vector valued function

Vector-Valued Functions The Calculus of Vector-Valued Functions Motion in Space Curvature Tangent and Normal Vectors Parametric Surfaces

CalculusVector-Valued Functions


Outline

1 Vector-Valued Functions

2 The Calculus of Vector-Valued Functions

3 Motion in Space

4 Curvature

5 Tangent and Normal Vectors

6 Parametric Surfaces


Vector Valued Functions: An Example

To specified the path of the plane indicated in the figure, wedescribe the planes location at any given time by the endpointof a vector called position vector

Figure: [9.1a] Airplanes flightpath.

Figure: [9.1b] Vectors indicatingplanes position at several times.


Definition of Vector-Valued FunctionsDefinition (1.1)A vector-valued function r(t) is a mapping from its domainD R to its range R V3, so that for each t in D, r(t) = v foronly one vector v V3. We can always write a vector-valuedfunction as

r(t) = f (t)i + g(t)j + h(t)k,

for some scalar functions f , g and h (called the componentfunctions of r).

we can likewise define a vector-valued function r(t) in V2 by

r(t) = f (t)i + g(t)j

for some scalar function f and g.


Sketching the Curve Defined by a Vector-ValuedFunction

Example (1.1)Sketch a graph of the curve traced out by the endpoint of thetwo-dimensional vector-valued function

r(t) = (t + 1)i + (t2 2)j.


The Trace of a Vector-Valued FunctionIn example 1.1, the curve traced out by the endpoint of thevector-valued function

r(t) = (t + 1)i + (t2 2)jis identical to the curve described by the parametric equations

x(t) = t + 1 and y(t) = t2 1

Figure: [9.2a] Some values ofr(t) = (t + 1)i + (t2 2)j.

Figure: [9.2b] Curve defined byr(t) = (t + 1)i + (t2 2)j.


A Vector-Valued Function Defining an Ellipse

Example (1.2)Sketch a graph of the curve traced out by the endpoint of thevector-valued function r(t) = 4 cos t i 3 sin t j, t R.

Figure: [9.3] Curve defined by r(t) = 4 cos t i 3 sin t j.


A Vector-Valued Function Defining an Elliptical HelixExample (1.3)Plot the curve traced out by the vector-valued functionr(t) = sin t i 3 cos t j + 2t k, for t 0.

Figure: [9.4a] Elliptical helix,r(t) = sin t i 3 cos t j + 2t k.

Figure: [9.4b] Computer sketch:r(t) = sin t i 3 cos t j + 2t k.


A Vector-Valued Function Defining a Line

Example (1.4)Plot the curve traced out by the vector-valued function

r(t) = 3 + 2t, 5 3t, 2 4t, t R.

Figure: [9.5] Straight line: r(t) = 3 + 2t, 5 3t, 2 4t.


Matching a Vector-Valued Function to Its Graph (I)

Example (1.5)Match each of the vector-valued functionsr1(t) = cos t, ln t, sin t, r2(t) = t cos t, t sin t, t,r3(t) = 3 sin 2t, t, t and r4(t) = 5 sin3 t, 5 cos3 t, t with thecorresponding computer-generated graph.


Matching a Vector-Valued Function to Its Graph (II)

r1(t) = cos t, ln t, sin t r2(t) = t cos t, t sin t, tr3(t) = 3 sin 2t, t, t r4(t) = 5 sin3 t, 5 cos3 t, t


Matching a Vector-Valued Function to Its Graph (III)r1(t) = cos t, ln t, sin t r2(t) = t cos t, t sin t, tr3(t) = 3 sin 2t, t, t r4(t) = 5 sin3 t, 5 cos3 t, t


Arc Length in R2

Recall from section 5.3 that if f and f are continuous on theinterval [a, b], then the arc length of the curve y = f (x) on thatinterval is given by

s = b

a

1 + (f (x))2 dx

Consider a curve defined parametrically by x = f (t) andy = g(t), where f , f , g and g are all continuous for t [a, b]. Ifthe curve is traversed exactly once as t increases from a to b,then the arc length is given

s = b

a

(f (x))2 + (g(x))2 dt


Arc Length in R3

Suppose that a curve in three dimensions is traced out by theend point of the vector-valued function

r(t) =< f (t), g(t), h(t) >,

where f , f , g, g, h and h are all continuous for t [a, b] andwhere the curve is traversed exactly once as t increases from ato b. For this situation, the arc length is given by

s = b

a

(f (x))2 + (g(x))2 + (h(x))2 dt

The integral can only rarely be computed exactly and we musttypically be satisfied with a numerical approximation.


Approximating the Arc Length of a Curve in R3We approximate the curve by a number of line segments anduse the distance formula to yield an approximation to the arclength. As the number of segments increases without bound,the sum of the distances approaches the actual arc length.

Figure: [9.7a]Approximate arclength in R2.

Figure: [9.7b]Approximate arclength in R3.

Figure: [9.7c]Improved arc lengthapproximation.


Computing Arc Length in R3

Example (1.6)Find the arc length of the curve traced out by the endpoint ofthe vector-valued function r(t) = 2t, ln t, t2, for 1 t e.

Figure: [9.8] The curve defined by r(t) = 2t, ln t, t2.


Approximating Arc Length in R3

Example (1.7)Find the arc length of the curve traced out by the endpoint ofthe vector-valued function r(t) = e2t, sin t, t, for 0 t 2.


Finding Parametric Equations for an Intersection of Surfaces

Example (1.8)Find parametric equations for the curve determined by theintersection of the cone z =

x2 + y2 and the plane y + z = 2.

Figure: [9.9a] z =x2 + y2 andy + z = 2.

Figure: [9.9b] x = t, y = 1 t24 ,z = 1 + t

2

4 .


Limit of a Vector-Valued Function (I)For a vector-valued function

r(t) =< f (t), g(t), h(t) >,

If we writelimta r(t) = u,

we mean that as t gets closer and closer to a, the vector r(t) isgetting closer and closer to the vector u. If we writeu =< u1, u2, u3 >, this means that

limta r(t) = limta < f (t), g(t), h(t) >= u =< u1, u2, u3 >

Notice that for this to occur, we must have that

f (t) u1, g(t) u2, and h(t) u3.


Limit of a Vector-Valued Function (II)

Definition (2.1)For a vector-valued function r(t) = f (t), g(t), h(t), the limit ofr(t) as t approaches a is given by

limta r(t) = limta f (t), g(t), h(t) =

limta f (t), limta g(t), limta h(t)

,

provided all of the indicated limits exist. If any one of the limitsindicated on the right-hand side of (2.1) fails to exist, thenlimta r(t) does not exist.


Finding the Limit of a Vector-Valued Function

Example (2.1)Find lim

t0t2 + 1, 5 cos t, sin t.


A Limit That Does Not Exist

Example (2.2)Find lim

t0

e2t + 5, t2 + 2t 3, 1

t

.


Continuity of a Vector-Valued Function (I)Recall that for a scalar function f , we say that f is continuous ata if and only if

limta f (t) = f (a)

We defined the continuity of vector-valued functions in thesame way.

Definition (2.2)The vector-valued function r(t) = f (t), g(t), h(t), iscontinuous at t = a whenever

limta r(t) = r(a)

(i.e., whenever the limit and the value of the vector-valuedfunction are the same).


Continuity of a Vector-Valued Function (III)Notice that in terms of the components of r, the definition saysthat t is continuous at a whenever

limta < f (t), g(t), h(t) >=< f (a), g(a), h(a) >

Further since

limta < f (t), g(t), h(t) >=< limta f (a), limta g(a), limta h(a) >

It follows that r(t) is continuous at t = a if and only if

< limta f (t), limta g(t), limta h(t) >=< f (a), g(a), h(a) >

Note that this occurs if and only if

limta f (t) = f (a), limta g(t) = g(a), and limta h(t) = f (a).

The above result is summarized in Theorem 2.1.


Continuity of a Vector-Valued Function (IV)

Theorem (2.1)A vector-valued function r(t) = f (t), g(t), h(t) is continuous att = a if and only if all of f , g and h are continuous at t = a.Notice that Theorem 2.1 says that if you want to determinewhether or not a vector-valued function is continuous, you needonly check the continuity of each component function.


Determining Where a Vector-Valued Function IsContinuous

Example (2.3)Determine for what values of t the vector-valued functionr(t) = e5t, ln(t + 1), cos t is continuous.


A Vector-Valued Function with Infinitely ManyDiscontinuities

Example (2.4)Determine for what values of t the vector-valued functionr(t) = tan t, |t + 3|, 1t2 is continuous.


The Derivative of a Vector-Valued Function (I)Recall that we defined the derivative of a scalar function f to be

f (t) = limt0

f (t +t) f (t)t

Similarly, we define the derivative of a vector-valued function asfollows.

Definition (2.3)The derivative r(t) of the vector-valued function r(t) is definedby

r(t) = limt0

r(t +t) r(t)t

,

for any values of t for which the limit exists. When the limitexists for t = a, we say that r is differentiable at t = a.


The Derivative of a Vector-Valued Function (II)

Theorem (2.2)Let r(t) = f (t), g(t), h(t) and suppose that the componentsf , g and h are all differentiable for some value of t. Then r isalso differentiable at that value of t and its derivative is given by

r(t) = f (t), g(t), h(t).Theorem 2.2 says that the derivative of a vector-valued functionis found directly from the derivatives of the individualcomponents.


Finding the Derivative of a Vector-Valued Function

Example (2.5)Find the derivative of r(t) = sin(t2), ecos t, t ln t.


Rules for Computing Derivatives of Vector-ValuedFunctions

Theorem (2.3)Suppose that r(t) and s(t) are differentiable vector-valuedfunctions, f (t) is a differentiable scalar function and c is anyscalar constant. Then

1ddt [r(t) + s(t)] = r

(t) + s(t)

2ddt [cr(t)] = cr

(t)

3ddt [ f (t)r(t)] = f

(t)r(t) + f (t)r(t)

4ddt [r(t) s(t)] = r

(t) s(t) + r(t) s(t) and

5ddt [r(t) s(t)] = r

(t) s(t) + r(t) s(t).


A Graphical Interpretation of the Derivative of aVector-Valued Function (I)

Recall that the derivative of r(t) att = a is given by

r(a) = limt0

r(a +t) at

Notice that the endpoint of thevector-valued function r(t) traces outa curve C in R3 as shown in thefigure.Observe that the vectorr(a +t) r(t)

tpoints in the same

direction as r(a +t) r(t).


A Graphical Interpretation of the Derivative of aVector-Valued Function (II)

If we take smaller and smaller value of t, r(a +t) r(t)t

willapproach r(t).


A Graphical Interpretation of the Derivative of aVector-Valued Function (III)

As t 0, notice that the vectorr(a +t) r(t)

tapproaches a

vector that is tangent to the curve Cat the terminal point of r(t). We referto r(a) as the tangent vector to thecurve C at the point correspondingto t = a.Observe that r(a) lies along thetangent line to the curve at t = a andpoints in the direction of theorientation of C.


A Graphical Interpretation of the Derivative of aVector-Valued Function (IV)


Drawing Position and Tangent VectorsExample (2.6)For r(t) = cos 2t, sin 2t, plot the curve traced out by theendpoint of r(t) and draw the position vector and tangent vectorat t = pi4 .

Figure: [9.11] Position and tangent vectors.


||r(t)|| = constant r(t) r(t)

Theorem (2.4)r(t) = constant if and only if r(t) and r(t) are orthogonal, forall t.Theorem 2.4 implies that:

1 in two (three) dimensions, if ||r(t)|| = c, then the curvedtraced out by the position r(t) must lie on the circle(sphere) of radius c, centered at the origin, and

2 the path traced out by r(t) lies on a circle (sphere) centeredat the origin if and only if the tangent vector is orthogonalto the position vector at every point on the curve.


Antiderivatives of the Vector-Valued Functions

Definition (2.4)The vector-valued function R(t) is an antiderivative of thevector-valued function r(t) whenever R(t) = r(t).

Notice that if r(t) =< f (t), g(t), h(t) > and f , g and h haveantiderivatives F, G and H, respectively, then

ddt < F(t),G(t),H(t) >=< F

(t),G(t),H(t) >=< f (t), g(t), h(t) >

That is, < F(t),G(t),H(t) > is an antiderivative of r(t). In fact,< F(t) + c1,G(t) + c2,H(t) + c3 > is also an antiderivative of r(t),for any choice of constants, c1, c2 and c3.


Indefinite Integral of a Vector-Valued Function

Definition (2.5)If R(t) is any antiderivative of r(t), the indefinite integral of r(t)is defined to be

r(t)dt = R(t) + c,

where c is an arbitrary constant vector.

As in the scalar case, R(t) + c is the most general antiderivativeof r(t). Notice that this says that

r(t) dt =

< f (t), g(t), h(t) >=

f (t) dt,

g(t) dt,

h(t) dt

That is, you integrate a vector-valued function by integratingeach of the individual components.


Evaluating the Indefinite Integral of a Vector-ValuedFunction

Example (2.7)Evaluate the indefinite integral

t2 + 2, sin 2t, 4tet2

dt.


Definite Integral of a Vector-Valued Function

Similarly, we defined the definite integral of a vector-valuedfunction in the obvious way.

Definition (2.6)For the vector-valued function r(t) = f (t), g(t), h(t), we definethe definite integral of r(t) by b

a

r(t)dt = b

a

f (t), g(t), h(t)dt = b

a

f (t)dt, b

a

g(t)dt, b

a

h(t)dt.

Notice that this says that the definite integral of a vector-valuedfunction r(t) is simply the vector whose components are thedefinite integrals of the corresponding components of r(t).


Fundamental Theorem of Calculus for Vector-ValuedFunctions

Theorem (2.5)Suppose that R(t) is an antiderivative of r(t) on the interval[a, b]. Then, b

a

r(t)dt = R(b) R(a).


Evaluating the Definite Integral of a Vector-ValuedFunction

Example (2.8)Evaluate

10

sinpit, 6t2 + 4t

dt.


Motion in Space (I)

Suppose that an object moves alonga curve described parametrically by

C : x = f (t), y = g(t), z = h(t)

where t [a, b]. We can think of thecurve as being traced out by theendpoint of the vector-valuedfunction

r(t) =< f (t), g(t), h(t) > .Figure: [9.12] Position,velocity and accelerationvectors.


Motion in Space (II)

Differentiating r(t), we have

r(t) =< f (t), g(t), h(t) >

and the magnitude of thisvector-valued function is

||r(t)|| =[f (t)]2 + [g(t)]2 + [h(t)]2

Figure: [9.12] Position,velocity and accelerationvectors.


Motion in Space (III)

Recall that the arc length of theportion of the curve from u = t0 up tou = t is given by

s(t) = t

t0

[f (u)]2 + [g(u)]2 + [h(u)]2 dt.

If we differentiate both sides of theabove equation, we get

s(t) =[f (t)]2 + [g(t)]2 + [h(t)]2

= ||r(t)||. Figure: [9.12] Position,velocity and accelerationvectors.


Motion in Space (IV)

Since s(t) represent arc length,s(t) gives the instantaneousrate of change of arc lengthwith respect to time, i.e., thespeed of the object as it movesalong the curve.



Motion in Space (V)

For any given t, r(t) is atangent vector pointing in thedirection of the orientation of Cand whose magnitude gives thespeed of the object. We call r(t)the velocity vector, usuallydenoted v(t).



Motion in Space (VI)

We refer to the derivative of thevelocity vector v(t) = r(t) asthe acceleration vector,denoted a(t).



Finding Velocity and Acceleration Vectors

Example (3.1)Find the velocity and acceleration vectors if the position of anobject moving in the xy-plane is given by r(t) = t3, 2t2.

Figure: [9.13] Position, velocity and acceleration vectors.


Finding Velocity and Position from AccelerationExample (3.2)Find the velocity and position of an object at any time t, giventhat its acceleration is a(t) = 6t, 12t + 2, et, its initial velocity isv(0) = 2, 0, 1 and its initial position is r(0) = 0, 3, 5.

Figure: [9.14] Position, velocity and acceleration vectors.


Newtons Second Law of Motion

The Newtons second law of motion states that the net forceacting on an object equals the product of the mass and theacceleration.Using vector notation, we have the vector form of Newtonssecond law:

F = ma

Here, m is the mass, a is the acceleration vector and F is thevector representing the net force acting on the object.


Newtons Second Law of Motion and LinearMomentum

When a force F(t) applies on an object of constant mass m, themotion of the object obeys the Newtons second law of motion,

F(t) = ma(t)

where a is the acceleration of the object.Now, integrating the Newtons second law with respect to timewe have t2

t1

F(t) dt = m t2

t1

a dt

= mv(t)|t2t1 = mv(t2) mv(t1)

In the last expression, the term mv(t) is referred to as the linearmomentum of the object.


Rotational Version of Newtons Second Law of Motion

In the case of an object rotating in two dimensions, the primaryvariable that we track is an angle of displacement, denoted by. For a rotating body, the angle measured from fixed raychanges with time t, so that the angle is a function (t). Wedefine the angular velocity to be (t) = (t) and the angularacceleration to be

(t) = (t) = (t)

The equation of rotation motion is then

= I

where I is the moment of initial of a body and is the torquecausing the rotation.


Rotational Version of Newtons Second Law of Motionand Angular Momentum

For rotational motion in three dimensions, the calculations aresomewhat more complicated. Recall that we had defined thetorque due to a force F applied at position r to be

= r F

Integrating the above equation with respect to time, we yield t2t1

dt = t2

t1

r F dt = t2

t1

v mv + r ma dt

= t2

t1

v mv + r mv dt = t2

t1

(r mv) dt

= (r mv)|t2t1The term rmv is referred to as the angular linear momentum


Finding the Force Acting on an ObjectExample (3.3)Find the force acting on an object moving along a circular pathof radius b centered at the origin, with constant angular speed.

Figure: [9.15a] Motion along acircle.

Figure: [9.15b] Centripetal force.


Analyzing the Motion of a ProjectileExample (3.4)A projectile is launched with an initial speed of 140 feet persecond from ground level at an angle of pi4 to the horizontal.Assuming that the only force acting on the object is gravity (i.e.,there is no air resistance, etc.), find the maximum altitude, thehorizontal range and the speed at impact of the projectile.

Figure: [9.16a] Initial velocityvector.

