Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
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.
A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.
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.
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.
A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.
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.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems. We started to develop ways to enhance students IQ. We started to leave an indelible mark on the students who have undergone APEX training. That is why APEX INSTITUTE is very well known of its quality of education
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems. We started to develop ways to enhance students IQ. We started to leave an indelible mark on the students who have undergone APEX training. That is why APEX INSTITUTE is very well known of its quality of education
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
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.
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.
Lesson 11: Implicit Differentiation (Section 41 slides)Mel Anthony Pepito
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.
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.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
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1. Section 2.6
Implicit Differentiation
V63.0121.034, Calculus I
October 7, 2009
Announcements
Midterm next , covering §§1.1–2.4.
.
.
Image credit: Telstar Logistics
. . . . . .
2. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
3. 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).
. . . . . .
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). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
. . . . . .
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 −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
. . . . . .
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
dy 3 /5 3/5 3
Evaluate: =√ = = .
dx x=3/5 1 − (3 /5 )2 4/5 4
. . . . . .
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. 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)
. . . . . .
11. 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” and is . x
.
differentiable
The chain rule then
applies for this local .
choice.
. . . . . .
12. 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” and is . x
.
differentiable
The chain rule then
applies for this local .
choice.
. . . . . .
13. 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” and is . x
.
differentiable
The chain rule then
applies for this local .
choice.
l
.ooks like a function
. . . . . .
14. 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” and is . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .
15. 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” and is . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .
16. 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
So f(x) is defined
“locally” and is . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .
17. 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” and is . . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .
18. 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” and is . . x
.
differentiable
The chain rule then
applies for this local
choice.
. . . . . .
19. 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” and is . . x
.
differentiable .
The chain rule then does not look like a
applies for this local function, but that’s
choice. OK—there are only
two points like this
. . . . . .
20. Problem
Find the slope of the line which is tangent to the curve
x2 + y2 = 1 at the point (3/5, −4/5).
Solution (Implicit, with Leibniz notation)
Differentiate. Remember y is assumed to be a function of x:
dy
2x + 2y = 0,
dx
dy
Isolate :
dx
dy x
=− .
dx y
Evaluate:
dy 3 /5 3
= = .
dx ( 3 ,− 4 ) 4/5 4
5 5
. . . . . .
21. Summary
If a relation is given between x y
.
and y,
“Most of the time”, i.e., “at
most places” y can be
.
assumed to be a function of
x
.
we may differentiate the
relation as is
dy
Solving for does give the
dx
slope of the tangent line to
the curve at a point on the
curve.
. . . . . .
22. Mnemonic
Explicit Implicit
y = f(x) F(x, y) = k
. . . . . .
23. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
25. 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 11
= =− .
dx (3,−6) 2(−6) 4
. . . . . .
26. 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 11
= =− .
dx (3,−6) 2(−6) 4
11
Thus the equation of the tangent line is y + 6 = − (x − 3).
4
. . . . . .
30. Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We solve for dy/dx = 0:
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
The possible solution x = 0 leads to y = 0, which is not a
smooth point of the function (the denominator in dy/dx
becomes 0).
. . . . . .
31. Example
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solution
We solve for dy/dx = 0:
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
The possible solution x = 0 leads to y = 0, which is not a
smooth point of the function (the denominator in dy/dx
becomes 0).
The possible solution x = − 2 yields y = ± 3√3 .
3
2
. . . . . .
33. 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
2y = 3x2 + 2x , so
dy dy
dx 2y
= 2
dy 3x + 2x
. . . . . .
34. 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
2y = 3x2 + 2x , so
dy dy
dx 2y
= 2
dy 3x + 2x
This is 0 only when y = 0.
. . . . . .
35. 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
2y = 3x2 + 2x , so
dy dy
dx 2y
= 2
dy 3x + 2x
This is 0 only when y = 0.
We get the false solution x = 0 and the real solution x = −1.
. . . . . .
37. 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 + 2x cos(x2 )
Now factor and divide:
2x(y3 + cos x2 )
y′ =
5y4 + 3x2 y2
. . . . . .
38. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
. . . . . .
39. 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
. . . . . .
40. 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.
. . . . . .
41. 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
. . . . . .
42. .
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
.
Image credit: Neil Better
. . . . . .
43. .
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
. The smaller the β , the “harder” the fluid.
Image credit: Neil Better
. . . . . .
45. 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.
. . . . . .
46. 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.
. . . . . .
47. 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
. . . . . .
53. 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).
. . . . . .
54. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
58. 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.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y
. . . . . .
59. 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.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y
Now yq−1 = x(p/q)(q−1) = xp−p/q so
x p −1
= xp−1−(p−p/q) = xp/q−1
y q −1
. . . . . .