lesson 11: implicit differentiation (slides)
DESCRIPTION
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.TRANSCRIPT
..
Sec on 2.6Implicit Differen a on
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
February 28, 2011
Music Selection
“The Curse of Curves”by Cute is What WeAim For
Announcements
I Quiz 2 in recita on thisweek. Covers §§1.5, 1.6,2.1, 2.2
I Midterm next week.Covers §§1.1–2.5
Objectives
I Use implicit differenta onto find the deriva ve of afunc on definedimplicitly.
Outline
The big idea, by example
ExamplesBasic ExamplesVer cal and Horizontal TangentsOrthogonal TrajectoriesChemistry
The power rule for ra onal powers
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
.. x.
y
.
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
.. x.
y
.
Motivating Example
ProblemFind the slope of the linewhich is tangent to thecurve
x2 + y2 = 1
at the point (3/5,−4/5).
.. x.
y
.
Motivating Example, SolutionSolu on (Explicit)
I Isolate:y2 = 1− x2 =⇒ y = −
√1− x2.
(Why the−?)
I Differen ate:dydx
= − −2x2√1− x2
=x√
1− x2I Evaluate:
dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
.. x.
y
.
Motivating Example, SolutionSolu on (Explicit)
I Isolate:y2 = 1− x2 =⇒ y = −
√1− x2.
(Why the−?)I Differen ate:
dydx
= − −2x2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
.. x.
y
.
Motivating Example, SolutionSolu on (Explicit)
I Isolate:y2 = 1− x2 =⇒ y = −
√1− x2.
(Why the−?)I Differen ate:
dydx
= − −2x2√1− x2
=x√
1− x2I Evaluate:
dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
.. x.
y
.
Motivating Example, SolutionSolu on (Explicit)
I Isolate:y2 = 1− x2 =⇒ y = −
√1− x2.
(Why the−?)I Differen ate:
dydx
= − −2x2√1− x2
=x√
1− x2I Evaluate:
dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
.. x.
y
.
Motivating Example, SolutionSolu on (Explicit)
I Isolate:y2 = 1− x2 =⇒ y = −
√1− x2.
(Why the−?)I Differen ate:
dydx
= − −2x2√1− x2
=x√
1− x2I Evaluate:
dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
.. x.
y
.
Motivating Example, another wayWe know that x2 + y2 = 1 does not define y as a func on of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differen ate this equa on to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
Motivating Example, another wayWe know that x2 + y2 = 1 does not define y as a func on of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differen ate this equa on to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
Motivating Example, another wayWe know that x2 + y2 = 1 does not define y as a func on of x, butsuppose it did.
I Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
I We could differen ate this equa on to get
2x+ 2f(x) · f′(x) = 0
I We could then solve to get
f′(x) = − xf(x)
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
..
looks like a func on
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
..looks like a func on
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
..
does not look like afunc on, but that’sOK—there are onlytwo points like this
.
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
..
looks like a func on
Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!
I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.
I So f(x) is defined “locally”,almost everywhere and isdifferen able
I The chain rule then applies forthis local choice.
.. x.
y
..
looks like a func on
Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solu on
I Differen ate: 2x+ 2ydydx
= 0Remember y is assumed to be a func on of x!
I Isolate:dydx
= −xy. Then evaluate:
dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solu on
I Differen ate: 2x+ 2ydydx
= 0
Remember y is assumed to be a func on of x!
I Isolate:dydx
= −xy. Then evaluate:
dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solu on
I Differen ate: 2x+ 2ydydx
= 0Remember y is assumed to be a func on of x!
I Isolate:dydx
= −xy. Then evaluate:
dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solu on
I Differen ate: 2x+ 2ydydx
= 0Remember y is assumed to be a func on of x!
I Isolate:dydx
= −xy.
Then evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).
Solu on
I Differen ate: 2x+ 2ydydx
= 0Remember y is assumed to be a func on of x!
I Isolate:dydx
= −xy. Then evaluate:
dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
SummaryIf a rela on is given between x and y which isn’t a func on:I “Most of the me”, i.e., “atmost places” y can beassumed to be a func on of x
I we may differen ate therela on as is
I Solving fordydx
does give theslope of the tangent line tothe curve at a point on thecurve.
.. x.
y
.
Outline
The big idea, by example
ExamplesBasic ExamplesVer cal and Horizontal TangentsOrthogonal TrajectoriesChemistry
The power rule for ra onal powers
Another ExampleExample
Find y′ along the curve y3 + 4xy = x2 + 3.
Solu onImplicitly differen a ng, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y =⇒ (3y2 + 4x)y′ = 2x− 4y =⇒ y′ =2x− 4y3y2 + 4x
Another ExampleExample
Find y′ along the curve y3 + 4xy = x2 + 3.
Solu onImplicitly differen a ng, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y =⇒ (3y2 + 4x)y′ = 2x− 4y =⇒ y′ =2x− 4y3y2 + 4x
Another ExampleExample
Find y′ along the curve y3 + 4xy = x2 + 3.
