Calculus 10 Extra Topic

•What is tested is the calculus of parametric equation and vectors.No dot product, no cross product. Books often go directly to 3Dvectors and do not have much on 2D vectors; the BC exam onlytests 2D vectors. On the BC exam, particles often move in theplane. Their position is defined by two parametric equations orthe equivalent vector (x(t),y(t)) . The velocity is the vector (x'(t),y'(t)) and the acceleration is the vector (x''(t), y''(t)). Any ofthese three may be given with initial conditions(s) and you may beasked to find the others. What you should know how to do: Initialvalue differential equation problems – given the velocity oracceleration vector with initial conditions, find the position and/orvelocity Find the speed at time t: speed = length of velocity vectorSlope = (dy/dt)/(dx/dt) Use the definite integral for arc length tofind the distance traveled - (Integral of speed). Vectors are given inordered pair form; answers may be in ordered pairs form, usingparentheses ( ) or pointed brackets < >. (From the moderator.)

Vector-Valued Functions

Vectors-Motion along a Curve

Velocity and Acceleration

12.112.212.3

Review

Any vector can be written as a linearcombination of two standard unit vectors.

The vector v is a linear combinationof the vectors i and j.

The scalar a is the horizontalcomponent of v and the scalar b isthe vertical component of v.

We can describe the position of a moving particle by avector, r(t).

If we separate r(t) into horizontal and vertical components,we can express r(t) as a linear combination of standardunit vectors i and j.

In three dimensions the component form becomes:

Is the position vector at any

time t. It’s component form is:

or:

Space Curves and Vector-ValuedFunctions

Space Curves and Vector-Valued Functions

By letting the parameter t represent time, you can use avector-valued function to represent motion along a curve.

The terminal point of the position vector r(t) coincides withthe point (x, y) or (x, y, z) on the curve. The arrowhead on thecurve indicates the curve’s orientation by pointing in thedirection of increasing values of t.

Space Curves and Vector-Valued Functions

Unless stated otherwise, the domain of a vector-valuedfunction r is considered to be the intersection of thedomains of the component functions f, g, and h.

For instance, the domain of

is the interval (0, 1].

Space Curves and Vector-Valued Functions

Sketch the plane curve represented by the vector-valuedfunction

r(t) = 2cos t i – 3sin t j, 0 ≤ t ≤ 2. Vector-valued function

Solution:

From the position vector r(t), you can write theparametric equations x = 2cos t and y = –3sin t.

Solving for cos t and sin t and using the identity

cos2 t + sin2 t = 1 produces

Rectangular equation

Example 1 – Sketching a Plane Curve

The graph of this rectangular equation is the ellipseshown in Figure 12.2.

The curve has a clockwise orientation.

That is, as t increases from 0 to 2,

the position vector r(t) moves clockwise,

and its terminal point traces the ellipse.

cont’d

Example 1 – Solution

Differentiation of Vector-ValuedFunctions

Differentiation of Vector-Valued Functions

Differentiation of vector-valued functions can be doneon a component-by-component basis.

Example 1 – Differentiation of Vector-Valued Functions

For the vector-valued function given by r(t) = ti + (t2 + 2)j, findr′(t). Then sketch the plane curve represented by r(t), and thegraphs of r(1) and r′(1).

Solution:

Differentiate on a component-by-component basis to obtain

r′(t) = i + 2t j.Derivative

From the position vector r(t), you can write the parametricequations x = t and y = t2 + 2.

The corresponding rectangular equation is y = x2 + 2.

When t = 1, r(1) = i + 3j and r′(1) = i + 2j.

Example 1 – Solution

In Figure 12.9, r(1) is drawn starting at the origin, and r′(1) isdrawn starting at the terminal point of r(1).

Figure 12.9

cont’d

y = x2 + 2.

i + 3j

i + 2j

Differentiation of Vector-Valued Functions

Example 3 – Using Properties of the Derivative

For the vector-valued functions given by

Find Dt[r(t) . u(t)] (The derivative of the dot product of the 2vector valued functions r(t) and u(t).)

