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Lesson 8

Differentials and

Parametric Equations

OBJECTIVES:

At the end of the lesson, the student should be able

to:

DIFFERENTIAL

FORMULAS

EXAMPLE 1: Find dy for y = x

3

+ 5 x −1.

( )

dy ( 3x^ 5 ) dx

3 x dx 5 dx

dy d x 5 x 1

2

2

3

2 2

2

3 x 1

2dx dy 3 x 1

6 x 2 6 x dy

3 x 1

3 x 1 2 2 x 3

3 x 1

2 x dy d

−

− ⇒∴ = 

 

 

 



−

− −

−

− −

 

  



−

=

multiply itby dx.

oftherightmemberoftheequationand

Note:Inpractice,wesimplygetthederivative

EXAMPLE 4: Use the local linear approximation to estimate

the value of to the nearest thousandth.

_3

26. 55_

( )

( ) ( )

( )

( )

26. 55 3 - 0.0167 2.

60

1 less that 27 3 ; therefore 26. 55 3

60

1 which implies that 26. 55 willbe approximately

1. 0167 60

1

100

45

27

1

1. 45 327

1 Hence, dy

Because x isdecreasing from 27 to26.55, then x dx 0.

dx 3 x

1 x dx 3

1 dy

dy f' x dx f x x

x 27. Thedifferentialdy is

value of x that is aperfect cube andisrelatively closeto26.55,namely

Because the function youare applyingisf x x, choose a convenient

3

3 3

3

3

2

3

2

3

2

3

1

3

≈ ≈

= ≈ −

= ⋅ − = ⋅− =− =−

∆ = =−

= =

= → =

=

=

−

EXAMPLE 5: If y = x

3

+ 2x

2

• 3, find the approximate value of

y when x = 2.01.

of x dx 0.01 to an originalvalueof x 2.

considering 2.01 as the result of applying an increment

y dy. Note that if we write 2.01 2 0.01,then we are

to find the approximate value,then we shall solve for

The exact value is y y but since we are simply asked

( )

y dy 13 0. 20 13. 20

therefore,the required approximation is

dy 12 8 0. 01 0. 20

and when x 2 and dx 0. 01 , then

when x 2 , then y 8 8 3 13

then dy 3 x 4 x dx

Since y x 2 x 3

2

3 2

• = + =

= + =

= =

= = + − =

= +

= + −

Derivative of Parametric Equations

2

2 3 2

dx

d y

2. If x=t − 1 , y=t +t, find

2 3 t

2 t 1

dt

dx

dt

dy

dx

dy + = =

( ) 2 5

2

4 2

2

2

2

2 2

2

9 t

2 t 1

dx

d y

3 t

1

9 t

3 t 2 2 t 1 6 t

dx

d y

dx

dt

3 t

2 t 1

dt

d

dx

d y

− + ∴ =

− +

 

  

 +

2

2

dx

d y

3. If x= 2 sinθ, y= 1 − 4 cosθ, find

= θ θ

θ

θ

θ = 2 tan 2 cos

4 sin

d

dx

d

dy

dx

dy

( )

∴ = θ

= θ• θ

θ

= θ•

θ θ θ

=

3 2

2

2 2

2

2 2

2

2

2

sec dx

d y

sec sec dx

d y

2 cos

1 2 sec dx

d y

dx

d 2 tan d

d

dx

d y