THE DIFFERENTIALS (Applications)

EXAMPLE 1:  Use differentials to approximate the change in  the area of a square  if  the  length of  its side  increases from 6 cm to 6.23 cm.Let x = length of the side of the square. The area may be expressed as a function of x, where  A= x2. The differential dA is  ( ) dxx2dA dxx'fdA ⋅=⇒⋅=

 Because x is increasing from 6 to 6.23,    you find that                          Δ x = dx = .23 cm; hence, 

( ) ( )2cm76.2dA

cm23.0cm62dA

=

=

( ) .cm 2.8129 is y area in increase

exact the that Note 6.23. to6 from increases length sideits as

cm2.76 ely approximatby increasewill squarethe of area The

2

2

∆

EXAMPLE 2: Use the local linear approximation to estimate      the value of                to the nearest thousandth.

3 55.26

( )

( ) ( )

( )( )

2.9830.0167-3   55.26                                             601

 355.26  therefore  ;327  that  less

601

  tely approxima  be  will  55.26  that  implies  which

0167.0601

10045

271

45.0273

1dy  , Hence

0.45dxx  then  26.55,  to  27  from  decreasing is  x  Because

dx  x3

1    dx  x

31

dy                                       

xxf     dx  xf'dy                                       

 is  dy  aldifferenti  The  27.x

 namely  26.55,  to close  relatively  is  and  cube  perfect  a  is  that  xof    value

 convenient  a  choose  ,xxf is applying  are  you  function  the  Because

3

33

3

3

2

3

23

2

3

1

3

≈≈

−≈=

−=−=−⋅=−⋅=

−==∆

==

=→=

=

=

−

EXAMPLE 3: If y = x3 + 2x2 – 3, find the approximate value of y when x = 2.01.

 2.xof   value original  an  to  0.01dxxof   

 increment  an  applyingof    result  the  as  2.01  gconsiderin

 are  we  then 0.01,22.01  write  weif    that  Note  dy.  y

for  solve  shall  we  then value,  eapproximat  the  find  to

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

===∆

+=+

∆+

( )

( ) ( )

20.1320.013dyy

is  ionapproximat  required  the,therefore

20.001.0812dy

then  ,01.0dx    and    2x      when  and

13388y    then     ,2x     when

dxx4x3dy       then

3x2xy        Since2

23

=+=+

=+===

=−+==+=

−+=

EXAMPLE 4: Use an appropriate local linear approximation                       to estimate the value of cos 310.

( )( )

( )

( )( ) ( )

( ) 8573.0008725.0866.0dyy

is  ionapproximat  required  the,therefore

008725.001745.05.0dy180

130 sindy

then  ,01745.0180

1dx    and    30x      when  and

866.030cosy    then     ,30x     when

dx  x  sindy       then

  x cosy            Let

0

00

0

00

00

=−+=+

−=−=

π•−=

=

π==

===−=

=

EXAMPLE 4: Use an appropriate local linear approximation                       to estimate the value of cos 310.

( )( )

( )

( )( ) ( )

( ) 8573.0008725.0866.0dyy

is  ionapproximat  required  the,therefore

008725.001745.05.0dy180

130 sindy

then  ,01745.0180

1dx    and    30x      when  and

866.030cosy    then     ,30x     when

dx  x  sindy       then

  x cosy            Let

0

00

0

00

00

=−+=+

−=−=

π•−=

=

π==

===−=

=