APPLICATION:(I) TANGENT LINE

(II) RELATED RATES(III) MINIMUM AND MAXIMUM VALUES

CHAPTER 4

Tangent line

Consider a function , with point lying on the graph:

Tangent line to the function at is the straight line that touches

at that point. Normal line is the

line that is perpendicular to the tangent line.

xfy

11,yx

11,yx

xfy

Tangent Line

Normal Line

0dx

dy

y

x42 xy

mdx

dy

• Tangent Line Equation:

or

• Normal Line Equation:

or

11)( yxxmy

cmxy

11)(1 yxx

my

cxm

y 1

Example 1:1. Find the slope of the curve at the given

points

2. Find the lines that are tangent and normal to the curve at the given point.

)1,2;(2)(

3;12422

2

xyxybxxxya

)3,2;(1)(

2;322

2

yxyxbxxya

• A process of finding a rate at which a quantity changes by relating that quantity to the other quantities.

• The rate is usually with respect to time, t.

RELATED RATES

Example 2Suppose that the radius, r and area, of a circle are differentiable functions of t. Write an equation that relates to .

Answer:

2rA

dtdA

dtdr

dtdrr

dtdA

rdtdA

dtd

rA

2

: wrt t(1) ateDifferenti2

2 1

Example 3How fast is the area of a rectangle changing from one side 10cm long and the side increase at a rate of 2cm/s and the other side is 8cm long and decrease at a rate of 3cm/s?

Solution:

Differentiate (1) wrt t:

x

y

1 :rectangle of Area

/3,8At ./2,10At

xyA

scmdt

dxyscm

dt

dxx

scm

dt

dxy

dt

dyx

dt

dA

xydt

dA

dt

d

/4

283102

Example 4A stone is dropped into a pond, the ripples forming concentric circles which expand. At what rate is the area of one of these circles increasing when the radius is 3m and increasing at the rate of 0.6ms-1?

Solution:

Differentiate wrt t :

r

12 :circle of Area

/6.0,3At

rA

scmdt

drr

sm

dt

drr

dt

Ad

rdt

dA

dt

d

/6.3

6.032

2

2

2

Exercise 1A 13ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5ft/s.

(a) How fast is the top of the ladder sliding down the wall?

(b)At what rate is the area of the triangle formed by the ladder, wall and ground changing

(c) At what rate is the angle between the ladder and the ground changing?

Exercise 2The length l of a rectangle is decreasing at the rate of 2cm/s, while the width w is increasing at the rate 2cm/s. When l=12cm and w=5cm find the rates of change

(a) The area(b)The perimeter

Exercise 3:

When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01cm/min. At what rate is the plate’s area increasing when the radius is 50cm?

• Use 1st derivative to locate and identify extreme values(stationary values) of a continuous function from its derivative

Definition: Absolute Maximum and Absolute Minimum• Let f be a function with domain D. Then f has an

ABSOLUTE MAXIMUM value on D at a point c if:

ABSOLUTE MINIMUM

MAXIMUM & MINIMUM

Dxcfxf ),(

Dxcfxf ),(

• A point on the graph of a function y = f(x) where the rate of change is zero.

Example 6Find stationary points:

0dxdy

STATIONARY POINT

33)2(

3413

2

xxy

xxy

Let f be a function defined on an interval I and let x1 and x2 be any two points in I

1) If f (x1)< f (x2) whenever x1 < x2, then f is said to be increasing on I

2) If f (x1)> f (x2) whenever x1 < x2, then f is said to be decreasing on I

INCREASING & DECREASING

Suppose that f is continuous on [a,b] and differentiable on (a,b).

1) If f’(x)>0 at each point , then f is said to be increasing on [a,b]

2) If f’(x)<0 at each point , then f is said to be decreasing on [a,b]

],[ bax

],[ bax

1st DERIVATIVE TEST

The graph of a differentiable function y=f(x)

1) Concave up on an open interval if f’ is increasing on I

2) Concave down on an open interval if f’ is decreasing on I

CONCAVITY

Let y=f(x) be twice-differentiable on an interval I

1) If f”(x)>0 on I, the graph of f over I is concave up2) If f”(x)<0 on I, the graph of f over I is concave down

2ND DERIVATIVE TEST:TEST FOR CONCAVITY

• If y is minimum

Therefore (x,y) is a minimum point.

• If y is maximum

Therefore (x,y) is a maximum point.

02

2

dxyd

02

2

dxyd

MAXIMUM POINT & MINIMUM POINT

A point where the graph of a function has a tangent line and where the concavity changes is a POINT OF INFLEXION.

02

2

dx

yd

CONCAVITY

Example 5:Find y’ and y” and then sketch the graph of y=f(x)

5823 xxxy

Solution:Step 1: Find the stationary point

Therefore, the stationary points are:

2,3

4

0243

0823

823

0 : valueStationary

2

2

x

xx

xx

xxdx

dydx

dy

17,2&27

41,

3

4

Step 2 : Find inflexion point

Therefore, the inflexion points is:3

1

026

26

0 :valueInflexion

2

2

2

2

x

x

xdx

yddx

yd

27

209,

3

1

Step 3: 1st and 2nd Derivative Test

IntervalTest

Value , x -3 -1 0 2

+ - - +

Increasing Decreasing Decreasing Increasing

- - + +

Concave Concave down

Concave down

Concave up Concave up

2,

dxdy

2

2

dxyd

3

1,2

3

4,

3

1

,

3

4

Step 4: Test for maximum and minimum

At

Therefore, (-2, 17) is a maximum point.

At

Therefore, (4/3, -41/27) is a minimum point.

010

;2

2

2

dx

yd

x

010

;3

4

2

2

dx

yd

x

Example 6:Find y’ and y” and then sketch the graph of y=f(x)

104 34 xxy