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Section	2.6
                                  Implicit	Differentiation

                                  V63.0121.006/016, Calculus	I



                                       February	23, 2010


        Announcements
                 Quiz	2	is	February	26, covering	§§1.5–2.3
                 Midterm	is	March	4, covering	§§1.1–2.5
                 On	HW 5, Problem	2.3.46	should	be	2.4.46

.       .
Image	credit: Telstar	Logistics
                                                           .     .   .   .   .   .
Announcements	on	white	background




  Announcements
     Quiz	2	is	February	26, covering	§§1.5–2.3
     Midterm	is	March	4, covering	§§1.1–2.5
     On	HW 5, Problem	2.3.46	should	be	2.4.46




                                           .     .   .   .   .   .
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogonal	Trajectories
     Chemistry


  The	power	rule	for	rational	powers




                                        .   .   .   .   .   .
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                           .                   x
                                                               .
            2     2
           x +y =1

at	the	point (3/5, −4/5).




                                   .   .       .   .   .   .
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                           .                   x
                                                               .
            2     2
           x +y =1

at	the	point (3/5, −4/5).




                                   .   .       .   .   .   .
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                           .                       x
                                                                   .
            2     2
           x +y =1

at	the	point (3/5, −4/5).                              .




                                   .   .       .   .       .   .
Motivating	Example                                  y
                                                    .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                                    .                       x
                                                                            .
            2     2
           x +y =1

at	the	point (3/5, −4/5).                                       .
   Solution	(Explicit)
                                     √
        Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why	the −?)




                                            .   .       .   .       .   .
Motivating	Example                                  y
                                                    .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                                    .                       x
                                                                            .
            2     2
           x +y =1

at	the	point (3/5, −4/5).                                       .
   Solution	(Explicit)
                                      √
        Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why	the −?)
                       dy       −2x        x
        Differentiate:    =− √        =√
                       dx     2 1−x 2    1 − x2




                                            .   .       .   .       .   .
Motivating	Example                                  y
                                                    .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                                    .                       x
                                                                            .
            2     2
           x +y =1

at	the	point (3/5, −4/5).                                       .
   Solution	(Explicit)
                                         √
        Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why	the −?)
                       dy       −2x             x
        Differentiate:    =− √           =√
                       dx     2 1−x    2      1 − x2
                   dy             3 /5          3/5   3
        Evaluate:           =√                =      = .
                   dx x=3/5     1 − ( 3 /5 )2   4/5   4


                                            .   .       .   .       .   .
Motivating	Example                                  y
                                                    .


Problem
Find	the	slope	of	the	line	which
is	tangent	to	the	curve
                                                    .                       x
                                                                            .
            2     2
           x +y =1

at	the	point (3/5, −4/5).                                       .
   Solution	(Explicit)
                                         √
        Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why	the −?)
                       dy       −2x             x
        Differentiate:    =− √           =√
                       dx     2 1−x    2      1 − x2
                   dy             3 /5          3/5   3
        Evaluate:           =√                =      = .
                   dx x=3/5     1 − ( 3 /5 )2   4/5   4


                                            .   .       .   .       .   .
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
      Suppose	we	had y = f(x), so	that

                            x2 + (f(x))2 = 1




                                               .   .   .   .   .   .
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
      Suppose	we	had y = f(x), so	that

                            x2 + (f(x))2 = 1


      We	could	differentiate	this	equation	to	get

                          2x + 2f(x) · f′ (x) = 0




                                               .    .   .   .   .   .
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
      Suppose	we	had y = f(x), so	that

                            x2 + (f(x))2 = 1


      We	could	differentiate	this	equation	to	get

                          2x + 2f(x) · f′ (x) = 0


      We	could	then	solve	to	get
                                              x
                              f′ (x ) = −
                                            f(x)

                                                   .   .   .   .   .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                                          .                       x
                                                                                  .


                                                                      .




                                                 .    .       .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                                          .                       x
                                                                                  .


                                                                      .




                                                 .    .       .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                                 y
                                                            .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                                            .                       x
                                                                                    .


                                                                        .

                                                          l
                                                          .ooks	like	a	function


                                                 .    .         .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the                                         .
      graph	of	a	function.


                                                          .                       x
                                                                                  .




                                                 .    .       .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the                                         .
      graph	of	a	function.


                                                          .                       x
                                                                                  .




                                                 .    .       .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                                 y
                                                            .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the                                           .
      graph	of	a	function.
                                                          l
                                                          .ooks	like	a	function
                                                            .                       x
                                                                                    .




                                                 .    .         .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                             .            .                   x
                                                                              .




                                                 .    .       .   .   .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                               y
                                                          .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                             .            .                   x
                                                                              .




                                                 .    .       .   .   .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                                 y
                                                            .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.


                                              .             .                   x
                                                                                .
                                .
                                    does not look like a
                                     function, but that’s
                                    OK—there are only
                                    two points like this


                                                  .    .        .   .   .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                                 y
                                                            .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.
      So f(x) is defined
      “locally”, almost
                                                            .                       x
                                                                                    .
      everywhere and	is
      differentiable

                                                                        .

                                                          l
                                                          .ooks	like	a	function


                                                 .    .         .   .       .   .
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                                 y
                                                            .
      the	curve x2 + y2 = 1,
      the	curve	resembles	the
      graph	of	a	function.
      So f(x) is defined
      “locally”, almost
                                                            .                       x
                                                                                    .
      everywhere and	is
      differentiable
      The	chain	rule	then
                                                                        .
      applies	for	this	local
      choice.
                                                          l
                                                          .ooks	like	a	function


                                                 .    .         .   .       .   .
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
   x2 + y2 = 1 at	the	point (3/5, −4/5).




                                                .    .   .    .   .   .
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
   x2 + y2 = 1 at	the	point (3/5, −4/5).

   Solution
                                dy
       Differentiate: 2x + 2y      =0
                                dx




                                                .    .   .    .   .   .
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
   x2 + y2 = 1 at	the	point (3/5, −4/5).

   Solution
                          dy
       Differentiate: 2x + 2y =0
                          dx
       Remember y is	assumed	to	be	a	function	of x!




                                                .    .   .    .   .   .
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
   x2 + y2 = 1 at	the	point (3/5, −4/5).

   Solution
                          dy
       Differentiate: 2x + 2y =0
                          dx
       Remember y is	assumed	to	be	a	function	of x!
                dy    x
       Isolate:    =− .
                dx    y




                                                .    .   .    .   .   .
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
   x2 + y2 = 1 at	the	point (3/5, −4/5).

   Solution
                                dy
       Differentiate: 2x + 2y      =0
                                dx
       Remember y is	assumed	to	be	a	function	of x!
                dy       x
       Isolate:    =− .
                dx       y
                  dy             3 /5  3
       Evaluate:               =      = .
                  dx ( 3 ,− 4 ) 4/5    4
                       5   5




                                                .    .   .    .   .   .
Summary


  If	a	relation	is	given	between x and y which	isn’t	a	function:
   “Most	of	the	time”, i.e., “at
   most	places” y can	be                           y
                                                   .
   assumed	to	be	a	function	of                                 .
   x
   we	may	differentiate	the                        .                           x
                                                                               .
   relation	as	is
                dy
   Solving	for     does	give	the
                dx
   slope	of	the	tangent	line	to
   the	curve	at	a	point	on	the
   curve.



                                               .       .   .       .   .   .
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogonal	Trajectories
     Chemistry


  The	power	rule	for	rational	powers




                                        .   .   .   .   .   .
Example
Find y′ along	the	curve y3 + 4xy = x2 + 3.




                                             .   .   .   .   .   .
Example
Find y′ along	the	curve y3 + 4xy = x2 + 3.

Solution
Implicitly	differentiating, we	have

                   3y2 y′ + 4(1 · y + x · y′ ) = 2x




                                               .      .   .   .   .   .
Example
Find y′ along	the	curve y3 + 4xy = x2 + 3.

Solution
Implicitly	differentiating, we	have

                       3y2 y′ + 4(1 · y + x · y′ ) = 2x

Solving	for y′ gives

                         3y2 y′ + 4xy′ = 2x − 4y
                         (3y2 + 4x)y′ = 2x − 4y
                                         2x − 4y
                              =⇒ y′ = 2
                                        3y + 4x



                                                   .      .   .   .   .   .
Example
Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).




                                          .   .   .   .   .   .
Example
Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).

Solution
Differentiating	implicitly:

      5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x)

Collect	all	terms	with y′ on	one	side	and	all	terms	without y′ on
the	other:

        5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 )

Now	factor	and	divide:
                                2xy(cos x2 − y2 )
                      y′ =
                              5y4 + 3x2 y2 − sin x2


                                                 .    .    .    .     .   .
Example
Find	the	equation	of	the	line
tangent	to	the	curve
                                   .
  y 2 = x 2 (x + 1 ) = x 3 + x 2

at	the	point (3, −6).
                                                       .




                                       .   .   .   .   .   .
Example
Find	the	equation	of	the	line
tangent	to	the	curve
                                                 .
  y 2 = x 2 (x + 1 ) = x 3 + x 2

at	the	point (3, −6).
                                                                     .
 Solution
 Differentiating	the	expression	implicitly	with	respect	to x gives
    dy                  dy    3x2 + 2x
 2y    = 3x2 + 2x, so      =           , and
    dx                  dx       2y

               dy                3 · 32 + 2 · 3    33   11
                             =                  =−    =− .
               dx   (3,−6)           2(−6)         12   4



                                                     .   .   .   .   .   .
Example
Find	the	equation	of	the	line
tangent	to	the	curve
                                                 .
  y 2 = x 2 (x + 1 ) = x 3 + x 2

at	the	point (3, −6).
                                                                           .
 Solution
 Differentiating	the	expression	implicitly	with	respect	to x gives
    dy                  dy    3x2 + 2x
 2y    = 3x2 + 2x, so      =           , and
    dx                  dx       2y

               dy                3 · 32 + 2 · 3    33   11
                             =                  =−    =− .
               dx   (3,−6)           2(−6)         12   4

                                                             11
 Thus	the	equation	of	the	tangent	line	is y + 6 = −             (x − 3).
                                                             4
                                                     .   .      .    .     .   .
Line	equation	forms



      slope-intercept	form

                                y = mx + b

      where	the	slope	is m and (0, b) is	on	the	line.
      point-slope	form

                             y − y0 = m(x − x0 )

      where	the	slope	is m and (x0 , y0 ) is	on	the	line.