Solu onImplicitly differen a ng, we have
3y2y′ + 4(1 · y+ x · y′) = 2x
Solving for y′ gives
3y2y′ + 4xy′ = 2x− 4y =⇒ (3y2 + 4x)y′ = 2x− 4y =⇒ y′ =2x− 4y3y2 + 4x
Yet Another ExampleExample
Find y′ if y5 + x2y3 = 1+ y sin(x2).
Solu on
Yet Another ExampleExample
Find y′ if y5 + x2y3 = 1+ y sin(x2).
Solu onDifferen a ng implicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Yet Another ExampleExample
Find y′ if y5 + x2y3 = 1+ y sin(x2).
Solu onDifferen a ng implicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collect all terms with y′ on one side and all terms without y′ on theother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Yet Another ExampleExample
Find y′ if y5 + x2y3 = 1+ y sin(x2).
Solu onCollect all terms with y′ on one side and all terms without y′ on theother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Now factor and divide: y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
Finding tangents with implicit differentitiation
Example
Find the equa on of the linetangent to the curve
y2 = x2(x+ 1) = x3 + x2
at the point (3,−6).
..
SolutionSolu on
I Differen ate: 2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
I Thus the equa on of the tangent line is y+ 6 = −114(x− 3).
SolutionSolu on
I Differen ate: 2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
I Thus the equa on of the tangent line is y+ 6 = −114(x− 3).
Finding tangents with implicit differentitiation
Example
Find the equa on of the linetangent to the curve
y2 = x2(x+ 1) = x3 + x2
at the point (3,−6).
..
Recall: Line equation formsI slope-intercept form
y = mx+ b
where the slope ism and (0, b) is on the line.I point-slope form
y− y0 = m(x− x0)
where the slope ism and (x0, y0) is on the line.
Horizontal Tangent LinesExample
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solu onWe have to solve these two equa ons:
I y2 = x3 + x2 [(x, y) is on the curve]
I3x2 + 2x
2y= 0 [tangent line is horizontal]
Horizontal Tangent LinesExample
Find the horizontal tangent lines to the same curve: y2 = x3 + x2
Solu onWe have to solve these two equa ons:
I y2 = x3 + x2 [(x, y) is on the curve]
I3x2 + 2x
2y= 0 [tangent line is horizontal]
Solution, continuedI Solving the second equa on gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.
I Subs tu ng x = 0 into the first equa on gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down thatroad.
Solution, continuedI Solving the second equa on gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.
I Subs tu ng x = 0 into the first equa on gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down thatroad.
Solution, continuedI Solving the second equa on gives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Subs tu ng x = 0 into the first equa on gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down thatroad.
Solution, continued
I Subs tu ng x = −2/3 into the first equa on gives
y2 =(−23
)3
+
(−23
)2
=427
=⇒ y = ±√
427
= ± 23√3,
so there are two horizontal tangents.
Horizontal Tangents
...
(−2
3 ,2
3√3
)..(
−23 ,−
23√3
)
..
node
..(−1, 0)
Horizontal Tangents
...
(−2
3 ,2
3√3
)..(
−23 ,−
23√3
)..node
..(−1, 0)
Example
Find the ver cal tangent lines to the same curve: y2 = x3 + x2
Solu on
I Tangent lines are ver cal whendxdy
= 0.
I Differen a ng x implicitly as a func on of y gives2y = 3x2
dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(no ce this is the
reciprocal of dy/dx).I We must solve y2 = x3 + x2 [(x, y) is on the curve] and
2y3x2 + 2x
= 0 [tangent line is ver cal]
Example
Find the ver cal tangent lines to the same curve: y2 = x3 + x2
Solu on
I Tangent lines are ver cal whendxdy
= 0.
I Differen a ng x implicitly as a func on of y gives2y = 3x2
dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(no ce this is the
reciprocal of dy/dx).I We must solve y2 = x3 + x2 [(x, y) is on the curve] and
2y3x2 + 2x
= 0 [tangent line is ver cal]
Example
Find the ver cal tangent lines to the same curve: y2 = x3 + x2
Solu on
I Tangent lines are ver cal whendxdy
= 0.
I Differen a ng x implicitly as a func on of y gives2y = 3x2
dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(no ce this is the
reciprocal of dy/dx).
I We must solve y2 = x3 + x2 [(x, y) is on the curve] and2y
3x2 + 2x= 0 [tangent line is ver cal]
Example
Find the ver cal tangent lines to the same curve: y2 = x3 + x2
Solu on
I Tangent lines are ver cal whendxdy
= 0.
I Differen a ng x implicitly as a func on of y gives2y = 3x2
dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(no ce this is the
reciprocal of dy/dx).I We must solve y2 = x3 + x2 [(x, y) is on the curve] and
2y3x2 + 2x
= 0 [tangent line is ver cal]
Solution, continuedI Solving the second equa on gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).
I Subs tu ng y = 0 into the first equa on gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 is not allowed by the first equa on, but x = −1 is.
Solution, continuedI Solving the second equa on gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).I Subs tu ng y = 0 into the first equa on gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.
