ME 2304: 3D Geometry & Vector Calculus

Calculus with Vector Functions

The Limit of a Vector Valued Function

So, all that we do is take the limit of each of the components functions and leave it as a vector.

Example: Limits

Notice that we had to use L’Hospital’s Rule on the y component

L'Hopital's rule Revisited

Uses derivatives to help evaluate limits involving indeterminate forms.

Application (or repeated application) of the rule often converts an indeterminate form to a determinate form, allowing easy evaluation of the limit.

L’Hopital’s Rule: Examples

2

2

4lim

2x

x

x

limx a

f x

g x

2

2

4lim

2x

dx

dxd

xdx

2

2lim

1x

x

4

11

1

1

)1(

)(ln

1

lnlimlimlimlim

1111

x

x

xdxd

xdxd

x

x

xxxx

Remember:

20

1 coslimx

x

x x

0

sinlim

1 2x

x

x

0

If it’s no longer indeterminate, then STOP differentiating!

If we try to continue with L’Hopital’s rule:

0

sinlim

1 2x

x

x

0

coslim

2x

x

1

2

which is wrong!

Example # 1

xx

xxFind

x sin

2sinsin2lim

0

Now if we put x=0, it will be 0/0 form i.e. undefined. Therefore, by applying L’Hopital’s Rule (LHR)

form

0

0

cosx1

2cos2x2cosxlim

sinxx

sin2x2sinxlim

0x

LHR

0x

ng LHR,so applyi0

0again

x

xxx

LHR

)sin(0

)2sin4(sin2lim

0

61

1.81.2

cos

2cos8cos2lim

0

x

xxx

LHR

The Derivative of a Vector Valued Function

So, all that we do is take the derivative of each of the components functions and leave it as a vector.

Most of the basic facts that we know about derivatives still hold however, just to make it clear here are some facts about derivatives of vector functions.

Example

Example 2: Let A(t) = cos ti + sin tj + tk. Find the derivative of A(t).Solution:

A′(t) = -sin ti + cos tj + k

Rules of Vector Differentiation 0,

d

dt

A if A = constant.

( )d d d

dt dt dt

A BA B

( )d d d

dt dt dt

B AA . B A . . B

( )d d d

dt dt dt

B AA B A B

( )d d dp

p pdt dt dt

A

A A

Velocity & Acceleration

• Recall that if r(t) is a position function– r'(t) is the velocity function– r''(t) is the acceleration function

Example• Given parametric equations which describe a

vector-valued position function– x = t3 – t– y = 4t – 3t2

• What is the velocity vector?

• What is the acceleration vector?

2( ) 3 1 4 6t t t v i j

( ) 6 6t t a i j

Example• For the same vector-valued function

– x = t3 – t and y = 4t – 3t2

• What is the magnitude of v(t) when t = 1?

• The direction?

2

2 2

( ) 3 1 4 6

(1) 2 2

2 2 8 2 2

t t t

v i j

v i j

v

1 2 7tan

2 4 4

Integration of vector valued functions• Using both limits and derivatives as a guide it shouldn’t be too

surprising that we also have the following for integration for

indefinite integrals

and the following for definite integrals.

Integration of vector valued functions

• With the indefinite integrals we put in a vector sign on constant of integration to make sure that it was clear that the constant in this case needs to be a vector instead of a regular constant.

Example: 1

All we need to do is integrate each of the components and be done with it

In this case all that we need to do is reuse the result from the previous example and then do the evaluation.

Example: 2

Let A(t) = cos ti + sin tj + tk. Find 2

0( )t dt

A

Solution:

2 2 2 2

1 2 30 0 0 0

2 2 2

0 0 0

22 2 2

0 0 0

2

( ) ( ) ( ) ( )

cos sin

1sin cos

2

2

t dt A t dt A t dt A t dt

tdt tdt tdt

t t t

A i j k

i j k

i j k

k

Tangent, Normal and Binormal Vectors

• Three vectors play an important role when

studying the motion of an object along a space

curve. These vectors are:

• the unit tangent vector,

• the unit normal vector and

• the binormal vector.

The Slope of a Tangent to a Curve

The slope of a curve y = f(x) at the point P means the slope of the tangent at the point P. We need to find this slope to solve many applications since it tells us the rate of change at a particular instant.

