Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
1. Section 2.6
Implicit Differentiation
V63.0121.021, Calculus I
New York University
October 12, 2010
Announcements
Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2
Midterm next week. Covers §§1.1–2.5
. . . . . .
2. Announcements
Quiz 2 in recitation this
week. Covers §§1.5, 1.6,
2.1, 2.2
Midterm next week.
Covers §§1.1–2.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 2 / 34
3. Objectives
Use implicit differentation
to find the derivative of a
function defined implicitly.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 3 / 34
4. Outline
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 4 / 34
5. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 1
at the point (3/5, −4/5).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
6. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 1
at the point (3/5, −4/5).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
7. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 1
.
at the point (3/5, −4/5).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
8. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 1
.
at the point (3/5, −4/5).
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
9. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
10. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
11. Motivating Example
y
.
Problem
Find the slope of the line
which is tangent to the
curve . x
.
x2 + y2 = 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
14. 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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
15. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
. x
.
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
16. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
. x
.
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
17. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
. x
.
.
l
.ooks like a function
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
18. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the .
curve resembles the graph
of a function.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
19. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the .
curve resembles the graph
of a function.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
20. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the .
curve resembles the graph
of a function. l
.ooks like a function
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
21. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
22. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
23. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
24. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
So f(x) is defined “locally”,
. x
.
almost everywhere and is
differentiable
.
l
.ooks like a function
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
25. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
y
.
“Near” most points on the
curve x2 + y2 = 1, the
curve resembles the graph
of a function.
So f(x) is defined “locally”,
. x
.
almost everywhere and is
differentiable
The chain rule then applies
.
for this local choice.
l
.ooks like a function
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
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).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
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!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
30. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
31. Summary
If a relation is given between x and y which isn’t a function:
“Most of the time”, i.e., “at y
.
most places” y can be
.
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 9 / 34
32. Outline
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 10 / 34
33. Another Example
Example
Find y′ along the curve y3 + 4xy = x2 + 3.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
34. Another Example
Example
Find y′ along the curve y3 + 4xy = x2 + 3.
Solution
Implicitly differentiating, we have
3y2 y′ + 4(1 · y + x · y′ ) = 2x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
35. Another Example
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
36. Yet Another Example
Example
Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
37. Yet Another Example
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
38. Finding tangent lines with implicit differentitiation
.
Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6). .
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
39. Finding tangent lines with implicit differentitiation
.
Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6). .
Solution
dy dy 3x2 + 2x
Differentiate: 2y = 3x2 + 2x, so = , and
dx dx 2y
dy 3 · 32 + 2 · 3 33 11
= =− =− .
dx (3,−6) 2(−6) 12 4
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
40. Finding tangent lines with implicit differentitiation
.
Example
Find the equation of the line
tangent to the curve
.
y2 = x2 (x + 1) = x3 + x2
at the point (3, −6). .
Solution
dy dy 3x2 + 2x
Differentiate: 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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
41. Recall: 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 14 / 34
42. Horizontal Tangent Lines
Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
43. Horizontal Tangent Lines
Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We have to solve these two equations:
.
. 3x2 + 2x
2 3 2 = 0
1 y = x. + x
. [(x, y) is on the curve] 2
.
2y
[tangent line
is horizontal]
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
46. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
47. 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 3 2 2 4 2
y = −
2
+ − = =⇒ y = ± √ ,
3 3 27 3 3
so there are two horizontal tangents.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
50. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
51. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
dx
Tangent lines are vertical when = 0.
dy
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
52. 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).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
53. 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 = x3 + x2 =0
3x2 + 2x
1
. [(x, y). is on
the curve]
2
. [tangent line
. . .
is vertical]
. . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
54. Solution, continued
Solving the second equation gives
2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x
(as long as 3x2 + 2x ̸= 0).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
55. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
56. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
58. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 21 / 34
59. Orthogonal Families of Curves
y
.