Figure: [9.16b] Path of aprojectile.


The Rotational Motion of a Merry-Go-Round

Example (3.5)A stationary merry-go-round of radius 5 feet is started in motionby a push consisting of a force of 10 pounds on the outsideedge, tangent to the circular edge of the merry-goround, for 1second. The moment of inertia of the merry-go-round is I = 25.Find the resulting angular velocity of the merry-go-round.


Relating Torque and Angular Momentum

Example (3.6)Show that torque equals the derivative of angular momentum.


Analyzing the Motion of a Projectile in ThreeDimensions (I)

Example (3.7)A projectile of mass 1 kg islaunched from ground leveltoward the east at 200meters/second, at an angle ofpi6 to the horizontal. If a gustingnortherly wind applies asteady force of 2 newtons tothe projectile, find the landinglocation of the projectile andits speed at impact.

Figure: [9.17a] The initial velocityand wind velocity vectors.


Analyzing the Motion of a Projectile in ThreeDimensions (II)

Figure: [9.17b] Path of theprojectile.

Figure: [9.17c] Projection of pathonto the xz-plane.


Arc Length of a CurveRecall that for the curve traced out by the endpoint of thevector-valued function

r(t) =< f (t), g(t), h(t) >, for a t b,we define the arc length parameter s(t) to be the arc length ofthat portion of the curve from u = a up to u = t, i.e.,

s(t) = t

0

[f (u)]2 + g(u)]2 + h(u)]2 du

Recognizing that[f (u)]2 + g(u)]2 + h(u)]2 = ||r(u)||,

we can write this more simply as

s(t) = t

a

||r(u)|| du


Parameterizing a Curve in Terms of Arc Length

Example (4.1)Find an arc length parameterization of the circle of radius 4centered at the origin.


Unit Tangent Vector (I)

Consider the curve C traced out bythe endpoint of the vector-valuedfunction r(t). Recall that for each t,v(t) = r(t) can be thought of as boththe velocity vector and a tangentvector, pointing in the direction ofmotion (i.e., the orientation of C).

Figure: [9.18] Unit tangentvectors.


Unit Tangent Vector (II)

Notice that

T(t) =r(t)||r(t)||

is also a tangent vector, but haslength one (||T(t)|| = 1). We call T isa tangent vector of length onepointing in the direction of theorientation of C. Figure: [9.18] Unit tangent

vectors.


Finding a Unit Tangent Vector

Example (4.2)Find the unit tangent vector to the curve determined byr(t) = t2 + 1, t.

Figure: [9.18] Unit tangent vectors.


Tangent Vectors and "Sharpness" of Curves (I)In the following figures, we show two curves both connectingthe points A and B. The curve in Figure 9.19b indicates a muchsharper turn than the curve in Figure 9.19a. The questionbefore us is to see how to mathematically describe this degreeof "sharpness".

Figure: [9.19a] Gentle curve. Figure: [9.19b] Sharp curve.


Tangent Vectors and "Sharpness" of Curves (II)In the same figures, we have draw in a number of unit tangentvectors at equally spaced points on the curve. Notice that theunit tangent vectors change very slowly along the gentle curvein Figure 9.19c, but twist and turn quite rapidly in the vicinity ofthe sharp curve in Figure 9.19d.

Figure: [9.19c] Unit tangentvectors.

Figure: [9.19d] Unit tangentvectors.


Tangent Vectors and "Sharpness" of Curves (III)

Based on the analysis, the rate of change of the unit tangentvectors with respect to the arc length along the curve will giveus a measure of sharpness.

Figure: [9.19c] Unit tangentvectors.

Figure: [9.19d] Unit tangentvectors.


Curvature of a Curve (I)

Definition (4.1)The curvature of a curve is the scalar quantity

=dTds

.Computing the curvature of given curve by the definitionis not a simple matter. To do so, we would need to first findthe arc length parameter and the unit tangent vector T(t),rewrite T(t) in terms of the arc length parameter s and thendifferentiate with respect to s


Curvature of a Curve (II)Observe that by the chain rule

T(t) =dTdt =

dTds

dsdt ,

so that =

dTds = T(t) ds

dt

Recall thats(t) =

ta

||r(u)||du

By the Fundamental Theorem of Calculus,dsdt = ||r

(t)||

Hence, we have =

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


Finding the Curvature of a Straight Line

Example (4.3)Find the curvature of a straight line.


Finding the Curvature of a Circle

Example (4.4)Find the curvature for a circle of radius a.


Another Way of Computing the Curvature of a Curve

Theorem (4.1)The curvature of the curve traced out by the vector-valuedfunction r(t) is given by

=r(t) r(t)

r(t)3 .

