangent and principal normal:

ˆ ˆ ˆ B T N

ˆ constant B

the curve lies entirely in one plane (the plane defined by ˆ ˆandT N ).

The three vectors form an orthonormal set (they are mutually perpendicular and

each one is a unit vector; that is, magnitude = 1). All three unit vectors are

undefined where the curve suddenly reverses direction, (such as at a cusp).

Curvature:

The derivative of the unit tangent with respect to distance travelled along a curve is the principal

normal vector, ˆd

ds

TN . Its direction is the unit principal normal N . Its magnitude is a

measure of how rapidly the curve is turning and is defined to be the curvature:

ˆ ˆd d d

ds dt dt

T T rN

ˆ N N

The radius of curvature is 1

.

The radius of curvature at a point on a curve is the radius of the circle which best fits the curve at

that point.

Example 2.01 (continued)

(c) Find the curvature s at any point for which 0s , for the curve

T

1 14 4

2 sin 2 cos2 sins s s s s r

where s is the arc length from the point 14

0, , 0 .

From part (b):

2sin

ˆ sin cos

cos

s

s s

s

T

ENGI 4430 Parametric Vector Functions Page 2-07

Example 2.01 (c) (continued)

2 2

2sin cos sin 2

cos sin cos 2

sin sin

s s sd

s s sds

s s

TN

21 sind

s sds

T

2

1 1

1 sin s

It then follows that the unit principal normal vector at every point is

2

sin 21

cos 21 sin

sin

sd d

sds ds s

s

T TN N

A Maple plot of this curve is available from the demonstration files section of the web site,

www.engr.mun.ca/~ggeorge/4430/demos/index.html.

^

^

^ ^

ENGI 4430 Parametric Vector Functions Page 2-08

Another formula for curvature:

Let and , thend ds d

sdt dt dt

r r

r r

d ds

dt dt

r rT r T

ds s

dt

Tr T (product rule of differentiation)

Butd ds

sds dt

d

dt

TTN

2 3s s s s r r T T N 0 B

3 , but . Therefores s r r r

3

r r

r

If the parameter t is time, then the curvature formula is also

3v

v a

where v v is the speed (scalar), the vector v is the velocity and the vector a is the

acceleration.

^ ^

^^

^ ^^

^ ^ ^ ^

ENGI 4430 Parametric Vector Functions Page 2-09

Example 2.04

Find the curvature and the radius of curvature for the helix, given in parametric form by

x = cos t , y = sin t , z = t . Assume SI units.

Let c = cos t , s = sin t .

1 0

c s c

s c s

t

r v r a r

2 2

ˆ

ˆ

ˆ 11 0

s c s s

c s c c

s c

i

r r j

k

2 2 1 2s c r r

2 2 1 2s c r

3

2 1

22 2

r r

r

Therefore

112

m constant

and

12m

ENGI 4430 Parametric Vector Functions Page 2-10

Surfaces of Revolution

Consider a curve in the x-y plane, defined by the equation y = f (x).

If it is swept once around the line y = c , then it will generate a surface of revolution.

At any particular value of x, a thin cross-section through that surface, parallel to the y-z plane,

will be a circular disc of radius r, where

r f x c

Let us now view the circular disc face-on, (so that the x axis and the axis of rotation are both

pointing directly out of the page and the page is parallel to the y-z plane).

Let (x, y, z) be a general point on the surface of revolution.

From this diagram, one can see that

22 2y cr z

Therefore, the equation of the surface generated, when the

curve y = f (x) is rotated once around the axis y = c, is

22 2y c z f x c

Special case: When the curve y = f (x) is rotated once around the x axis, the equation of the

surface of revolution is

22 2y z f x or 2 2y z f x

ENGI 4430 Parametric Vector Functions Page 2-11

Example 2.05

Find the equation of the surface generated, when the parabola 2 4y ax is rotated once around

the x axis.

The solution is immediate: 2

4f x ax

the equation of the surface is

2 2 4y z ax ,

which is the equation of an elliptic paraboloid,

(actually a special case, a circular paraboloid).

A Maple worksheet for this surface is available from the

demonstration files section of the ENGI 4430 web site.

The Curved Surface Area of a Surface of Revolution

For a rotation around the x axis,

the curved surface area swept out by the element of arc length

s is approximately the product of the circumference of a

circle of radius y with the length s.

2A y s

Integrating along a section of the curve y = f (x) from x = a to

x = b, the total curved surface area is

2x b

x aA f x ds

For a rotation of y = f (x) about the axis y = c, the curved surface area is

2 2bx b

x a a

dsA f x c ds f x c dx

dx

and

2 2

1ds dy

dx dx

Therefore

2

2 1b

aA f x c f x dx

ENGI 4430 Parametric Vector Functions Page 2-12

Example 2.06

Find the curved surface area of the circular paraboloid generated by rotating the portion of the

parabola 2 4 0y cx c from 0x a to x b a about the x axis.

