## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 13: Plane Curves – Section 13.1.4: Vector Study Of Motion In The Plane

13.1.4
Vector Study Of Motion In The Plane

 1. Plane Vectors

A vector.

### Fig. 1.2

Vectors As Ordered Pairs Of Their Components.

### Fig. 1.3

Magnitudes Of Vectors.

Equal Vectors.

### Fig. 1.5

Addition of 2 vectors where head of first coincides with tail of second.

### Fig. 1.6

Addition of 2 vectors where tail of one coincides with tail of the other.

### Fig. 1.7

Scalar Multiplication Of A Vector.

### Fig. 1.8

Multiplying A Vector By –1.

### Fig. 1.9

Subtraction Of Vectors.

### Fig. 1.10

Any Vector As A Linear Combination Of Standard Basis Vectors.

### Fig. 1.11

 2. Position Vector, Velocity Vector, And Acceleration Vector

### Fig. 2.1

Position Vector.

 The position vector at time t of the object moving in the xy-plane according to the parametric equations x = f(t), y = g(t) is this vector with tail at the origin and head at the object:     The components of the position vector are the corresponding coordinates of the object. The object is at the head of the position vector.

Differentiation Of Vector-Valued Functions

To differentiate a vector-valued function, we differentiate its components. The derivative of a vector-valued function is a
vector-valued function whose components are the derivatives of the corresponding components of the original vector-valued
function. For example:

##### Velocity Vector

The horizontal and vertical components of the velocity vector are the derivatives of the horizontal and vertical components of
the position vector respectively.

The magnitude of the velocity vector of course is:

### Fig. 2.2

Velocity Vector.

 The velocity vector of the object at time t is:

##### Differentiation Of The Dot Product Of Vector Functions

 [2.6]

The product rule for differentiation of the product of normal functions holds for differentiation of the dot product of vector
functions. An easier way to remember Eq. [2.6] is: (u . v)' = u' . v + u . v', just like ( fg)' = f 'g + fg'.

##### Acceleration Vector

The horizontal and vertical components of the acceleration vector are the derivatives of the horizontal and vertical components
of the velocity vector respectively.

The acceleration vector, like the velocity vector, has its tail at the object. Its magnitude of course is:

### Fig. 2.3

Direction Of Acceleration Vector.

 The acceleration vector of the object at time t is:

For the properties of speed, we here give a proof of the increasingness property. The proofs of the decreasingness and
constancy properties are similar to it.

If the speed is increasing then the square of the speed is increasing, then the derivative of the square of the speed is positive,
then:

Given The General Position Vector

The general position vector is the position vector at any time t.

##### Example 2.1

An object moves in the plane such that its position vector at any time t is given by:

Solution

The sketch is shown in Fig. 2.4.

### Fig. 2.4

Sketch For Example 2.2.

EOS

The general position vector is given and we're to find other characteristics of the motion.

Integration Of Vector-Valued Functions

Equal Vectors And The Zero Vector

Of course 2 vectors are equal iff their corresponding components are equal. So:

##### Example 2.2

Solution

EOS

Other characteristics of the motion other than the general position vector are given and we're to find the general position
vector.

# Problems & Solutions

1. Use vectors to show that the triangle with vertices (–1, 1), (2, 5), and (10, –1) is a right triangle.

Solution

Let points A = (–1, 1), B = (2, 5), and C = (10, –1). We have:

Thus triangle ABC is a right triangle.

Note

The graph shows that the sides AB and BC are perpendicular, so we pick them to show that their corresponding vectors are
perpendicular.

Solution

3. The position vector of a point moving in the plane is:

Sketch the path of the motion. Show on the sketch the velocity and acceleration vectors at times t = 1 and t = 2.

Solution

Solution

5. As discussed Section 6.1.7 Part 2, the general solution of the differential equation y'' + k2y = 0, where y is a function of
t and k is a constant, is y = A cos kt + B sin kt, where A and B are arbitrary constants.

Solution