This is a relatively simple matter to use this formula tocompute the curvature for nearly any three-dimensionalcurve.


Finding the Curvature of a HelixExample (4.5)Find the curvature of the helix traced out byr(t) = 2 sin t, 2 cos t, 4t.

Figure: [9.20] Circular helix.


Curvature for the plane curve y = f (x)For a plane curve y = f (x), we can derive a simple formula forthe curvature. Notice that such a curve is traced out by thevector-valued function r(t) =< t, f (t), 0 > in the xy-plane.Further,

r(t) =< 1, f (t), 0 > and r(t) =< 0, f (t), 0 >From Theorem 4.1, we have

=r(t) r(t)

r(t)3 =< 1, f (t), 0 > < 0, f (t), 0 >

< 1, f (t), 0 >3

=|f (t)|

{1 + [f (t)]2}3/2Replacing t by x, we write the curvature as

=|f (x)|

{1 + [f (x)]2}3/2


Finding the Curvature of a Parabola

Example (4.6)Find the curvature of the parabola y = ax2 + bx + c. Also, findthe limiting value of the curvature as x .


Stationary and Moving Frame of Reference

Up to this point, we have used a stationary frame of referencefor all of our work with vectors. That is, all the vector are interms of the standard unit basis vectors i, j and k. In thissection we introduce a moving frame of reference which issuitable for describing the motion of a moving object.


TNB Frame of Reference (I)

Consider an object moving along thecurved traced out by a vector-valuedfunction r =< f (t), g(t), h(t) >. Todefine a reference frame that moveswith the object, we will need to havethree mutually orthogonal unitvectors at each point on the curve.

Figure: [9.21] Principal unitnormal vectors.


TNB Frame of Reference (II)

One of these vectors should bepointing in the direction of motion,i.e., in the direction of the orientationof the curve. Since

T =r

ris a vector pointing in the directionof motion, T can be used as one ofthe three mutually orthogonal unitvectors.



TNB Frame of Reference (III)

Recall that T is a unit vector, i.e.,T = 1. Hence, from Theorem 2.4,we have

T T = 0,implying that T and T areorthogonal. This gives us a secondunit vector in our moving frame ofreference, as in Definition 5.1.



TNB Frame of Reference (IV)Principle Unit Normal Vector

Definition (5.1)The principal unit normal vector N(t) is a unit vector havingthe same direction as T(t) and is defined by

N(t) = T(t)

T(t) .


TNB Frame of Reference (V)

Recall that for a given curve traced out by r, the arc lengthparameter s(t) is given by

s(t) = t

a

r(u)du

Note thatdsdt = r

(t) > 0This implies that dsdt

= dsdt


TNB Frame of Reference (VI)

From the chain, we have

T (t) =dTdt =

dTds

dsdt .

This gives us

N(t) = T(t)

T(t) =dTds

dsdt

dTds dsdt

=dTds

dTds =

1

dTds

where we have used the definition of curvature.


TNB Frame of Reference (VII)

The expression

N(t) = 1

dTds

is not a practically useful formula forcomputing N. However, it can beused to determine the direction ofN(t). Since > 0, N has the samedirection as dTds . Notice that

dTds is

the instantaneous rate of change ofthe unit tangent vector with respectto the arc length. Figure: [9.21] Principal unit

normal vectors.


TNB Frame of Reference (VIII)

This says that dTds points in thedirection in which T is turning as arclength increases. That is, N(t) willalways point to the concave side ofthe curve.



TNB Frame of Reference (IX)

To get a third unit vector orthogonal to both T and N, we simplytake their cross product. This leads to the following definition.

Definition (5.2)We define the binormal vector B(t) to be

B(t) = T(t) N(t).Notice that since T and N are unit vectors and orthogonal toeach other, the magnitude of B is 1.


TNB Frame of Reference (X)

This triple of three unit vectors T(t),N(t) and B(t) forms a frame ofreference, called the TNB frame (orthe moving trihedral), that movesalong the curve defined as r(t). Thismoving frame of reference hasparticular importance in a branch ofmathematics call differentialgeometry.

Figure: [9.24] The TNBframe.


Finding Unit Tangent and Principal Unit Normal Vectors

Example (5.1)Find the unit tangent and principal unit normal vectors to thecurve defined by r(t) = t2, t.

Figure: [9.22] Unit tangent and principal unit normal vectors.


Finding Unit Tangent and Principal Unit Normal Vectors

Example (5.2)Find the unit tangent and principal unit normal vectors to thecurve determined by r(t) = sin 2t, cos 2t, t.

Figure: [9.23] Unit tangent and principal unit normal vectors.


Finding the Binormal Vector

Example (5.3)Find the binormal vector B(t) for the curve traced out byr(t) = sin 2t, cos 2t, t.