Dt[r(t) . u(t)]= r(t) . u'(t) + r '(t) . u(t)

Example 3 – Solution

Because and u′(t) = 2ti – 2j, you have

The dot product is a real-valued function, not a vector-valued function.

Integration of Vector-ValuedFunctions

Integration of Vector-Valued Functions

Integration of vector-valued functions can be done on acomponent-by-component basis.

Example 1 – Integrating a Vector-Valued Function

Find the indefinite integral ∫(t i + 3j) dt.

Solution:

Integrating on a component-by-component basis produces

Example 2– Integrating a Vector-Valued Function

Evaluate the integral:

Example 3 – Integrating a Vector-Valued Function

Find the anti-derivative of :

that satisfies the initial condition:

Example 7 – You try:

Find v(t) given the following conditions:

Velocity and Acceleration

Velocity and Acceleration

As an object moves along a curve in the plane, the coordinates xand y of its center of mass are each functions of time t.

Rather than using the letters f and g to represent these twofunctions, it is convenient to write x = x(t) and y = y(t).

So, the position vector r(t) takes the form

r(t) = x(t)i + y(t)j.

If x and y are twice differentiable functions of t, and r is a vector-valued function given by r(t) = x(t)i + y(t)j then,

Velocity = v(t) = r'(t) = x'(t)i + y'(t)j

Acceleration = a(t) = r''(t) = x''(t)i + y''(t)j

Speed =

Velocity and Acceleration

For motion along a space curve, the definitions are similar.

That is, if r(t) = x(t)i + y(t)j + z(t)k, you have

Velocity = v(t) = r'(t) = x'(t)i + y'(t)j + z'(t)k

Acceleration = a(t) = r''(t) = x''(t)i + y''(t)j + z''(t)k

Speed =

Velocity and Acceleration

Velocity and Acceleration

The velocity vector is the tangent vector to the curve atpoint P.

The magnitude of the velocity vector r'(t) gives thespeed of the object at time t.

Velocity and Acceleration

The arc length of the curve gives the distance traveledover an interval of time.

Find the velocity vector, speed, and acceleration vector of aparticle that moves along the plane curve C described by

Solution:

The velocity vector is

The speed (at any time) is

Example 1 – Finding Velocity and Acceleration Along a PlaneCurve

The acceleration vector is

Example 1 – Solution

Note that the velocity and accelerationvectors are orthogonal at any point intime. This is characteristic of motionat a constant speed.

Example 1, con’t

By eliminating the parameter, you can

obtain the rectangular equation for r(t).

Because the velocity vector

has a constant magnitude,

but a changing direction as t increases, the particle moves

around the circle at a constant speed.

Example 2 (you try):

a) Find the velocity and acceleration vectors.

b) Find the velocity, acceleration, speed and direction

of motion at .

Example 2:

b) Find the velocity, acceleration, speed and direction

of motion at .

velocity:

acceleration:

Example 2:

b) Find the velocity, acceleration, speed and direction

of motion at .

speed:

direction:

Example 3:

a) Write the equation of the line tangent where .

At :

position:

slope:

Tangent line:

The horizontal component of the velocity is .

Example 3:

b) Find the coordinates of each point on the path where

the horizontal component of the velocity is 0.

Example 4:

An object starts from rest atthe point (1,2,0) and

moves with an acceleration of .

Find the position of the objectafter 2 seconds.

Example 5 (you try) :

Use the given acceleration function to find thevelocity and position vectors. Then find the positionat time t = 2.

Homework:

Section 12.1 through 12.3

Day 1: Vector Module Day 4 WS

Day 2: Vector Module Day 5 WS

Day 3: Vector Module Day 5 WS

HWQ 3/19/15

Find v(t) given the following conditions: