What is Calculus ?

alculus is the mathematics for motion and change . Calculus was first invented to meet the mathematical needs of the scientists of the sixteenth and seventeenth centuries . There are two primary branches of calculus : C

Differential Calculus – dealt with the problem of calculating rate of change .

Calculus Integral Calculus – dealt with the problem of determining a function from information about its rate of change .

Calculus is a very important branch of mathematics and has been applied widely in the scientific research and commercial fields . Examples : An architect will use integration to determined the amount of materials necessary to construct a curved dome over a sports area , as well as calculate the weight of dome and determine the type of support structure required , biologists use differential calculus to determine the exact rate of growth in a bacterial culture when different variables such as temperature and food source are changed . This research can help increase the rate of growth of necessary bacteria , or decrease the rate of growth for harmful and potentially threatening bacteria . A graphics artist uses calculus to determine how different three-dimensional models will behave when subjected to rapidly changing conditions . This can create a realistic environment for movies or video games. Doctors can use calculus to help build the discipline necessary for solving complex problems , such as diagnosing patients . The above examples show that calculus has widespread application in science , economics , engineering and can solve many problems for which algebra alone is insufficient .

1. Introduction To Limits To approach the subject of calculus , we examine the idea of limits .

We want to study the behaviour of f(x) when x is near a . In other words , when x is close to a ,what happens to f(x) ?

Page 1 of 16

Suppose f(x) is a real-valued function and a is a real number . The expression

means that f(x) can be made to be as close to L as desired by making

sufficiently close to a . Note that this statement can be true even if f(a) L . Indeed , the function f(x) need not even be defined at a .

Reading Material

1.1 The formal definition of limit(Optional)

Definition :

For every > 0 , there exists a corresponding number > 0 such that for all ,

0 < | - c | < | f ( x ) - L | <

* The proven of limit is not included in our syllabus !

Example 1 :

(a) Find . (b)

[solution:]

Page 2 of 16

(c)

Page 3 of 16

is the value of f(x) when x approaches the value of a

1.2 Determine t he value of a function when its variable approaches a certain value

Page 4 of 16

Example 2:

Find the limit for each question below

(a) lim (0.2) x =x→0 (b) (c) (d)

Example 3 :

(a)

=

(b) (c)

(d) (e) (f)

Page 5 of 16

Note: x a

2. Chords , Normals , Tangent Consider any two points , A and B , on any curve .

The line joining A and B is called a chord .The line that touches the curve at A is called the tangent at A .The point A (touch point) is called the point of contact.

The line perpendicular to the tangent at A is called the normal at A. Figure 1

The Gradient Of A CurveFrom figure 1 , since the curve is not a straight line , and hence its gradient is not constant . In fact , the gradient of a curve changes continuously from point to point . So , what is the gradient of a curve ?The gradient of a curve at a point A is defined as the gradient of the tangent drawn at the point A .

2.1 Find the grad i ent of a ch o rd joining two p oints on a curve

Example 4 : Determine the gradient of the chord AB as shown below

(a) yy=2x2

(b) yy=x2+1

(c) yy = x 2 − 4

32-------------- B. 10--------------- -F 5-------------------- B

8------- A

0 2 4

2----------Ex

0 1 3 -4

From the definition , the gradient of a curve at any point is found by drawing a tangent to the curve at the particular point , but it is just a approximate values . To obtain more accurate value , the following method is needed .

Page 6 of 16

A

x

A

B

2.2 Finding the first derivative of a function y = f(x ), as the gradient of tangent to its graph.

y = x2

y

P(3 9)

Q (3.05 , 9.30)

Tangent at P

Complete the table below and hence determine the first derivative of the function y = x2

Point Q (x2, y2) x2 -x1 y2 - y1

Gradient PQ

x2 y2

3.05 9.3025 0.05 0.3025 6.053.01 9.0601 0.01 0.0601 6.013.001 9.006001 0.001 0.006001 6.0013.0001 9.0006 0.0001 0.0006 6.0001

When Q P , gradient PQ 6, then the gradient at point P = 6

In general ,

Let P(x,y) be a point on the curve y = x2 .

Consider a point Q on the curve whose coordinate is

( )

Since P(x,y) is on the curve , then y = x2 -------------- (1)

Since Q( ) lies on the curve , then

----------- (2)

As Q approaches P , becomes smaller and smaller , ie. 0 . The gradient of PQ thus tend to the gradient of the tangent at P .

Hence , as QP , gradient of PQ =

Page 7 of 16

The first principles of differentiation

2.3 How to find the first derivative of polynomia l s u s ing the first pri n cipl e s dy

a) Let δ x be a small increment in x and δ y be the corresponding small increment in y.

b) Substitute x and y in the equation y = f(x) with x + δ x and y + δ y respectively

c) Express δ y in terms of x and δ x

d) Find and .

E x ample 5 :

1. Given y = 3x2

+ 5,find

Solution

dy by using the first principle

dx2. Determine the first derivative of y =

first principle .

Solution :

4 − 3 by using

x

Page 6 of 16

3. Differentiate the following functions from first principles :

(i) y = x3 – 2x

3. What is Differentiation ?

3.1 Determining the first derivative of the function y = ax n

using formula

Page 7 of 16

where Important :

We call (pronounced “dee y by dee x “) the gradient function ,

derived function or differential coefficient of y with respect to x .

The process of finding the dy/dx is called differentiation .

where Important :

We call (pronounced “dee y by dee x “) the gradient function ,

derived function or differential coefficient of y with respect to x .