                                                .    .      .   .   .   .
Example
Find	the	horizontal	tangent	lines	to	the	same	curve: y2 = x3 + x2




                                            .    .   .    .   .     .
Example
Find	the	horizontal	tangent	lines	to	the	same	curve: y2 = x3 + x2

Solution
We	have	to	solve	these	two	equations:


                                         .
     .     2     3     2                     3x2 + 2x
         y = x +x                                      = 0
    1
    .     [(x, y). is on
           the curve]
                                        2
                                        .
                                                2y
                                             [tangent line
                                             is horizontal]




                                             .    .   .       .   .   .
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
             2y

      (as	long	as y ̸= 0). So x = 0 or 3x + 2 = 0.




                                               .     .   .   .   .   .
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
             2y

      (as	long	as y ̸= 0). So x = 0 or 3x + 2 = 0.
      Substituting x = 0 into	the first equation	gives

                      y2 = 03 + 02 = 0 =⇒ y = 0

      which	we’ve	disallowed. So	no	horizontal	tangents	down
      that	road.




                                               .     .   .   .   .   .
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
             2y

      (as	long	as y ̸= 0). So x = 0 or 3x + 2 = 0.
      Substituting x = 0 into	the first equation	gives

                      y2 = 03 + 02 = 0 =⇒ y = 0

      which	we’ve	disallowed. So	no	horizontal	tangents	down
      that	road.
      Substituting x = −2/3 into	the	first	equation	gives
               (    )  (    )
             2    2 3     2 2    4          2
            y = −     + −     =    =⇒ y = ± √ ,
                  3       3     27         3 3
      so	there	are	two	horizontal	tangents.
                                               .     .   .   .   .   .
Horizontal	Tangents




    (           )
    . − 2 , 3√3
        3
             2

           .
                      .
           .
   (              )
   . − 2 , − 3 √3
       3
               2




                          .   .   .   .   .   .
Horizontal	Tangents




    (           )
    . − 2 , 3√3
        3
             2

           .
                      .
           .
   (              )
   . − 2 , − 3 √3
       3
               2          n
                          . ode




                                  .   .   .   .   .   .
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2




                                             .   .    .   .       .   .
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2
Solution
                                      dx
    Tangent	lines	are	vertical	when      = 0.
                                      dy




                                             .   .    .   .       .   .
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2
Solution
                                       dx
    Tangent	lines	are	vertical	when       = 0.
                                       dy
    Differentiating x implicitly	as	a	function	of y gives
              dx       dx      dx        2y
    2y = 3x2      + 2x , so        = 2         (notice	this	is	the
              dy       dy      dy     3x + 2x
    reciprocal	of dy/dx).




                                              .    .    .    .       .   .
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2
Solution
                                       dx
    Tangent	lines	are	vertical	when       = 0.
                                       dy
    Differentiating x implicitly	as	a	function	of y gives
              dx       dx      dx        2y
    2y = 3x2      + 2x , so        = 2         (notice	this	is	the
              dy       dy      dy     3x + 2x
    reciprocal	of dy/dx).
    We	must	solve



           .                                      .
                                                          2y
               y2 = x 3 + x2                                     = 0
           1
           .    [(x, y). is on
                 the curve]
                                                  2
                                                  .
                                                      3x2 + 2x
                                                      [tangent line
                                                        is vertical]

                                              .       .    .   .       .   .
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y = 0 =⇒ y = 0
                  3x2 + 2x

      (as	long	as 3x2 + 2x ̸= 0).




                                          .   .   .   .   .   .
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y = 0 =⇒ y = 0
                  3x2 + 2x

      (as	long	as 3x2 + 2x ̸= 0).
      Substituting y = 0 into	the first equation	gives

                         0 = x3 + x2 = x2 (x + 1)

      So x = 0 or x = −1.




                                              .     .   .   .   .   .
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y = 0 =⇒ y = 0
                  3x2 + 2x

      (as	long	as 3x2 + 2x ̸= 0).
      Substituting y = 0 into	the first equation	gives

                         0 = x3 + x2 = x2 (x + 1)

      So x = 0 or x = −1.
      x = 0 is	not	allowed	by	the	first	equation, but

                               dx
                                             = 0,
                               dy   (−1,0)

      so	here	is	a	vertical	tangent.
                                                    .   .   .   .   .   .
Tangents




          (           )
          . − 2 , 3√3
              3
                   2

                 .
. −1 , 0 ) .
(                        .
                 .
        (              )
        . − 2 , − 3 √3
             3
                    2      n
                           . ode




                                   .   .   .   .   .   .
Examples
  Example
  Show	that	the	families	of	curves

                   xy = c                  x2 − y2 = k

  are	orthogonal, that	is, they	intersect	at	right	angles.




                                                 .    .      .   .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




 xy = c
 x2 − y2 = k                    .                           x
                                                            .




                                    .   .   .   .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                        .xy
                                             =
                                              1
 xy = c
 x2 − y2 = k                    .                                x
                                                                 .




                                    .    .       .   .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                             .xy
                                                 =
                                        .xy

                                                     2
                                             =
                                              1
 xy = c
 x2 − y2 = k                    .                                    x
                                                                     .




                                    .    .       .       .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                                 .xy
                                             .xy
                                                      =
                                                       3
                                                 =
                                        .xy

                                                      2
                                             =
                                              1
 xy = c
 x2 − y2 = k                    .                                     x
                                                                      .




                                    .    .        .       .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                                    .xy
                                                .xy
                                                         =
                                                          3
                                                    =
                                        .xy

                                                         2
                                                =
                                                 1
 xy = c
 x2 − y2 = k                    .                                        x
                                                                         .




                                            1
                                        −
                                        =
                                .xy

                                    .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                                    .xy
                                                .xy
                                                         =
                                                          3
                                                    =
                                         .xy

                                                         2
                                                =
                                                 1
 xy = c
 x2 − y2 = k                    .                                        x
                                                                         .




                                            1
                                         −
                                                2
                                        =
                                                −
                                .xy
                                            =
                                        .xy
                                    .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                    y
                                    .




                                                    .xy
                                                .xy
                                                         =
                                                          3
                                                    =
                                         .xy

                                                         2
                                                =
                                                 1
 xy = c
 x2 − y2 = k                    .                                        x
                                                                         .




                                            1
                                         −
                                                − 2
                                        =
                                                  −
                                .xy



                                                 3
                                            =
                                        .xy
                                                =
                                         .xy
                                    .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                               y
                                               .




                                                               .xy
                                                           .xy
                                                                    =
                                                                     3
                                                               =
                                                    .xy

                                                                    2
                                                           =
                            . 2 − y2 = 1




                                                            1
 xy = c
 x2 − y2 = k                               .                                        x
                                                                                    .




                                                       1
                                                    −
                                                           − 2
                                                   =
                            x




                                                             −
                                           .xy



                                                            3
                                                       =
                                                   .xy
                                                           =
                                                    .xy
                                               .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                             y
                                             .




                                                             .xy
                                                         .xy
                                                                  =
                                                                   3
                                                             =
                                                  .xy

                                                                  2
                                                         =
                          . 2 − y2 = 2
                          . 2 − y2 = 1




                                                          1
 xy = c
 x2 − y2 = k                             .                                        x
                                                                                  .




                                                     1
                                                  −
                                                         − 2
                                                 =
                          x
                          x




                                                           −
                                         .xy



                                                          3
                                                     =
                                                 .xy
                                                         =
                                                  .xy
                                             .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                            y
                                            .




                                                            .xy
                                                        .xy
                                                                 =
                                                                  3
                                                            =
                                                 .xy

                                                                 2
                                                        =
                         . 2 − y2 = 3
                         . − y2 = 2
                         . 2 − y2 = 1




                                                         1
 xy = c
 x2 − y2 = k                            .                                        x
                                                                                 .




                                                    1
                                                 −
                         x2




                                                        − 2
                                                =
                         x
                         x




                                                          −
                                        .xy



                                                         3
                                                    =
                                                .xy
                                                        =
                                                 .xy
                                            .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                            y
                                            .




                                                            .xy
                                                        .xy
                                                                 =
                                                                  3
                                                            =
                                                 .xy

                                                                 2
                                                        =
                         . 2 − y2 = 3
                         . − y2 = 2
                         . 2 − y2 = 1




                                                         1
 xy = c
 x2 − y2 = k                            .                                        x
                                                                                 .




                                                    1
                                                 −
                         x2      . 2 − y2 = −1




                                                        − 2
                                 x




                                                =
                         x
                         x




                                                          −
                                        .xy



                                                         3
                                                    =
                                                .xy
                                                        =
                                                 .xy
                                            .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                            y
                                            .




                                                            .xy
                                                        .xy
                                                                 =
                                                                  3
                                                            =
                                                 .xy

                                                                 2
                                                        =
                         . 2 − y2 = 3
                         . − y2 = 2
                         . 2 − y2 = 1




                                                         1
 xy = c
 x2 − y2 = k                            .                                        x
                                                                                 .




                                                    1
                                                 −
                         x2      . 2 − y2 = −1




                                                        − 2
                                 x




                                                =
                         x
                         x




                                                          −
                                        .xy



                                                         3
                                 . 2 − y2 = −2
                                 x




                                                    =
                                                .xy
                                                        =
                                                 .xy
                                            .       .        .       .   .   .
Orthogonal	Families	of	Curves
                                            y
                                            .




                                                            .xy
                                                        .xy
                                                                 =
                                                                  3
                                                            =
                                                 .xy

                                                                 2
                                                        =
                         . 2 − y2 = 3
                         . − y2 = 2
                         . 2 − y2 = 1




                                                         1
 xy = c
 x2 − y2 = k                            .                                        x
                                                                                 .




                                                    1
                                                 −
                         x2      . 2 − y2 = −1




                                                        − 2
                                 x




                                                =
                         x
                         x




                                                          −
                                        .xy



                                                         3
                                 . 2 − y2 = −2
                                 x2




                                                    =
                                 . − y2 = −3



                                                .xy
                                 x




                                                        =
                                                 .xy
                                            .       .        .       .   .   .
Examples
  Example
  Show	that	the	families	of	curves

                   xy = c                  x2 − y2 = k

  are	orthogonal, that	is, they	intersect	at	right	angles.
  Solution
  In	the	first	curve,
                                                 y
                       y + xy′ = 0 =⇒ y′ = −
                                                 x




                                                 .    .      .   .   .   .
Examples
  Example
  Show	that	the	families	of	curves

                   xy = c                  x2 − y2 = k

  are	orthogonal, that	is, they	intersect	at	right	angles.
  Solution
  In	the	first	curve,
                                                 y
                        y + xy′ = 0 =⇒ y′ = −
                                                 x
  In	the	second	curve,
                                                     x
                       2x − 2yy′ = 0 = =⇒ y′ =
                                                     y

  The	product	is −1, so	the	tangent	lines	are	perpendicular
  wherever	they	intersect.
                                                 .       .   .   .   .   .
Music	Selection




        “The	Curse	of	Curves”	by	Cute	is	What	We	Aim	For
                                          .   .    .   .   .   .
Ideal	gases




     The ideal	gas	law relates
     temperature, pressure, and
     volume	of	a	gas:

                      PV = nRT

     (R is	a	constant, n is	the
     amount	of	gas	in	moles)



                                                                 .
.
Image	credit: Scott	Beale	/	Laughing	Squid
                                             .   .   .   .   .   .
Compressibility


   Definition
   The isothermic	compressibility of	a	fluid	is	defined	by

                                     dV 1
                             β=−
                                     dP V
   with	temperature	held	constant.




                                              .    .   .   .   .   .
Compressibility


   Definition
   The isothermic	compressibility of	a	fluid	is	defined	by

                                     dV 1
                              β=−
                                     dP V
   with	temperature	held	constant.

   Approximately	we	have

                 ∆V   dV           ∆V
                    ≈    = −β V =⇒    ≈ −β∆P
                 ∆P   dP           V
   The	smaller	the β , the	“harder”	the	fluid.



                                                .   .   .   .   .   .
Example
Find	the	isothermic	compressibility	of	an	ideal	gas.




                                             .   .     .   .   .   .
Example
Find	the	isothermic	compressibility	of	an	ideal	gas.

Solution
If PV = k (n is	constant	for	our	purposes, T is	constant	because	of
the	word isothermic, and R really	is	constant), then

                dP      dV        dV    V
                   ·V+P    = 0 =⇒    =−
                dP      dP        dP    P
So
                              1 dV      1
                        β=−     ·    =
                              V dP      P
Compressibility	and	pressure	are	inversely	related.



                                             .   .     .   .   .      .
Nonideal	gasses
Not	that	there’s	anything	wrong	with	that


   Example
   The van	der	Waals	equation
   makes	fewer	simplifications:                                                H
                                                                              ..
   (         )                                                         O .
                                                                       . xygen . .
          n2                                                                   H
     P + a 2 (V − nb) = nRT,                                   .
          V
                                                      H
                                                      ..
   where P is	the	pressure, V the           O .
                                            . xygen            H
                                                               . ydrogen	bonds
   volume, T the	temperature, n                       H
                                                      ..


   the	number	of	moles	of	the                                  .
   gas, R a	constant, a is	a                                           O .
                                                                       . xygen . .
                                                                               H
   measure	of	attraction
   between	particles	of	the	gas,                                              H
                                                                              ..
   and b a	measure	of	particle
   size.
                                                .          .       .      .    .     .
Nonideal	gasses
Not	that	there’s	anything	wrong	with	that


     Example
     The van	der	Waals	equation
     makes	fewer	simplifications:
     (         )
            n2
       P + a 2 (V − nb) = nRT,
            V

     where P is	the	pressure, V the
     volume, T the	temperature, n
     the	number	of	moles	of	the
     gas, R a	constant, a is	a
     measure	of	attraction
     between	particles	of	the	gas,
     and b a	measure	of	particle
     size.                                                      .
.
Image	credit: Wikimedia	Commons
                                            .   .   .   .   .   .
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
function	of P gives
         (         )                 (             )
               an2 dV                      2an2 dV
           P+ 2          + (V − bn) 1 − 3             = 0,
                V     dP                    V dP




                                            .    .   .    .     .   .
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
function	of P gives
         (         )                 (             )
               an2 dV                      2an2 dV
           P+ 2          + (V − bn) 1 − 3             = 0,
                V     dP                    V dP
so
                     1 dV       V2 (V − nb)
               β=−        =
                     V dP   2abn3 − an2 V + PV3




                                            .    .   .    .     .   .
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
function	of P gives
         (         )                 (             )
               an2 dV                      2an2 dV
           P+ 2          + (V − bn) 1 − 3             = 0,
                V     dP                    V dP
so
                     1 dV       V2 (V − nb)
               β=−        =
                     V dP   2abn3 − an2 V + PV3

     What	if a = b = 0?




