Vector-Valued Functions

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Vector-Valued FunctionsSection 10.3a

Standard Unit VectorsAny vector in the plane can be written as a linear combinationof the two standard unit vectors: i 1,0 j 0,1

as follows: ,v a b ,0 0,a b

1,0 0,1a b i ja b The vector v is a linear combination of the vectors i and j; thescalar a is the horizontal component of v and the scalar b is thevertical component of v:

Standard Unit VectorsAny vector in the plane can be written as a linear combinationof the two standard unit vectors: i 1,0 j 0,1

as follows: ,v a b ,0 0,a b

1,0 0,1a b i ja b

Quick Practice: Given P(–1, 5) and Q(3, 2), write the followingvector as a linear combination of i and j: v PQ

v 3 1 ,2 5 Component form: 4, 3

As a linear combination… 4i 3j

Planar CurvesWhen a particle moves through the plane during a time interval I,we think of the particle’s coordinates as functions defined on I:

,x f t ,y g t .t IThe points , make up the curve inthe plane that is the particle’s path. The equations and intervalin the equation above parametrize the curve. The vector

, , ,x y f t g t t I

r , i jt OP f t g t f t g t

from the origin to the particle’s position attime t is the particle’s position vector. The functions f and g arethe component functions (components) of the position vector.we think of the particle’s path as the curve traced by r duringthe time interval I.

,P f t g t

Planar CurvesIn the next example, r is defined as a vector function of the realvariable t on the interval I. More generally, a vector function orvector-valued function on a domain D is a rule that assigns avector in the plane to each element in D. The curve traced by avector function is its graph.

We refer to real-valued functions as scalar functions todistinguish them from vector functions. The components of r arescalar functions of t. When we define a vector-valued functionby giving its component functions, we assume the vectorfunction’s domain to be the common domain of the components.

Planar CurvesGraph the vector function

We can graph parametrically using

cos sinr t t t i t t j 0t

cos ,x t t sin ,y t t 0t Window settings: Tmin = 0, Tmax = 50, Tstep = 0.1, Xmin = –30,Xmax = 30, Xscl = 1, Ymin = –20, Ymax = 20, Yscl = 1

As the curve graphs, imagine how this curve isbeing defined as traced by a changing vector…

Limits and ContinuityDefinition: LimitLet . If r i jt f t g t

1limt cf t L

and 2lim

t cg t L

Then the limit of r(t) as t approaches c is

1 2lim r L i jt c

Another quick example: If , then r cos i sin jt t t

lim rt

lim cos i lim sin jt t

Limits and ContinuityDefinition: LimitLet . If r i jt f t g t

1limt cf t L

and 2lim

t cg t L

Then the limit of r(t) as t approaches c is

1 2lim r L i jt c

lim rt

lim cos i lim sin jt t

cos i sin j4 4

2 2i j

Limits and ContinuityDefinition: Continuity at a PointA vector function r(t) is continuous at a point t = c in itsdomain if lim r r

t ct c

Component Test for Continuity at a PointThe vector function is continuousat t = c if and only if f and g are continuous at t = c.

A vector function r(t) is continuous if it is continuous atevery point in its domain.

r i jt f t g t

Derivatives and MotionSuppose that is the position vectorof a particle moving along a curve in the plane and that f and gare differentiable functions of t. Then the difference betweenthe particle’s position at time and time t is

r i jt f t g t

r r rt t t

Direction ofincreasing tO

During this time interval,the particle moves alongthe path, undergoingdisplacement rPQ

The vector sum r rt gives the new position: r t t

Derivatives and MotionSuppose that is the position vectorof a particle moving along a curve in the plane and that f and gare differentiable functions of t. Then the difference betweenthe particle’s position at time and time t is

r i jt f t g t

r r rt t t Written in terms of components…

i j i jf t t g t t f t g t

i jf t t f t g t t g t

Now, what happens as approaches zero???t

Derivatives and Motion r i jf t t f t g t t g t

Now, what happens as approaches zero???t

Direction ofincreasing tO

1. Q approaches P along the curve.

2. The secant line PQ seems to approach a limiting position tangent to the curve at P.

3. The quotient approaches the following limit:

rlimt t

Derivatives and Motion r i jf t t f t g t t g t

Now, what happens as approaches zero???t3. The quotient approaches the following limit:

rlimt t

lim i lim jt t

f t t f t g t t g tt t

i jdf dgdt dt

Do these limits look familiar???

Derivatives and MotionDefinition: Derivative at a PointThe vector function has aderivative (is differentiable) at t if f and g have derivativesat t. The derivative is the vector

r i jt f t g t

r rr lim i jt

t t td df dgdt t dt dt

A vector function r is differentiable if it is differentiable at everypoint of its domain. The curve traced by r is smooth if dr/dt iscontinuous and never 0, that is, if f and g have continuous firstderivatives that are not simultaneously zero. On a smooth curvethere are no sharp corners or cusps.

Derivatives and MotionDefinitions: Velocity, Speed, Acceleration,Direction of Motion

rv dtdt

If r is the position vector of a particle moving along a smoothcurve in the plane, then at any time t,

1. is the particle’s velocity vector and is tangent

to the curve.

v t2. , the magnitude of velocity, is the particle’s speed.

Derivatives and MotionDefinitions: Velocity, Speed, Acceleration,Direction of Motion

v ra d dtdt dt

If r is the position vector of a particle moving along a smoothcurve in the plane, then at any time t,

3. , the derivative of velocity and the

second derivative of position, is the particle’s accelerationvector.

v v4. , a unit vector, is the direction of motion.

Velocity = = (speed)(direction)vvv

Studying Motion r 3cos i 3sin jt t t The vector gives the position

of a moving particle at time t. Find(a) the velocity and acceleration vectors.

rv ddt

3sin i 3cos jt t

va ddt

3cos i 3sin jt t

of a moving particle at time t. Find(b) the velocity, acceleration, speed, and direction of motion at

3sin i 3cos j4 4

3 3i j2 2

Velocity:

3cos i 3sin j4 4

3 3i j2 2

Acceleration:

of a moving particle at time t. Find(b) the velocity, acceleration, speed, and direction of motion at

2 23 32 2

3Speed:

v 4v 4

Direction:3 2 3 2i j

1 1i j2 2

of a moving particle at time t. Find(c) . Interpret this result graphically.v a

v a 3sin 3cos 3cos 3sint t t t

9sin cos 9sin cost t t t 0If the dot product is zero, then the vectors are orthogonal(graphically, they are perpendicular).

Let’s interpret all of these resultswith a graph on the calculator…

Studying Motion 3 2 3r 2 3 i 12 jt t t t t The vector gives the

position of a moving particle at time t.(a) Write an equation for the line tangent to the path of theparticle at the point where t = –1.

2 2v 6 6 i 3 12 jt t t t

At the time in question: r 1 5i 11j v 1 12i 9 j

The line passes through (–5, 11) and has slope –9/12 = –3/4:

311 54

y x 3 294 4

Studying Motion 3 2 3r 2 3 i 12 jt t t t t The vector gives the

position of a moving particle at time t.(b) Find the coordinates of each point on the path where thehorizontal component of the velocity is 0.

The horizontal component of velocity: 26 6t t

Again, support these results graphically…

6 1t t Which equals zero when t = 0 and t = 1.

Coordinates at t = 0: r 0 0 i 0 j 0,0

Coordinates at t = 1: r 1 1 i 11 j 1, 11

Vector-Valued Functions

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