The process of finding the dy/dx is called differentiation .

dy d1. If y = k, where k is a constant then or = (k ) = 0

dx dx2. If y = axn where k is a constant and n is positive and negative integer

dy d nthen = (ax ) = nax n-1

dx dx3. If f(x) = ax

n, then f ′( x) = anx n −1

. Notation f ′( x) is read as f prime x

Example 6: Differentiate the following with respect to x :

(a) y = (b) y = (x-1)( ) (c) y =

(d) y = (e) y= (f) y =

Example 7 :Find the coordinates of the points on the curve y = 2x3 – 4x2 + x + 1 at which the gradient is -1 .[Solution] y = 2x3 – 4x2 + x + 1

As gradient = -1 , then

When x = 1 , y = 2(1)3 – 4(1)2 + 1 +1 = 0 .

Page 8 of 16

When x = the required points are ( ) and ( 1, 0).

Example 8: Given that the curve y = has gradient 4 at the point A(1,5) , calculate the value of a and of b .

Example 9: Find the gradient of the curve y = at the point where the curve crosses the x-axis .

Example 10: Find the coordinates of the points on the curve y = where the tangent makes an angle of 450 with the positive direction of the x-axis .

Page 9 of 16

dya) represents the gradient of the tangent of the curve y = f(x)

dxb) The gradient of the tangent at a point A (p,q) on a curve y = f(x) can be determined by

dy substituting x = p

dx

Tangents

c) The equation of the tangent at point P(x1,y1 ) on the curve y =f(x) can be determined by Finding the gradient, m, of the tangent at point P Use the formula y – y1 = m (x – x1) to find the equation of the tangent.

Example 11 :Find the coordinates of the point on the curve y = 2x –x2 at which the tangent is perpendicular to theline y-2x= 3 .

Example 12:Find the equations of the tangents to the curve y=x2-3x-4 at its points of intersection with the x-axis , and find the co-ordinates of the point of intersection of these tangents ?

Example 13:The tangent at the point A on the curve y = x(x-1)(x-2) whose x –coordinate is equal to p meets the axis of y in T .Prove the length OT = 3p2 – 2p3 where O is origin .

Page 9 of 16

The equation of the normal to the curve y = f(x) at point P( x1 ,y1), can be determined by

Finding the gradient ,m1,of the tangent at point P, Finding the gradient, m2, of the normal at point P for which m1m2 = -1 using the formula y – y1 = m2 (x – x1) to find the equation of the normal

Example 14:

Find the equation of the tangent to the curve y = , where c is a constant , at the point ( ) . If the tangent

meets the axes of x and y in A and B , prove that the triangle AOB is of constant area , where O is origin.

Normal To the a curve

Example 15 : Find the equation of the normal for each of following equations and the corresponding points.

(a) y = x2- 4x + 1 at point (3,-2) (b) y = 2x + at point (1,10)

Example 16:Find the equation of the normal to the curve y = 8x – 3x2 at the point x = 2 and find where this normal meets the line x = 2y + 1 .

Page 10 of 16

Example 17:

P is the point (4,7) on the curve y = x2 -6x + 15 . Find the gradient of the curve at P and the equation of the normal at this point . Find the coordinates of the point where this normal cuts the curve again. The tangent at another pointQ is perpendicular to the tangent at P . Find the x-coordinate of Q . [ Ans : 2 ; -1/2 ; x+2y-18=0 ; (3/2 , 33/4 ) ; 11/4 ]

Example 18:

The tangent to the curve y=ax2+bx + 2 at (1, ) is parallel to the normal to the curve y = x2 + 6x + 10 at (-2,2) . Find the

values of a and b .

Example 19:

The tangent at the point P(a,b) on the curve y = meets the x-axis and y-axis at points Q and R respectively. Show that

PQ = RP .

Page 11 of 16

Miscellaneous Exercise

Problem 1: Find by using first principle of differentiation if (i) y = (ii)

Problem 2: Differentiate the following :

a) y = x4 + mx3 + n , where m,n are constant d)

b) e)

c) f) y =

Problem 3: Find the gradient of the curve y = at the point whose abscissa is 2 . Also find the coordinate of the

points where the tangent is horizontal . [Ans: 35/4 ; (1/3,6) ; (-1/3,-6) ]

Problem 4 : Given that the gradient of the tangent to the curve y = ax3 + bx2 + 3 at the point (1,4) is 7 , calculate the value of a and of b . [Ans: a=5 , b= -4 ]

Problem 5: Calculate the equation of tangent and normal to the curve at x = -1.

[Ans: y = 5x + 5 ; x + 5y +1 = 0 ] Problem 6: A curve is defined by the equation .

(a) Find the equations of the tangents to the curve which are perpendicular to the line (b) Find the equation of the normal to the curve at the point where the normal is given by the equation . [Ans; (a) y = 2x+4; y = 2x (b) 18x + 27y + 20 =0]

Problem 7: The line y = 5x + c is a tangent to the curve y = 7x2 – 5x + 3. (a) Find the value of the constant c. (b) If the tangent meets the x-axis at A and the y-axis at B, calculate the area of triangle

OAB, where O is the origin. [Ans: (a) (b) units2]

Problem 8: Show that if the line y = mx + c is a tangent to the curve y = x2 + x + 4, then m2 -2m + 4c – 15 = 0. Hence, find the equation(s) of the tangent to the curve which passes through the point (0, 3). [Ans: y = 3x + 3 ; y + x = 3]

Problem 9: The normal at the point P(12,4) on the curve xy = 48 meets the curve again in Q , find the length of PQ.

[Ans: units]

Problem 10: Find the equations of the normal to the curve y = at x =1 , x = 4 . Find the coordinates of the point

where these normal intersect . [Ans : x-6y+59=0 ; 2x+3y-38=0; (17/5 , 52/5 ) ]

Page 12 of 16