Figure: [9.25] The TNB frame for r(t) = sin 2t, cos 2t, t.


Normal Plane and Osculating Plane

For each value of t, the plane determined by N and B iscalled the normal plane. By definition, the normal plane toa curve at a point contains all of the lines that areorthogonal to the tangent vector at the given point on thecurve.

For each value of t, the plane determined by T and N iscalled the osculating plane. For a two-dimensional curve,the osculating plane is simply the xy-plane.


Osculating Circle (I)

For a given value of t, say t = t0, ifthe curvature of the curve at thepoint P corresponding to t0 isnonzero, then the circle of radius =

1

lying completely in theosculating plane and whose centerlies a distance of 1

from P along the

normal N(t) is called the osculatingcircle (or the circle of curvature).

Figure: [9.26] Osculatingcircle.


Osculating Circle (II)

Since the curvature of a circle is thereciprocal of its radius, it implies thatthe osculating circle has the sametangent and curvature at P as thecurve. Further, because the normalvector always points to the concaveside of the curve, the osculatingcircle lies on the concave side of thecurve.



Osculating Circle (III)

In this sense, the osculating circle isthe circle that "best fits" the curve atpoint P. The radius of the osculatingcircle is called the radius ofcurvature and the center of thecircle is called the center ofcurvature.



Finding the Osculating CircleExample (5.4)Find the osculating circle for the parabola defined byr(t) = t2, t at t = 0.

Figure: [9.27] Osculating circle.


Tangential/Normal Component of Acceleration (I)Suppose that the position of an object at time t is given by theterminal point of the vector-valued function r(t). Recall that

T =r(t)r(t) and r

(t) = dsdt ,

where s represents arc length. Then the velocity of the object isgiven by

v(t) = r(t) = r(t)T(t) = dsdt T(t).Further, we have the acceleration given by

a(t) = v(t) =ddt

(dsdt T(t)

)=

d2sdt2 T(t) +

dsdt T

(t)


Tangential/Normal Component of Acceleration (II)Recall that

N(t) = T(t)

T(t) T(t) = T(t)N(t)

Further by the chain rule,

T(t) =dTdt

= dTds dsdt = dsdt

dTds = dsdt

where we have used the definition of the curvature and thefact that dsdt > 0. So

T(t) = T(t)N(t) = dsdt N(t)

anda(t) =

d2sdt2 T(t) +

(dsdt

)2N(t)


Tangential/Normal Component of Acceleration (III)We have obtained the accelerationvector as

a(t) =d2sdt2 +

(dsdt

)2N(t)

= aTT(t) + aNN(t)

Notice that since a(t) is written as asum of a vector parallel to T(t) and avector parallel to N(t), the vector a(t)always lies in the osculating plane.The coefficient of T is the tangentialcomponent of acceleration aT andthe coefficient of N(t) is the normalcomponent of acceleration aN .

Figure: [9.28] Tangentialand normal components ofacceleration.


A Strategy for Keeping the Car on the Road (I)

From Newtons second law ofmotion, the net force acting on acar at any time t is F(t) = ma.Hence,

F(t) = ma(t)

= md2sdt2 + m

(dsdt

)2N(t)

where m is mass of the car.

Figure: [9.29] Driving arounda curve.


A Strategy for Keeping the Car on the Road (II)Observe that:

1 Since T points in the directionof the path of motion, we wantthe component of the forceacting in the direction of T tobe as large as possiblecompared to the component ofthe force acting in the directionof the normal N.

2 If the normal component of theforce is too large, it mayexceed the normal componentof the force of friction betweenthe tires and the highway,causing the car to skid off theroad.



A Strategy for Keeping the Car on the Road (III)

1 To minimize the force appliedin the direction of N, we needto make ds/dt small (reducingspeed).

2 To maximize the tangentialcomponent of the force, weneed to make d2s/dt2 as largeas possible. It implies that weneed to accelerate while in thecurve.

3 To make a turn it is better toslow down before you enterthe curve and then acceleratewhile in the curve.



A Strategy for Keeping the Car on the Road (IV)

If we wait until we are in the curveto slow down then

d2sdt2 < 0

in the curve and so the tangentialcomponent of the force is negative(acting in the opposite direction ofT), making it harder to get throughthe curve.

Figure: [9.30] Net force:d2sdt2 < 0.


Finding Tangential and Normal Components ofAcceleration

Example (5.5)Find the tangential and normal components of acceleration foran object with position vector r(t) = 2 sin t, 2 cos t, 4t.


Computing aT and aN (I)

1 It is simple to compute

aT =d2sdt2

We must only calculatingds/dt = r(t) and thendifferentiate the result.

