What is tested is the calculus of parametric equation and vectors. No dot product, no cross product. Books often go directly to 3D vectors and do not have much on 2D vectors; the BC exam only tests 2D vectors. On the BC exam, particles often move in the plane. Their position is defined by two parametric equations or the 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 of these three may be given with initial conditions(s) and you may be asked to find the others. What you should know how to do: Initial value differential equation problems – given the velocity or acceleration vector with initial conditions, find the position and/or velocity Find the speed at time t: speed = length of velocity vector Slope = (dy/dt)/(dx/dt) Use the definite integral for arc length to find the distance traveled - (Integral of speed). Vectors are given in ordered pair form; answers may be in ordered pairs form, using parentheses ( ) or pointed brackets. (From the moderator.)

Vector-Valued Functions 12 Copyright © Cengage Learning. All rights reserved.

Vectors-Motion along a Curve Velocity and Acceleration

Review

Copyright © 2011 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Any vector can be written as a linear combination of two standard unit vectors. The vector v is a linear combination of the vectors i and j. The scalar a is the horizontal component of v and the scalar b is the vertical component of v.

We can describe the position of a moving particle by a vector, r ( t ). If we separate r ( t ) into horizontal and vertical components, we can express r ( t ) as a linear combination of standard unit 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-Valued Functions

A

By letting the parameter t represent time, you can use a vector-valued function to represent motion along a curve. The terminal point of the position vector r(t) coincides with the point (x, y) or (x, y, z) on the curve. The arrowhead on the curve indicates the curve’s orientation by pointing in the direction of increasing values of t. Space Curves and Vector-Valued Functions

Unless stated otherwise, the domain of a vector-valued function r is considered to be the intersection of the domains 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-valued function r(t) = 2cos t i – 3sin t j, 0 ≤ t ≤ 2 . Vector-valued function Solution: From the position vector r(t), you can write the parametric equations x = 2cos t and y = –3sin t. Solving for cos t and sin t and using the identity cos 2 t + sin 2 t = 1 produces Rectangular equation Example 1 – Sketching a Plane Curve

The graph of this rectangular equation is the ellipse shown in Figure 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-Valued Functions

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

Example 1 – Differentiation of Vector-Valued Functions For the vector-valued function given by r(t) = ti + (t 2 + 2)j, find r′(t). Then sketch the plane curve represented by r(t), and the graphs 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 parametric equations x = t and y = t The corresponding rectangular equation is y = x 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) is drawn starting at the terminal point of r(1). Figure 12.9 cont’d y = x i + 3j i + 2j

Differentiation of Vector-Valued Functions

Example 3 – Using Properties of the Derivative For the vector-valued functions given by Find D t [r(t). u(t)] (The derivative of the dot product of the 2 vector valued functions r(t) and u(t).) D t [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-Valued Functions

Integration of vector-valued functions can be done on a component-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

As an object moves along a curve in the plane, the coordinates x and y of its center of mass are each functions of time t. Rather than using the letters f and g to represent these two functions, 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

The velocity vector is the tangent vector to the curve at point P. The magnitude of the velocity vector r'(t) gives the speed of the object at time t.

Velocity and Acceleration The arc length of the curve gives the distance traveled over an interval of time.

Find the velocity vector, speed, and acceleration vector of a particle 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 Plane Curve

The acceleration vector is Example 1 – Solution Note that the velocity and acceleration vectors are orthogonal at any point in time. This is characteristic of motion at 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 at the point (1,2,0) and moves with an acceleration of. Find the position of the object after 2 seconds.

Example 5 (you try) : Use the given acceleration function to find the velocity and position vectors. Then find the position at 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: