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Section	2.6
                                  Implicit	Differentiation

                                  V63.0121.006/016...
Announcements	on	white	background




  Announcements
     Quiz	2	is	February	26, covering	§§1.5–2.3
     Midterm	is	March...
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogona...
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	...
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	...
Motivating	Example                         y
                                           .


Problem
Find	the	slope	of	the	...
Motivating	Example                                  y
                                                    .


Problem
Find...
Motivating	Example                                  y
                                                    .


Problem
Find...
Motivating	Example                                  y
                                                    .


Problem
Find...
Motivating	Example                                  y
                                                    .


Problem
Find...
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
    ...
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
    ...
Motivating	Example, another	way
  We	know	that x2 + y2 = 1 does	not	define y as	a	function	of x,
  but	suppose	it	did.
    ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Yes, we	can!

   The	beautiful	fact	(i.e., deep	theorem)	is	that	this	works!
      “Near”	most	points	on                  ...
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
...
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
...
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
...
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
...
Motivating	Example, again, with	Leibniz	notation


   Problem
   Find	the	slope	of	the	line	which	is	tangent	to	the	curve
...
Summary


  If	a	relation	is	given	between x and y which	isn’t	a	function:
   “Most	of	the	time”, i.e., “at
   most	places...
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogona...
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...
Example
Find y′ along	the	curve y3 + 4xy = x2 + 3.

Solution
Implicitly	differentiating, we	have

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



      slope-intercept	form

                                y = mx + b

      where	the	slope	is m ...
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:
...
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ ...
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ ...
Solution, continued
      Solving	the	second	equation	gives

          3x2 + 2x
                   = 0 =⇒ 3x2 + 2x = 0 =⇒ ...
Horizontal	Tangents




    (           )
    . − 2 , 3√3
        3
             2

           .
                      .
 ...
Horizontal	Tangents




    (           )
    . − 2 , 3√3
        3
             2

           .
                      .
 ...
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
...
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2
Solution
                                       dx...
Example
Find	the vertical tangent	lines	to	the	same	curve: y2 = x3 + x2
Solution
                                       dx...
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y ...
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y ...
Solution, continued
      Solving	the	second	equation	gives

                     2y
                           = 0 =⇒ 2y ...
Tangents




          (           )
          . − 2 , 3√3
              3
                   2

                 .
. −1 ,...
Examples
  Example
  Show	that	the	families	of	curves

                   xy = c                  x2 − y2 = k

  are	ortho...
Orthogonal	Families	of	Curves
                                    y
                                    .




 xy = c
 x2 ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                    y
                                    .




            ...
Orthogonal	Families	of	Curves
                                               y
                                           ...
Orthogonal	Families	of	Curves
                                             y
                                             ...
Orthogonal	Families	of	Curves
                                            y
                                            .
...
Orthogonal	Families	of	Curves
                                            y
                                            .
...
Orthogonal	Families	of	Curves
                                            y
                                            .
...
Orthogonal	Families	of	Curves
                                            y
                                            .
...
Examples
  Example
  Show	that	the	families	of	curves

                   xy = c                  x2 − y2 = k

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

                   xy = c                  x2 − y2 = k

  are	ortho...
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:

                    ...
Compressibility


   Definition
   The isothermic	compressibility of	a	fluid	is	defined	by

                                 ...
Compressibility


   Definition
   The isothermic	compressibility of	a	fluid	is	defined	by

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


   Example
   The van	der	Waals	equation
   makes	fewer	simpli...
Nonideal	gasses
Not	that	there’s	anything	wrong	with	that


     Example
     The van	der	Waals	equation
     makes	fewer	...
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
functi...
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
functi...
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
functi...
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
functi...
Let’s	find	the	compressibility	of	a	van	der	Waals	gas.
Differentiating	the	van	der	Waals	equation	by	treating V as	a
functi...
Nasty	derivatives




       dβ    (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
          =−
       db              ...
Outline


  The	big	idea, by	example


  Examples
     Basic	Examples
     Vertical	and	Horizontal	Tangents
     Orthogona...
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
 ...
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = x...
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = x...
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = x...
The	power	rule	for	rational	powers
   Theorem
                                                         p p/q−1
   If y = x...
What	have	we	learned	today?




      Implicit	Differentiation	allows	us	to	pretend	that	a	relation
      describes	a	func...
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Lesson 11: Implicit Differentiation

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With implicit differentiation we can find the rate of change of a relation which isn't necessarily a function

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Transcript of "Lesson 11: Implicit Differentiation"

  1. 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. 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. 3. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  4. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 31. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  32. 32. Example Find y′ along the curve y3 + 4xy = x2 + 3. . . . . . .
  33. 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. 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. 35. Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). . . . . . .
  36. 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. 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. 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. 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. 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. . . . . . .
  41. 41. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 . . . . . .
  42. 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. 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. 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. 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. 46. Horizontal Tangents ( ) . − 2 , 3√3 3 2 . . . ( ) . − 2 , − 3 √3 3 2 . . . . . .
  47. 47. Horizontal Tangents ( ) . − 2 , 3√3 3 2 . . . ( ) . − 2 , − 3 √3 3 2 n . ode . . . . . .
  48. 48. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 . . . . . .
  49. 49. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy . . . . . .
  50. 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. 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. 52. Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). . . . . . .
  53. 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. 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. 55. Tangents ( ) . − 2 , 3√3 3 2 . . −1 , 0 ) . ( . . ( ) . − 2 , − 3 √3 3 2 n . ode . . . . . .
  56. 56. Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. . . . . . .
  57. 57. Orthogonal Families of Curves y . xy = c x2 − y2 = k . x . . . . . . .
  58. 58. Orthogonal Families of Curves y . .xy = 1 xy = c x2 − y2 = k . x . . . . . . .
  59. 59. Orthogonal Families of Curves y . .xy = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . .
  60. 60. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . .
  61. 61. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − = .xy . . . . . .
  62. 62. Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − 2 = − .xy = .xy . . . . . .
  63. 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. 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. 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. 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. 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. 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. 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. 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. 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. 72. Music Selection “The Curse of Curves” by Cute is What We Aim For . . . . . .
  73. 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. 74. Compressibility Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . . . . . .
  75. 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. . . . . . .
  76. 76. Example Find the isothermic compressibility of an ideal gas. . . . . . .
  77. 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. 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. 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. 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. 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. 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. 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. 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. 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. 86. Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .
  87. 87. Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx . . . . . .
  88. 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. 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. 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. 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. 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. 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. . . . . . .
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