                                            .    .   .    .     .   .
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
function	of P gives
         (         )                 (             )
               an2 dV                      2an2 dV
           P+ 2          + (V − bn) 1 − 3             = 0,
                V     dP                    V dP
so
                      1 dV       V2 (V − nb)
               β=−         =
                      V dP   2abn3 − an2 V + PV3

     What	if a = b = 0?
                                                              dβ
     Without	taking	the	derivative, what	is	the	sign	of          ?
                                                              db




                                             .    .       .      .   .   .
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
function	of P gives
         (         )                 (             )
               an2 dV                      2an2 dV
           P+ 2          + (V − bn) 1 − 3             = 0,
                V     dP                    V dP
so
                      1 dV       V2 (V − nb)
               β=−         =
                      V dP   2abn3 − an2 V + PV3

     What	if a = b = 0?
                                                        dβ
     Without	taking	the	derivative, what	is	the	sign	of    ?
                                                        db
                                                        dβ
     Without	taking	the	derivative, what	is	the	sign	of    ?
                                                        da


                                             .    .       .   .   .   .
Nasty	derivatives




       dβ    (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
          =−
       db                  (2abn3 − an2 V + PV3 )2
                     ( 2        )
                 nV3 an + PV2
          = −(                    )2 < 0
               PV3 + an2 (2bn − V)

                   dβ   n2 (bn − V)(2bn − V)V2
                      = (                    )2 > 0
                   da     PV3 + an2 (2bn − V)
      (as	long	as V > 2nb, and	it’s	probably	true	that V ≫ 2nb).




                                             .    .   .    .       .   .
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogonal	Trajectories
     Chemistry


  The	power	rule	for	rational	powers




                                        .   .   .   .   .   .
Using	implicit	differentiation	to	find	derivatives



   Example
          dy       √
   Find      if y = x.
          dx




                                      .   .   .   .   .   .
Using	implicit	differentiation	to	find	derivatives



   Example
          dy       √
   Find      if y = x.
          dx
   Solution
         √
   If y =   x, then
                                  y2 = x,
   so
                           dy        dy   1    1
                      2y      = 1 =⇒    =    = √ .
                           dx        dx   2y  2 x




                                             .   .   .   .   .   .
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = xp/q , where p and q are	integers, then y′ =     x     .
                                                         q




                                                 .   .      .    .   .   .
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = xp/q , where p and q are	integers, then y′ =     x     .
                                                         q
   Proof.
   First, raise	both	sides	to	the qth	power:

                         y = xp/q =⇒ yq = xp




                                                 .   .      .    .   .   .
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = xp/q , where p and q are	integers, then y′ =     x     .
                                                         q
   Proof.
   First, raise	both	sides	to	the qth	power:

                          y = xp/q =⇒ yq = xp

   Now, differentiate	implicitly:

                         dy            dy  p xp−1
                 qyq−1      = pxp−1 =⇒    = · q −1
                         dx            dx  q y




                                                 .   .      .    .   .   .
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = xp/q , where p and q are	integers, then y′ =     x     .
                                                         q
   Proof.
   First, raise	both	sides	to	the qth	power:

                          y = xp/q =⇒ yq = xp

   Now, differentiate	implicitly:

                         dy            dy  p xp−1
                 qyq−1      = pxp−1 =⇒    = · q −1
                         dx            dx  q y

   Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so

                x p −1   xp−1
                       = p−p/q = xp−1−(p−p/q) = xp/q−1
                y q −1  x

                                                 .   .      .    .   .   .
What	have	we	learned	today?




      Implicit	Differentiation	allows	us	to	pretend	that	a	relation
      describes	a	function, since	it	does, locally, “almost
      everywhere.”
      The	Power	Rule	was	established	for	powers	which	are
      rational	numbers.




                                               .    .   .    .    .   .

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Lesson 11: Implicit Differentiation