2 Computing aN is a bit morecomplicated, since it requires youto first compute the curvature . Figure: [9.31]

Components of a(t).


Computing aT and aN (II)An easier way to compute aN can beobtained by using the orthogonality ofthe two vectors T and N. Notice that

a(t) =d2sdt2 +

(dsdt

)2N(t)

= aTT(t) + aNN(t)

SinceT N = 0.

we have

a2 = a2T + a2Nfollowed by the PythagoreanTheorem.

Figure: [9.31]Components of a(t).


Computing aT and aN (III)

Solving

a2 = a2T + a2Nfor aN , we get

aN =a2 a2T

where we have taken the positive rootsince

aN = (

dsdt

)2Figure: [9.31]Components of a(t).


Finding Tangential and Normal Components of Acceleration

Example (5.6)Find the tangential and normal components of acceleration foran object whose path is defined by r(t) = t, 2t, t2. In particular,find these components at t = 1. Also, find the curvature.

Figure: [9.32] Tangential and normal components of acceleration att = 1.


Proving = r(t) r(t)

r(t)3Recall that

a(t) =d2sdt2 T(t) +

(dsdt

)2N(t)

Taking the cross product of both sides of this equation with Tgives us

T a = d2s

dt2 T T + (

dsdt

)2T N =

(dsdt

)2B

HenceT a =

(dsdt

)2B =

(dsdt

)2Recalling that T = r/r, a = r and ds/dt = r gives us

r(t) r(t)r(t) = r

2 = r(t) r(t)

r(t)3


KEPLERS LAWS OF PLANETARY MOTIONTheorem

1 Each planet follows an elliptical orbit, with the sun at onefocus.

2 The line segment joining the sun to a planet sweeps outequal areas in equal times.

3 If T is the time required for a given planet to make one orbitof the sun and if the length of the major axis of its ellipticalorbit is 2a, then T2 = ka3, for some constant k (i.e., T2 isproportional to a3).Keplers laws are based on a careful analysis of a massivenumber of astronomical observations.Using Newtons second law of motion and Newtons law ofuniversal gravitation and Vector Calculus, one can derivethe Keplers Laws. A detail derivation can be found in thebook.


Parametric Surfaces

In this section, we extend the notion of parametricequations to those with two independent parameters.We will be working with the simple cases of functions oftwo variables, which are developed in more detail inChapter 10.


Parametric Equations with One Variable: An Helix

Consider the helix defined by the parametric equations

x = cos ty = sin t andz = t

This curve winds around the cylinder

x2 + y2 = 1


Parametric Equations with Two Variables: An Cylinder

Suppose that we wanted to obtain parametric equations thatdescribed the entire cylinder

x2 + y2 = 1

we can use the parameters u and v to obtain the correspondingparametric equations

x = cos uy = sin u andz = v

Parametric equations with two independent parameterscorrespond to a two-dimensional surface.


Graphing a Parametric SurfaceExample (6.1)Identify and sketch a graph of the surface defined by theparametric equations x = 2 cos u sin v, y = 2 sin u sin v andz = 2 cos v.

Figure: [9.35] x2 + y2 + z2 = 4. Figure: [9.36] z =

4 x2 y2.


Parametric Equations for Hyperboloids and HyperbolicParaboloids

For parametric equations of hyperboloids and hyperbolicparaboloids, it is convenient to use the hyperbolic functions

cosh x = ex + ex

2

sinh x = ex ex

2

Notice thatcosh2 x sinh2 x = 1


Graphing a Parametric SurfaceExample (6.2)Sketch the surface defined parametrically byx = 2 cos u cosh v, y = 2 sin u cosh v and z = 2 sinh v, 0 u 2piand < v


Parametric Equations in Two Dimensions

In two dimensions, certain curves are more easily described inpolar coordinates than in rectangular coordinates. For example,polar coordinates are essentially the parametric equations forcircles. In particular, the polar coordinates r and are related tox and y by

x = r cos , y = r sin and r =

x2 + y2


Finding a Parametric Representation of a HyperbolicParaboloid

Example (6.3)Find parametric equations for the hyperbolic paraboloidz = x2 y2.


Finding Parametric Representations of SurfacesExample (6.4)Find a parametric representation of each surface:

1 the portion of z =

x2 + y2 inside x2 + y2 = 4 and2 the portion of z = 9 x2 y2 above the xy-plane with y 0.

Figure: [9.39a] Portion ofz =

x2 + y2 inside x2 + y2 = 4.

Figure: [9.39b] Portion ofz = 9 x2 y2 above thexy-plane, with y 0.

Vector-Valued FunctionsThe Calculus of Vector-Valued FunctionsMotion in SpaceCurvatureTangent and Normal VectorsParametric Surfaces

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