The Slope of a Tangent to a Curve (cont)

By definition, the slope is given by:

12

12

xx

yy

x

y

xinchange

yinchangem

We use this relationship to find a numerical solution to the slope of a curve i.e. using derivative dy/dx.

Example: 1Find the slope of the tangent at the point (2, -1) for the the curve 052 32 yxyImplicit differentiation will give:

0322 2 dx

dyyx

dx

dy232

2

y

x

dx

dy

Now, when x=2, y= -1

4)1(32

)2(22

1,2

yxdx

dy

So, the slope of the tangent at (2, -1) is -4.

Unit tangent vector• In the past we’ve used the fact that the derivative of

a function was the slope of the tangent line.• With vector functions we get exactly the same result

Tangent vector measures the change of "distance" and thus gives the speed of a moving point.

• There is a nice geometric description of the

derivative r'(t).

• The derivative r'(t) is tangent to the space curve

r(t).

• This is shown in the figure below, where the

derivative vector r'(t)=<-2sin(t),cos(t)> is

plotted at several points along the curve

r(t)=<2cos(t),sin(t)> with 0<=t<=2*pi.

Unit tangent vector

Unit tangent vector: Graphical Representation

We could have used the unit tangent vector (from previous example) had we wanted to for the parallel vector. However, that would have made for a more complicated equation for the tangent line.

Expressing the Acceleration in Terms of the unit Tangent and the Unit Normal

• We have seen that given• then the velocity vector is

• and the velocity vector is tangent to the curve at each point along the trajectory.

• Further the acceleration vector is

jtyitxtr

)()()(

jdt

dyi

dt

dx

dt

rdtv

)(

jdt

ydi

dt

xd

dt

rdta

2

2

2

2

2

2

)(

• There are 2 ways in which a particle can accelerate: either due to a change in speed or a change in direction.

• We want to express the acceleration in terms of a tangential component aT and a normal component aN.

• The tangential component will tell us the change in speed and the normal component will give us the change in direction.

• In this discussion we concentrate on finding • and i.e. the unit tangent and unit normal

vectors

T

N

Let be the unit tangent . Since we know the velocity is tangent to the curve it follows:

Whereas, unit normal vector is given by:

T

)('

)('

tr

tr

dtrd

dtrd

T

dtTd

dtTd

dt

rddt

rd

N

2

2

2

2

Summary• The derivative of a vector valued function gives a

new vector valued function that is tangent to the defined curve.

• The analogue to the slope of the tangent line is the direction of the tangent line.

• Since a vector contains a magnitude and a direction, the velocity vector contains more information than we need.

• We can strip a vector of its magnitude by dividing by its magnitude.

Summary

• Let r(t) be a differentiable vector valued function and

v(t) = r'(t) be the velocity vector. Then we define the

unit tangent vector as the unit vector in the direction

of the velocity vector i.e.

)(

)()(

tv

tvtT

Exercise: 1 Find unit tangent vector T(t) and T(0) of the

curve x = t, y = et, z = -3t2

Solution: The position vector is Now computing velocity vector i.e. tangent vector to

the curve

And now the speed

Now computing the unit tangent vector i.e. T(t)

ktjetitr t 23)(

tkjeitrtv t 6)(')(

222 361)(' tetrv t

222 361

6

)(

)()(

te

tkjei

tv

tvtT

t

t

Exercise: 1 (Contd.)

Now to find T(0), plugging in t = 0 yields

222 361

6

)(

)()(

te

tkjei

tv

tvtT

t

t

jiji

v

vT

e

kjei

v

vT

2

1

2

1

11)0(

)0()0(

0361

06

)0(

)0()0(

2022

0

Exercise: 2 Find unit tangent vector T(t) and T(pi/2) of the

curve x = t – sint , y = 1 - costAnswer: The unit tangent vector i.e. T(t) is

2

1,

2

1)

2(

cos22

sin,

cos22

cos1

)(

)()(

T

t

t

t

t

tv

tvtT

Navigation

Principal Unit Normal Vectors• The unit normal is orthogonal (or normal, or

perpendicular) to the unit tangent vector and hence to

the curve as well. We’ve already seen normal vectors

when we were dealing with Equations of Planes.

• Given a vector v in the space, there are infinitely many

perpendicular vectors. Our goal is to select a special

vector that is normal to the unit tangent vector.

Unit Normal Vectors• At a given point on a smooth space curve r(t) i.e. if r’(t)

is defined, there are many vectors that are orthogonal

to the unit tangent vector T(t).

• By definition T is a unit vector, that is, . Note that

T(t) · T(t) = 1 implies by differentiation T’(t) · T(t) = 0

• Therefore T'(t) is orthogonal to T(t), thus N (t) is

orthogonal to T(t). Notice that T'(t) is itself not a unit

vector. Therefore, we can define the unit normal

vector N(t) as

1T

Illustration

Summary

Example: 1

Example: 2

Binormal Vector

• So far we have a pair of orthogonal vectors

defined along the path of motion. We can get

the third orthogonal vector from the cross

product, i.e. the binormal vector is defined to

be:)()()( tNtTtB

Binormal Vector• Because the binormal vector is defined to be

the cross product of the unit tangent and unit

normal vector we then know that the

binormal vector is orthogonal to both the

tangent vector and the normal vector.

• The triple of unit vectors T(t), N(t), and B(t)

forms a moving frame of reference called the

TNB frame or moving trihedral.

Example: 3

Example: 3 (contd.)

Example: 4

Arc Length with Vector Functions• In this section we’ll recast an old formula into terms

of vector functions. We want to determine the length of a vector function,

• Recall that we can write the vector function into the parametric form

• Also, recall that with two dimensional parametric curves the arc length is given by,

)(),(),()( thtgtftr

bta interval the on

h(t) zg(t) yf(t) x

dttgtfLb

a 22 )(')('

Arc Length with Vector Functions• There is a natural extension of this to three

dimensions. So, the length of the curve

• Notice that the integrand (the function we’re integrating) is nothing more than the magnitude of the tangent vector, i.e.

• Therefore, the arc length can be written as,

bta interval the )( ontr

dtthtgtfLb

a 222 )(')(')('

dtthtgtftrb

a 222 )(')(')(')(

b

a

dttrL )('

– For plane curves r(t) = f(t) i + g(t) j

– For space curves r(t) = f(t) i + g(t) j + h(t) k

2 2'( ) '( ) '( ) '( ) '( )t f t g t f t g t r i j

2 2 2

'( ) '( ) '( ) '( )

'( ) '( ) '( )

t f t g t h t

f t g t h t

r i j k

ARC LENGTH

• We need to take a quick look at another concept here. We define the arc length function as,

• Before we look at why this might be important let’s work a quick example.

Arc Length with Vector Functions

t

duurts0

)(')(

• It is often useful to parametrize a curve with respect to arc length i.e. s.

– This is because arc length arises naturally

from the shape of the curve and does not

depend on a particular coordinate system.

PARAMETRIZATION

• If a curve r(t) is already given in terms of

a parameter t and s(t) is the arc length function

given by Equation,

• then we may be able to solve for t as a function

of s; i.e. t = t(s)

PARAMETRIZATION

t

duurts0

)(')(

• Then, the curve can be reparametrized in terms of s by substituting for t:

r = r(t(s))• Thus, if s = 3 for instance, r(t(3)) is

the position vector of the point 3 units of length along the curve from its starting point.

REPARAMETRIZATION

Okay, just why would we want to do this? Well let’s take

the result of the example above and solve it for t.

Example:2 (Contd.)

Now, taking this and plugging it into the original vector

function and we can reparameterize the function into the

form For our function this is,

• So, why would we want to do this? Well with the reparameterization we can now tell where weare on the curve after we’ve traveled a distance of s along the curve. • Note as well that we will start the measurement of distance from where we are at t = 0

))(( str

Curvature• The curvature measures how fast a curve is

changing direction at a given point.• It is natural to define the curvature of a

straight line to be identically zero. • The curvature of a circle of radius R should be

large if R is small and small if R is large. Thus the curvature of a circle is defined to be the reciprocal of the radius, i.e.

R

1

CURVATURE (contd.)• Given any curve C and a point P on it, there is

a unique circle or line which most closely approximates the curve near P, the osculating circle at P.

The curvature of C at P is then defined to be the curvature of that circle or line. The radius of curvature is defined as the reciprocal of the curvature.

Curvature of curve• Suppose that a particle moves along the curve with

unit speed.

• Taking the time s as the parameter for C, this provides a natural parametrization for the curve.

• The unit tangent vector T (which is also the velocity vector, since the particle is moving with unit speed) also depends on time.

• The curvature is then the magnitude of the rate of change of T, i.e.

ds

dT

T is the unit tangent vector.

• The curvature is easier to compute if it is expressed in terms of the parameter t instead of s.

• So, we use the Chain Rule to write:

• However, ds/dt = |r’(t)|. So,

CURVATURE

/and

/

d d ds d d dt

dt ds dt ds ds dt

T T T T

'( )( )

'( )

tt

t

T

r

• Geometrically, this measures how fast the unit

tangent vector to the curve rotates.

• If a curve keeps close to the same direction,

the unit tangent vector changes very little and

the curvature is small; where the curve

undergoes a tight turn, the curvature is large.

Curvature of curve (contd.)

Curvature of a curve• There are several formulas for determining the

curvature for a curve.

3)('

)('')('or

)('

)('

tr

trtr

tr

tT

Example: 1• Determine the curvature for ttttr cos3,sin3,)(

Example: 1 (contd.)

The curvature is then

In this case the curvature is constant. This means that the curve is changing direction at the same rate at every point along it.

Exercise: 1

Determine the curvature for tkittr 2)(

In this case the second form of the curvature would probably be easiest. Here are the first couple of derivatives

Velocity and Acceleration

• In this section we will be looking at the velocity

and acceleration of a moving object.

• From Calculus course, we know that given the

position function of an object that the velocity

of the object is the first derivative of the position

function and the acceleration of the object is

the second derivative of the position function.

• Suppose a particle moves through

space so that its position vector at

time t is r(t).

VELOCITY

• Notice from the figure that, for small values of h, the vector

approximates the direction of the particle moving along the curve r(t).

VELOCITY

( ) ( )t h t

h

r r

Vector 1

• Its magnitude measures the size of the displacement vector per unit time.

• The vector 1 on previous slide gives the average velocity over a time interval of length h.

VELOCITY

• Its limit is the velocity vector v(t) at time t :

• Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line

VELOCITY VECTOR

0

( ) ( )( ) lim

'( )

h

t h tt

ht

r rv

r

Equation 1

• The speed of the particle at time t is the

magnitude of the velocity vector, that is, |v(t)|.

• This is appropriate because, from equation1, we

have:

SPEED

| ( ) | | '( ) |ds

t tdt

v r

= rate of change of distance with respect to time

• As in the case of one-dimensional motion,

the acceleration of the particle is defined as the

derivative of the velocity:

a(t) = v’(t) = r”(t)

ACCELERATION

• The position vector of an object moving in a plane is given by:

r(t) = t3 i + t2 j

– Find its velocity, speed, and acceleration when t = 1 and illustrate geometrically.

VELOCITY & ACCELERATION (Example:1)

• The velocity and acceleration at time t are:

• v(t) = r’(t) = 3t2 i + 2t j• a(t) = r”(t) = 6t I + 2 j

• The speed at t is:

Example 1 (Contd.)

2 2 2

4 2

| ( ) | (3 ) (2 )

9 4

t t t

t t

v

• When t = 1, we have:

v(1) = 3 i + 2 j

a(1) = 6 i + 2 j

|v(1)| = 13

Example 1 (Contd.)

• These velocity and acceleration vectors are shown here.

Example 1 (Contd.)

• The vector integrals that were introduced in

earlier section can be used to find position

vectors when velocity or acceleration vectors

are known, as in the next example.

VELOCITY & ACCELERATION

• A moving particle starts at an initial position r(0) = ‹1, 0, 0›

with initial velocity v(0) = i – j + k

• Its acceleration is a(t) = 4t i + 6t j + k

– Find its velocity and position at time t.

Example 3

• Since a(t) = v’(t), we have:v(t) = ∫ a(t) dt

= ∫ (4t i + 6t j + k) dt =2t2 i + 3t2 j + t k + C

• To determine the value of the constant vector C, we use the fact that

v(0) = i – j + k• The preceding equation gives v(0) = C.

So, C = i – j + k

Example 3 (Contd.)

• It follows: v(t) = 2t2 i + 3t2 j + t k + i – j + k v(t) = (2t2 + 1) i + (3t2 – 1) j + (t + 1) k

• Since v(t) = r’(t), we have: r(t) = ∫ v(t) dt

= ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt

= (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D• Putting t = 0, we find that D = r(0) = i.

• So, the position at time t is given by:

r(t) = (⅔t3 + t + 1) i + (t3 – t) j + (½t2 + t) k

Example 3 (Contd.)

• The expression for r(t) that we obtained in Example 3 was used to plot the path of the particle here for 0 ≤ t ≤ 3.

Example 3 (Contd.)

• So, given this it shouldn’t be too surprising that if the

position function of an object is given by the vector

function r (t ) then the velocity and acceleration of

the object is given by,

Summary: Velocity and Acceleration

);('')( );(')( trtatrtv

Example: 4 (Contd.)

• If the force that acts on a particle is known, then

the acceleration can be found from Newton’s

Second Law of Motion.• The vector version of this law states that if,

at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t), then

F(t) = ma(t)

Newton’s Second Law of Motion

• An object with mass m that moves in

a circular path with constant angular speed ω

has position vector

r(t) = a cos ωt i + a sin ωt j

Find the force acting on the object and

show that it is directed toward the origin.

Exercise: 1

• To find the force, we first need to know the acceleration: v(t) = r’(t) = –aω sin ωt i + aω cos ωt j

a(t) = v’(t) = –aω2 cos ωt i – aω2 sin ωt j

• Therefore, Newton’s Second Law gives the force as: F(t) = ma(t)

= –mω2 (a cos ωt i + a sin ωt j)• Notice that; F(t) = –mω2r(t)

• This shows that the force acts in the direction opposite to the radius vector r(t)

Exercise: 1 (Contd.)

• Therefore, it points toward the origin.• Such a force is called a centripetal (center-seeking)

force.

Exercise: 1 (Contd.)

NORMAL AND TANGENTIAL COMPONENTSWhen a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used.

In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle).

The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion.The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued)

The positive n and t directions are defined by the unit vectors un and ut, respectively.

The center of curvature, O’, always lies on the concave side of the curve.The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point.

The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.

SummaryWhen we study the motion of a particle along a

curve, it is often useful to resolve the

acceleration into two components:

• Tangential (in the direction of the tangent)

• Normal (in the direction of the normal)

VELOCITY IN THE n-t COORDINATE SYSTEM

The velocity vector is always tangent to the path of motion (t-direction).

The magnitude is determined by taking the time derivative of the path function, s(t).

v = vut where v = s’ = ds/dt

Here v defines the magnitude of the velocity (speed) andut defines the direction of the velocity vector.

ACCELERATION IN THE n-t COORDINATE SYSTEM

Acceleration is the time rate of change of velocity:a = dv/dt = d(vut)/dt = vut + vut

. .

Here v represents the change in the magnitude of velocity and ut represents the rate of change in the direction of ut.

..

. a = vut + (v2/r)un = atut + anun.

After mathematical manipulation, the acceleration vector can be expressed as:

ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)

There are two components to the acceleration vector:

a = at ut + an un

• The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r

• The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.

at = v or at ds = v dv.

• The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5

• If we write v = |v| for the speed of the particle, then

– Thus,

v = vT– If we differentiate both sides of that equation with

respect to t, we get

Summary: Velocity - Acceleration

'( ) ( )( )

| '( ) | | ( ) |

t tt

t t v

r v vT

r v

' ' 'v v a v T T

• If we use the expression for the curvature given earlier, we have:

• The unit normal vector was defined as N = T’/ |T’| , therefore

• Then, Equation becomes

| ' | | ' |so | ' |

| ' |v

v

T TT

r

Summary: Velocity - Acceleration

' | ' | v T T N N

' ' 'v v a v T T

2'v v a T N

• Writing aT and aN for the tangential and normal components of acceleration, we have

a = aTT + aNN

whereaT = v’ and aN = ĸv2

Summary: Velocity - Acceleration

• This resolution is illustrated here.

ACCELERATION—COMPONENTS

• Let’s look at what does our acceleration formula says

• The first thing to notice is that the binormal vector B is absent.

– No matter how an object moves through space, its acceleration always lies in the plane of T and N

– Recall that T gives the direction of motion and N points in the direction the curve is turning.

ACCELERATION—COMPONENTS

2'v v a T N

• Next, we notice that:

– The tangential component of acceleration is v’,

the rate of change of speed i.e. it reflects change in

speed

– The normal component of acceleration is ĸv2,

the curvature times the square of the speed, i.e. it

reflects change in direction.

ACCELERATION—COMPONENTS

• This makes sense if we think of a passenger in a car.

– A sharp turn in a road means a large value of the curvature ĸ.

– So, the component of the acceleration perpendicular

to the motion is large and the passenger is thrown against a car door.

SPECIAL CASES OF MOTION

There are some special cases of motion to consider.

2) The particle moves along a curve at constant speed. at = v = 0 => a = an = v2/ r

.

The normal component represents the time rate of change in the direction of the velocity.

1) The particle moves along a straight line. r => an = v2/ r = 0 => a = at = v

.

The tangential component represents the time rate of change in the magnitude of the velocity.

• We have expressions for the tangential

and normal components of acceleration in

Equations.

• However, it’s desirable to have expressions that

depend only on r, r’, and r”.

ACCELERATION—COMPONENTS

2'v v a T N

• Thus, we take the dot product of v = vT with a as given

by Equation

v · a = vT · (v’ T + ĸv2N)

= vv’ T · T + ĸv3T · N

= vv’ (Since T · T = 1 and T · N = 0)

ACCELERATION—COMPONENTS

2'v v a T N

• Therefore,

ACCELERATION—COMPONENTS

'

'( ) "( )

| '( ) |

Ta vvt t

t

v a

r r

r

• Using the formula for curvature we have:

ACCELERATION—COMPONENTS

2 23

| '( ) "( ) || '( ) |

| '( ) |

| '( ) "( ) |

| '( ) |

N

t ta v t

t

t t

t

r rr

r

r r

r

• In the study of the motion of objects the

acceleration is often broken up into a tangential

component (aT), and a normal component (aN).

• The tangential component is the part of the

acceleration that is tangential to the curve and the

normal component is the part of the acceleration

that is normal (or orthogonal) to the curve. If we do

this we can write the acceleration as,

function.position for the normalunit andnt unit tange theare N & T ,

where

NaTaa NT

Components of Acceleration

• Tangential component of acceleration reflects

change of speed and normal component

reflects change of direction. For example:

• The particle moves along a curve at constant

speed will have at = v’ = 0 => a = an

• The particle moves along a straight line will

have an = 0 => a = at

• A particle moves with position

function

r(t) = ‹t2, t2, t3›

• Find the tangential aT and normal aN

components of acceleration.

Example: 1

2 2 3

2

2 4

( )

'( ) 2 2 3

"( ) 2 2 6

| ( ) | 8 9

t t t t

t t t t

t t

t t t

r = i + j+ k

r = i + j+ k

r = i + j+ k

r'

Example: 1 (contd.)

• Therefore, the tangential component is given by:

3

2 4

'( ) "( )

| '( ) |

8 18

8 9

T

t ta

t

t t

t t

r r

r

Example: 1 (contd.)

2

2 2

'( ) "( ) 2 2 3

2 2 6

6 6

t t t t t

t

t t

ji k

r r

i j

Example: 1 (contd.)

• Hence, the normal component will be :

2

2 4

'( ) "( )

| '( ) |

6 2

8 9

N

t ta

t

t

t t

r r

r

Example: 1 (contd.)

Example: 2 (Contd.)

Exercise: 1• If r(t) is the position vector of a moving

particle. Find aT and aN.

kttjitr 2)(

2

2

41

2

;41

4

:

ta

t

ta

Ans

N

T

Exercise: 2• If r(t) is the position vector of a moving

particle. Find aT and aN.

ktjttitr 32

3

1

2

1)(

42

24

42

3

1

14

;1

2

:

tt

tta

tt

tta

Ans

N

T