Calculus Of One Real Variable – By Pheng Kim
Ving


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1. Plane Vectors 

Fig. 1.1 A vector. 

Fig. 1.2 Vectors As Ordered Pairs Of Their Components. 

Fig. 1.3 Magnitudes Of Vectors. 

Fig. 1.4 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 
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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 xyplane
according to the parametric equations x = f(t), y = g(t) The components of the position vector are the
corresponding coordinates of the object. The object is at the head of the 
Differentiation
Of VectorValued Functions
To differentiate a vectorvalued function, we differentiate
its components. The derivative of a vectorvalued function is a
vectorvalued function whose components are the derivatives of the
corresponding components of the original vectorvalued
function. For example:
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: 

[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'.
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.
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
VectorValued Functions
Equal Vectors
And The Zero Vector
Of course 2 vectors are equal iff their corresponding components are equal. So:
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'' + k^{2}y = 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
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