.xy
=
1
xy = c
x2 − y2 = k . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
60. Orthogonal Families of Curves
y
.
.xy
=
.xy
2
=
1
xy = c
x2 − y2 = k . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
61. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
1
xy = c
x2 − y2 = k . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
62. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
1
xy = c
x2 − y2 = k . x
.
1
−
=
.xy
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
63. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
1
xy = c
x2 − y2 = k . x
.
1
−
2
=
−
.xy
=
.xy
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
64. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
1
xy = c
x2 − y2 = k . x
.
.xy = −1
= −2
.xy =
.xy
3
−
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
65. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
. 2 − y2 = 1
1
xy = c
x2 − y2 = k . x
.
.xy = −1
= −2
x
.xy =
.xy
3
−
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
66. Orthogonal Families of Curves
y
.
.xy
.xy
=
3
=
.xy
2
=
. 2 − y2 = 2
. 2 − y2 = 1
1
xy = c
x2 − y2 = k . x
.
.xy = −1
= −2
x
x
.xy =
.xy
3
−
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
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
.
.xy = −1
x2
= −2
x
x
.xy =
.xy
3
−
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
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
.
.xy = −1
x2
= −2
x
x
. 2 − y2 = −1
.xy =
x
.xy
3
−
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
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
.
.xy = −1
x2
= −2
x
x
. 2 − y2 = −1
.xy =
x
.xy
3
. 2 − y2 = −2
−
x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
70. 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
.
.xy = −1
x2
= −2
x
x
. 2 − y2 = −1
.xy =
x
.xy
3
. 2 − y2 = −2
−
x2
. − y2 = −3
x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
71. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
y
In the first curve, y + xy′ = 0 =⇒ y′ = −
x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
72. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
y
In the first curve, y + xy′ = 0 =⇒ y′ = −
x
x
In the second curve, 2x − 2yy′ = 0 =⇒ y′ =
y
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
73. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
y
In the first curve, y + xy′ = 0 =⇒ y′ = −
x
x
In the second curve, 2x − 2yy′ = 0 =⇒ y′ =
y
The product is −1, so the tangent lines are perpendicular
wherever they intersect.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
74. Music Selection
“The Curse of Curves” by Cute is What We Aim For . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 24 / 34
75. 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)
.
Ima
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 25 / 34
76. Compressibility
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
77. 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
≈ = −βV =⇒ ≈ −β∆P
∆P dP V
The smaller the β, the “harder” the fluid.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
78. Compressibility of an ideal gas
Example
Find the isothermic compressibility of an ideal gas.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
79. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
80. Nonideal gasses
Not that there's anything wrong with that
Example
The van der Waals equation H
..
makes fewer simplifications:
( ) O .
. xygen . .
H
n2 .
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 measure of O .
. xygen . .
H
attraction between particles of
the gas, and b a measure of H
..
particle size.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
81. 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.
. . ikimedia Commons
W
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
82. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
83. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
84. 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
Question
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
85. 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
Question
What if a = b = 0?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
86. 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
Question
What if a = b = 0?
dβ
Without taking the derivative, what is the sign of ?
db
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
87. 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
Question
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
88. Nasty derivatives
dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
=−
db (2abn3 − an2 V + PV3 )2
( )
nV3 an2 + PV2
= −( )2 < 0
PV3 + an2 (2bn − V)
dβ n2 (bn − V)(2bn − V)V2
=( )2 > 0
da 3 2 (2bn − V)
PV + an
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 30 / 34
89. Outline
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 31 / 34
90. Using implicit differentiation to find derivatives
Example
dy √
Find if y = x.
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
91. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
93. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
94. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
95. 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
xp−1 xp−1
= p−p/q = xp−1−(p−p/q) = xp/q−1
. . . . . .
V63.0121.021, Calculus I
yq−1
(NYU)
x Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
96. Summary
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 34 / 34