  • 1. Section 2.6 Implicit Differentiation V63.0121.006/016, Calculus I February 23, 2010 Announcements Quiz 2 is February 26, covering §§1.5–2.3 Midterm is March 4, covering §§1.1–2.5 On HW 5, Problem 2.3.46 should be 2.4.46 . . Image credit: Telstar Logistics . . . . . .
  • 2. Announcements on white background Announcements Quiz 2 is February 26, covering §§1.5–2.3 Midterm is March 4, covering §§1.1–2.5 On HW 5, Problem 2.3.46 should be 2.4.46 . . . . . .
  • 3. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  • 4. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .
  • 5. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .
  • 6. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . . .
  • 7. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) . . . . . .
  • 8. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 . . . . . .
  • 9. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 dy 3 /5 3/5 3 Evaluate: =√ = = . dx x=3/5 1 − ( 3 /5 )2 4/5 4 . . . . . .
  • 10. Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 dy 3 /5 3/5 3 Evaluate: =√ = = . dx x=3/5 1 − ( 3 /5 )2 4/5 4 . . . . . .
  • 11. Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 . . . . . .
  • 12. Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differentiate this equation to get 2x + 2f(x) · f′ (x) = 0 . . . . . .
  • 13. Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differentiate this equation to get 2x + 2f(x) · f′ (x) = 0 We could then solve to get x f′ (x ) = − f(x) . . . . . .
  • 14. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . . . . . . .
  • 15. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . . . . . . .
  • 16. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . l .ooks like a function . . . . . .
  • 17. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. . x . . . . . . .
  • 18. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. . x . . . . . . .
  • 19. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. l .ooks like a function . x . . . . . . .
  • 20. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . . . . . .
  • 21. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . . . . . .
  • 22. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . does not look like a function, but that’s OK—there are only two points like this . . . . . .
  • 23. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally”, almost . x . everywhere and is differentiable . l .ooks like a function . . . . . .
  • 24. Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally”, almost . x . everywhere and is differentiable The chain rule then . applies for this local choice. l .ooks like a function . . . . . .
  • 25. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). . . . . . .
  • 26. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx . . . . . .
  • 27. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! . . . . . .
  • 28. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! dy x Isolate: =− . dx y . . . . . .
  • 29. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! dy x Isolate: =− . dx y dy 3 /5 3 Evaluate: = = . dx ( 3 ,− 4 ) 4/5 4 5 5 . . . . . .
  • 30. Summary If a relation is given between x and y which isn’t a function: “Most of the time”, i.e., “at most places” y can be y . assumed to be a function of . x we may differentiate the . x . relation as is dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve. . . . . . .
  • 31. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  • 32. Example Find y′ along the curve y3 + 4xy = x2 + 3. . . . . . .
  • 33. Example Find y′ along the curve y3 + 4xy = x2 + 3. Solution Implicitly differentiating, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x . . . . . .
  • 34. Example Find y′ along the curve y3 + 4xy = x2 + 3. Solution Implicitly differentiating, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x Solving for y′ gives 3y2 y′ + 4xy′ = 2x − 4y (3y2 + 4x)y′ = 2x − 4y 2x − 4y =⇒ y′ = 2 3y + 4x . . . . . .
  • 35. Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). . . . . . .
  • 36. Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). Solution Differentiating implicitly: 5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x) Collect all terms with y′ on one side and all terms without y′ on the other: 5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 ) Now factor and divide: 2xy(cos x2 − y2 ) y′ = 5y4 + 3x2 y2 − sin x2 . . . . . .
  • 37. Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . . . . . . .
  • 38. Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives dy dy 3x2 + 2x 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4 . . . . . .
  • 39. Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives dy dy 3x2 + 2x 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4 11 Thus the equation of the tangent line is y + 6 = − (x − 3). 4 . . . . . .
  • 40. Line equation forms slope-intercept form y = mx + b where the slope is m and (0, b) is on the line. point-slope form y − y0 = m(x − x0 ) where the slope is m and (x0 , y0 ) is on the line. . . . . . .
  • 42. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We have to solve these two equations: . . 2 3 2 3x2 + 2x y = x +x = 0 1 . [(x, y). is on the curve] 2 . 2y [tangent line is horizontal] . . . . . .
  • 43. Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. . . . . . .
  • 44. Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. Substituting x = 0 into the first equation gives y2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road. . . . . . .
  • 45. Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. Substituting x = 0 into the first equation gives y2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road. Substituting x = −2/3 into the first equation gives ( ) ( ) 2 2 3 2 2 4 2 y = − + − = =⇒ y = ± √ , 3 3 27 3 3 so there are two horizontal tangents. . . . . . .
  • 46. Horizontal Tangents ( ) . − 2 , 3√3 3 2 . . . ( ) . − 2 , − 3 √3 3 2 . . . . . .
  • 47. Horizontal Tangents ( ) . − 2 , 3√3 3 2 . . . ( ) . − 2 , − 3 √3 3 2 n . ode . . . . . .
  • 49. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy . . . . . .
  • 50. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy Differentiating x implicitly as a function of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (notice this is the dy dy dy 3x + 2x reciprocal of dy/dx). . . . . . .
