Lesson 14 - Slope, Tangent and Normal Lines, Slides of Calculus for Engineers

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Lesson 3

Slope of a Curve,

Tangent Line and

Normal Line

OBJECTIVES:

At the end of this lesson, the students are

expected to accomplish the following:

• determine the slope of a curve and the

derivative of a function at a specified point;

• solve problems involving slope of a curve;
• determine the equations of tangent and normal

lines using differentiation; and

• solve problems involving tangent and normal

lines.

Consider a point on the curve

that is distinct from and

compute the slope of the secant line

through P and Q.

2 2

Q x f x

y  f (x), P(^ x 1 ,f (x 1 )),

PQ

m

x

f x f x

m PQ 





( ) ( ) 2 1

where :x^  x 2  x 1

x  x  x

and 2 1

x

f x x f x

m PQ 

  



( ) ( ) 1 1

If we let x

approach x

, then the point Q will

move along the curve and approach point P. As

point Q approaches P , the value of Δx

approaches zero and the secant line through P

and Q approaches a limiting position, then we

will consider that position to be the position of

the tangent line at P.

DEFINITION

The derivative of y = f(x) at point P on the curve

is equal to the slope of the tangent line at P , thus

the derivative of the function f given by y= f(x) with

respect to x at any x in its domain is defined as:

0 0

( ) ( )

lim lim x x

dy y f x x f x

dx  ^ x   x

   

 

 

provided the limit exists.

• Examples:

1. Findtheslopeof thecurvey 3 x 2 x 1 at 1 , 6 .

2    

m 6 x 2

y 6 x 2

Solution:

tangent line

'

 

m 6 ^1  2 8

at 1 , 6 :

tan gent line    

 4 , 4 .

8

3. tan 

  at x

Whatistheequationof the gentlineandnormallinetoy

 

     

 

 

x 2 y 12 0 eq'n of TL

2 y 8 x 4

x 4 2

1 y 4

line at 4,-4 is

andthe equationof thetangent

2

1 therefore m

2

1

8

4 y' 4 44 42

8 x 4 x 2

1 y' -

8 x then x

• 8 since y

Solution:

TL

3 2

3

2

3 2

3

2

1

  

  

  



   

  

 

 

 



 

 

2 x y 4 0 eq'n of NL

y 4 2 x 8

y 4 2 x 4

line at 4 , 4 is

andtheequation of thenormal

sin ce NL TL then mNL 2

  

  

  



 

// 8 3 0.

4. tan 2 3

2

  

 

thatis totheline x y

Find theequationof the gentlinetothecurve y x

mL

y 8 x 3

8 x y 3 0

Solution:

y' 4 x m , thus 4 x 8

y 2 x 3

the equation of thegiven curve

andby taking the derivative of

8 x y 3 then m m 8

sincethetangentlineis // to

TL

2

TL L

  

 

   

 

 

 

 

8x- y- 5 0 eq'n of TL

y- 11 8x- 16

y- 11 8 x- 2

tangentlineat 2,11 is

thereforetheequationof the

and y' 4 2 8

x 2 y 8 3 11

4 x 8 and y 2 2 3

2