2 2x b

a

b

axA y

dss dxd y

dx

2 24 4d d

y cx y cxdx dx

[Chain rule:]

22 4

cy y c y

y

2 2 2

2

4 41 1 1

4

ds dy c c x c

dx dx y cx x

2 2A c x x c

x

1/24

b b

aa

dx c x c dx

3/2

32

4

b

a

x cc

Therefore

3/2 3/28

3

cA b c a c

ENGI 4430 Parametric Vector Functions Page 2-13

Area Swept Out by a Polar Curve r = f (θ)

A Area of triangle

12

sinr r r

But the angle is small, so that sin

and the increment r is small compared to r.

Therefore

21

2A r

21

2A r d

Example 2.07

Find the area of a circular sector, radius r , angle .

r = constant,

2 2 21 1 12 2 2

A r d r r

Therefore

212

A r

Full circle: 2 2A r .

ENGI 4430 Parametric Vector Functions Page 2-14

Example 2.08

Find the area swept out by the polar curve r ae over ,

(where 0a and 2 ).

The condition 2

prevents the same area being swept

out more than once.

If 2 then one needs to

subtract areas that have been

counted more than once

[the red area in the diagram]

21

2A a e d

2

2 2 21 1

2 2 2

ea e d a

Therefore

2

2 2

4

aA e e

In general, the area bounded by two polar curves r f and r g and the radius vectors

and is

2 21

2A f g d

ENGI 4430 Parametric Vector Functions Page 2-15

Radial and Transverse Components of Velocity and Acceleration

At any point P (not at the pole), the unit radial vector r points directly away from the pole. The

unit transverse vector θ is orthogonal to r and points in the direction of increasing θ. These

vectors form an orthonormal basis for 2 .

Unlike the Cartesian i and j , if θ is not constant then r and θ will be variable unit vectors.

Examine r and θ at two nearby times t and t t :

The magnitudes of the unit vectors can never change.

By definition, the magnitude of any unit vector is always 1.

Only the directions can change.

Examine the difference in the unit radial vector at these two times.

The two longer sides of this isosceles triangle are both 1 ˆ r .

The two larger angles are each equal to / 2 .

Clearly the angle between ˆ tr and the incremental vector ˆr

approaches a right angle as t (and therefore ) approaches zero.

Therefore the derivative of ˆ tr is parallel to ˆ t .

Also, from triangle geometry (the sine rule),

ˆ ˆ

sin sin / 2

r r and ˆ 1r (unit vector)

As 0 , sin and sin / 2 sin / 2 1t

ˆ ˆ ˆ ˆ1 ˆ

1

d d d d

t t dt dt dt dt

r r r r

Sine rule:

sin sin sin

a b c

A B C

ENGI 4430 Parametric Vector Functions Page 2-16

Now examine the difference in the unit transverse vector at these two

times.

The two longer sides of this isosceles triangle are both 1.

Clearly the angle between ˆ t and the incremental vector ˆ

approaches a right angle as t (and therefore ) approaches zero.

Therefore the derivative of ˆ t points radially inwards

(ˆd

dt

is parallel to ˆ tr ).

Also, from triangle geometry (the sine rule),

ˆ ˆ

sin sin / 2

and ˆ 1

As 0 , sin and sin / 2 sin / 2 1t

ˆ ˆ ˆ ˆ1ˆ

1

d d d d

t t dt dt dt dt

r

Therefore the derivatives of the two [non-constant] polar unit vectors are

ˆ ˆd d

dt dt

r and

ˆˆ

d d

dt dt

r

Using the “overdot” notation to represent differentiation with respect to the parameter t, these

results may be expressed more compactly as

ˆ ˆˆ ˆand r r

The radial and transverse components of velocity and acceleration then follow:

ˆˆ ˆ ˆˆ r r r rr r rr r rv r

radial transverseandv r v r

ˆ ˆ ˆˆ ˆr r r r r r ra v r

2 2

transverseradial

ˆ ˆ ˆˆ ˆ ˆ2 2r r r r r r r r

aa

r r r

The transverse component of acceleration can also be written as 2

tr

1 da r

r dt

ENGI 4430 Parametric Vector Functions Page 2-17

Example 2.09

A particle follows the path r , where the angle at any time is equal to the time:

0t . Find the radial and transverse components of acceleration.

r t

1r

0r

ˆ ˆˆ ˆr r t v r r

21v t

2 ˆ ˆˆ ˆ2 2r r r r t a r r

24a t

Therefore

ra t and tr 2a

Example 2.10

For circular motion around the pole, with constant radius r and constant angular velocity ,

the velocity vector is purely tangential, ˆrv , and the acceleration vector is

2 2ˆˆ ˆ2r r r r r a r r

which matches the familiar result that “centrifugal” or “centripetal” acceleration 2r , directed

radially inward.

ENGI 4430 Parametric Vector Functions Page 2-18

Tangential and Normal Components of Velocity and Acceleration

From page 2.01, we know that the velocity vector is also the tangent vector, d

dt

rv T .