  • 51. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy Differentiating x implicitly as a function of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (notice this is the dy dy dy 3x + 2x reciprocal of dy/dx). We must solve . . 2y y2 = x 3 + x2 = 0 1 . [(x, y). is on the curve] 2 . 3x2 + 2x [tangent line is vertical] . . . . . .
  • 52. Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). . . . . . .
  • 53. Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Substituting y = 0 into the first equation gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1. . . . . . .
  • 54. Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Substituting y = 0 into the first equation gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1. x = 0 is not allowed by the first equation, but dx = 0, dy (−1,0) so here is a vertical tangent. . . . . . .
  • 55. Tangents ( ) . − 2 , 3√3 3 2 . . −1 , 0 ) . ( . . ( ) . − 2 , − 3 √3 3 2 n . ode . . . . . .
  • 56. Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. . . . . . .
  • 57. Orthogonal Families of Curves y . xy = c x2 − y2 = k . x . . . . . . .
  • 58. Orthogonal Families of Curves y . .xy = 1 xy = c x2 − y2 = k . x . . . . . . .
  • 59. Orthogonal Families of Curves y . .xy = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . .
  • 60. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . .
  • 61. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − = .xy . . . . . .
  • 62. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − 2 = − .xy = .xy . . . . . .
  • 63. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − − 2 = − .xy 3 = .xy = .xy . . . . . .
  • 64. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − − 2 = x − .xy 3 = .xy = .xy . . . . . .
  • 65. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − − 2 = x x − .xy 3 = .xy = .xy . . . . . .
  • 66. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − x2 − 2 = x x − .xy 3 = .xy = .xy . . . . . .
  • 67. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − x2 . 2 − y2 = −1 − 2 x = x x − .xy 3 = .xy = .xy . . . . . .
  • 68. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − x2 . 2 − y2 = −1 − 2 x = x x − .xy 3 . 2 − y2 = −2 x = .xy = .xy . . . . . .
  • 69. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . 1 − x2 . 2 − y2 = −1 − 2 x = x x − .xy 3 . 2 − y2 = −2 x2 = . − y2 = −3 .xy x = .xy . . . . . .
  • 70. Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy′ = 0 =⇒ y′ = − x . . . . . .
  • 71. Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy′ = 0 =⇒ y′ = − x In the second curve, x 2x − 2yy′ = 0 = =⇒ y′ = y The product is −1, so the tangent lines are perpendicular wherever they intersect. . . . . . .
  • 72. Music Selection “The Curse of Curves” by Cute is What We Aim For . . . . . .
  • 73. Ideal gases The ideal gas law relates temperature, pressure, and volume of a gas: PV = nRT (R is a constant, n is the amount of gas in moles) . . Image credit: Scott Beale / Laughing Squid . . . . . .
  • 74. Compressibility Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . . . . . .
  • 75. Compressibility Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. Approximately we have ∆V dV ∆V ≈ = −β V =⇒ ≈ −β∆P ∆P dP V The smaller the β , the “harder” the fluid. . . . . . .
  • 77. Example Find the isothermic compressibility of an ideal gas. Solution If PV = k (n is constant for our purposes, T is constant because of the word isothermic, and R really is constant), then dP dV dV V ·V+P = 0 =⇒ =− dP dP dP P So 1 dV 1 β=− · = V dP P Compressibility and pressure are inversely related. . . . . . .
  • 78. Nonideal gasses Not that there’s anything wrong with that Example The van der Waals equation makes fewer simplifications: H .. ( ) O . . xygen . . n2 H P + a 2 (V − nb) = nRT, . V H .. where P is the pressure, V the O . . xygen H . ydrogen bonds volume, T the temperature, n H .. the number of moles of the . gas, R a constant, a is a O . . xygen . . H measure of attraction between particles of the gas, H .. and b a measure of particle size. . . . . . .
  • 79. Nonideal gasses Not that there’s anything wrong with that Example The van der Waals equation makes fewer simplifications: ( ) n2 P + a 2 (V − nb) = nRT, V where P is the pressure, V the volume, T the temperature, n the number of moles of the gas, R a constant, a is a measure of attraction between particles of the gas, and b a measure of particle size. . . Image credit: Wikimedia Commons . . . . . .
  • 80. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP . . . . . .
  • 81. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 . . . . . .
  • 82. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 What if a = b = 0? . . . . . .
  • 83. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db . . . . . .
  • 84. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db dβ Without taking the derivative, what is the sign of ? da . . . . . .
  • 85. Nasty derivatives dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 ) =− db (2abn3 − an2 V + PV3 )2 ( 2 ) nV3 an + PV2 = −( )2 < 0 PV3 + an2 (2bn − V) dβ n2 (bn − V)(2bn − V)V2 = ( )2 > 0 da PV3 + an2 (2bn − V) (as long as V > 2nb, and it’s probably true that V ≫ 2nb). . . . . . .
  • 86. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  • 87. Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx . . . . . .
  • 88. Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx Solution √ If y = x, then y2 = x, so dy dy 1 1 2y = 1 =⇒ = = √ . dx dx 2y 2 x . . . . . .
  • 89. The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q . . . . . .
  • 90. The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp . . . . . .
  • 91. The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp Now, differentiate implicitly: dy dy p xp−1 qyq−1 = pxp−1 =⇒ = · q −1 dx dx q y . . . . . .
  • 92. The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp Now, differentiate implicitly: dy dy p xp−1 qyq−1 = pxp−1 =⇒ = · q −1 dx dx q y Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so x p −1 xp−1 = p−p/q = xp−1−(p−p/q) = xp/q−1 y q −1 x . . . . . .
  • 93. What have we learned today? Implicit Differentiation allows us to pretend that a relation describes a function, since it does, locally, “almost everywhere.” The Power Rule was established for powers which are rational numbers. . . . . . .