The [scalar] speed is d ds

vdt dt

r

v

The speed is therefore the derivative with respect to time of distance travelled along the curve.

It follows that vv T - the speed is also the tangential component of the velocity vector.

There are no components of velocity in the normal directions.

The acceleration vector is

d dv

dt d

dv dv

dt dtt

va T

TT

But andd d dt ds

vds dt ds dt

T T

N N

dv

dt

TN

2

T N

dvv a a

dt a T N T N

Therefore the tangential component of acceleration is the rate of change of speed (not velocity)

and is positive if the particle is speeding up, negative if it is slowing down and zero if the speed

is constant. It is also ˆTa

v

v aa T .

The normal component of acceleration (in the direction of the unit principal normal) is never

negative; both the curvature and 2v are non-negative at all times.

There is never any component of acceleration in the binormal direction.

Usually it is easiest to evaluate d d

adt dt

v v

a a and T

dva

dt first, then to find the

normal component (and hence the curvature) from

2 2

N Ta a a

^

^ ^ ^

^ ^^

^ ^

^ ^ ^ ^

ENGI 4430 Parametric Vector Functions Page 2-19

Example 2.04 (continued)

Find the tangential and normal components of velocity and acceleration for a particle travelling

along the helix, given in parametric form by x = cos t , y = sin t , z = t .

Hence find the curvature. Assume SI units.

From page 2.09,

cos sin cos

sin cos sin

1 0

t t td d

t t t tdt dt

t

r vr v a

2 2sin cos 1 2 0T T

dvv v t t a

dt

The acceleration is therefore purely normal.

2 2 2 2cos sin 0 1 1N Ta t t a a a a

Therefore

1 22 ms and 0 1 mst t v T a T N

2 111 and 2 m

2Na v v

^ ^

ENGI 4430 Parametric Vector Functions Page 2-20

Review of Lines [from MATH 2050]

Given a point o o o o, ,P x y z known to be on a line L ,

a non-zero vector T

a b cd parallel to the line,

and vectors T

x y zr and

T

o o o o oOP x y z r , the equations of the line

may be expressed in the vector form

o ,t t r r d

or, provided none of a, b, c are zero, in the symmetric Cartesian form

o o ox x y y z z

a b c

If, for example, b = 0, but a and c are non-zero, then the Cartesian form becomes

o oo ,

x x z zy y

a c

Example 2.11

Find the Cartesian equation of the line through the points (1, 3, 5) and (4, –1, 7).

The vector connecting the two points is an obvious choice for the line direction vector.

4 1 3

1 3 4

7 5 2

a

b

c

d

Either point may be chosen as Po .

o

o o

o

1

3

5

x

y

z

r

The equation of the line is

1 3 5

3 4 2

x y z

ENGI 4430 Parametric Vector Functions Page 2-21

Review of Planes [from MATH 2050]

A plane is defined by two vectors [ is seen here edge-on →].

Its orientation is determined by a

non-zero normal vector T

A B Cn .

Its location is determined by the position vector

T

o o o o oOP x y z r of any one point oP known to be on the plane.

The vector T

x y zr is then the position vector of a point on the plane if and only if

or n r n

or, equivalently,

0Ax By Cz D

where o o o oD Ax By Cz r n

If u and v are two non-zero vectors not parallel to each other and both parallel to the plane,

then a vector parametric form of the equation of the plane is

o , ,s t s t r r u v

The normal vector to the plane is n u v , from which the other two forms above follow.

Example 2.12

Find the Cartesian equation of the plane through the points (0, 3, 5), (2, 4, 5) and (4, –1, 7) and

find the equation of the normal line to that plane that passes through the origin.

The three points define three vectors, no pair of which are parallel to each other but all three of

which are parallel to the plane. Any two of them will serve as vectors u and v .

2 0 2 4 0 4 2

4 3 1 and 1 3 4 2 2

5 5 0 7 5 2 1

u v

ˆ 2 2 1 1

ˆ2 1 2 2 2 2 2

ˆ 6 60 1

i

n u v j

k

ENGI 4430 Parametric Vector Functions Page 2-22

Example 2.12 (continued)

Any non-zero multiple of this vector will also serve as a normal vector to the plane.

Therefore select T

1 2 6 n .

Any of the three points will serve as oP . Choose

T

o 0 3 5r

o

1 0

2 3 0 6 30 36

6 5

n r

The equation of the plane is o n r n r

2 6 36x y z

We can quickly check that all three given points satisfy this equation:

–0 + 2(3) + 6(5) = 6 + 30 = 36

–2 + 2(4) + 6(5) = –2 + 8 + 30 = 36

–4 + 2(–1) + 6(7) = –4 – 2 + 42 = 36

The direction vector for the normal line is

just the normal vector to the plane, T

1 2 6 d n

o r 0 (the line passes through the origin).

The equations of the normal line follow immediately

0 0 0

1 2 6

x y z

[End of Chapter 2]