I x = 0 is not allowed by the first equa on, but x = −1 is.
Solution, continuedI Solving the second equa on gives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x ̸= 0).I Subs tu ng y = 0 into the first equa on gives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 is not allowed by the first equa on, but x = −1 is.
Tangents
...
(−2
3 ,2
3√3
)..(
−23 ,−
23√3
)..node
..(−1, 0)
ExamplesExample
Show that the families of curves xy = c, x2 − y2 = k are orthogonal,that is, they intersect at right angles.
Solu on
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is−1, so the tangent lines are perpendicularwherever they intersect.
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
Orthogonal Families of Curves
xy = cx2 − y2 = k .. x.
y
.
xy=1
.
xy=2
.
xy=3
.
xy=−1
.
xy=−2
.
xy=−3
.
x2−
y2=
1
.
x2−
y2=
2
.
x2−
y2=
3
.
x2 − y2 = −1
.
x2 − y2 = −2
.
x2 − y2 = −3
ExamplesExample
Show that the families of curves xy = c, x2 − y2 = k are orthogonal,that is, they intersect at right angles.
Solu on
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is−1, so the tangent lines are perpendicularwherever they intersect.
ExamplesExample
Show that the families of curves xy = c, x2 − y2 = k are orthogonal,that is, they intersect at right angles.
Solu on
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is−1, so the tangent lines are perpendicularwherever they intersect.
ExamplesExample
Show that the families of curves xy = c, x2 − y2 = k are orthogonal,that is, they intersect at right angles.
Solu on
I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx
I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy
I The product is−1, so the tangent lines are perpendicularwherever they intersect.
Ideal gases
The ideal gas law relatestemperature, pressure, andvolume of a gas:
PV = nRT
(R is a constant, n is theamount of gas in moles)
..Image credit: Sco Beale / Laughing Squid
CompressibilityDefini onThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
Approximately we have
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
The smaller the β, the “harder” the fluid.
CompressibilityDefini onThe isothermic compressibility of a fluid is defined by
β = −dVdP
1V
Approximately we have
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
The smaller the β, the “harder” the fluid.
Compressibility of an ideal gasExample
Find the isothermic compressibility of an ideal gas.
Solu onIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
So β = −1V· dVdP
=1P. Compressibility and pressure are inversely
related.
Compressibility of an ideal gasExample
Find the isothermic compressibility of an ideal gas.
Solu onIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
So β = −1V· dVdP
=1P. Compressibility and pressure are inversely
related.
Nonideal gassesNot that there’s anything wrong with thatExample
The van der Waals equa onmakesfewer simplifica ons:(
P+ an2
V2
)(V− nb) = nRT,
where a is a measure of a rac onbetween par cles of the gas, and b ameasure of par cle size.
...Oxygen.
.H
.
.H
.
.
Oxygen
.
.
H.
.
H
.
.
Oxygen .
.
H.
.
H
... Hydrogen bonds
Nonideal gassesNot that there’s anything wrong with that
Example
The van der Waals equa onmakesfewer simplifica ons:(
P+ an2
V2
)(V− nb) = nRT,
where a is a measure of a rac onbetween par cles of the gas, and b ameasure of par cle size.
...
Wikimedia Commons
Compressibility of a van der Waals gas
Differen a ng the van der Waals equa on by trea ng V as afunc on of P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
soβ = −1
VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Compressibility of a van der Waals gas
Differen a ng the van der Waals equa on by trea ng V as afunc on of P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
soβ = −1
VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Nonideal compressibility,continued
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Ques on
I What if a = b = 0?I Without taking the deriva ve, what is the sign of
dβdb
?
I Without taking the deriva ve, what is the sign ofdβda
?
Nonideal compressibility,continued
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Ques on
I What if a = b = 0?
I Without taking the deriva ve, what is the sign ofdβdb
?
I Without taking the deriva ve, what is the sign ofdβda
?
Nonideal compressibility,continued
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Ques on
I What if a = b = 0?I Without taking the deriva ve, what is the sign of
dβdb
?
I Without taking the deriva ve, what is the sign ofdβda
?
Nonideal compressibility,continued
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
Ques on
I What if a = b = 0?I Without taking the deriva ve, what is the sign of
dβdb
?
I Without taking the deriva ve, what is the sign ofdβda
?
Nasty derivativesAnswer
I We get the old (ideal) compressibilityI We have
dβdb
= −nV3
(an2 + PV2
)(PV3 + an2(2bn− V)
)2 < 0
I We havedβda
=n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)
)2 > 0 (as long as
V > 2nb, and it’s probably true that V ≫ 2nb).
Outline
The big idea, by example
ExamplesBasic ExamplesVer cal and Horizontal TangentsOrthogonal TrajectoriesChemistry
The power rule for ra onal powers
Using implicit differentiation tofind derivatives
Example
Finddydx
if y =√x.
Solu onIf y =
√x, then
y2 = x,
so2y
dydx
= 1 =⇒ dydx
=12y
=1
2√x.
Using implicit differentiation tofind derivatives
Example
Finddydx
if y =√x.
Solu onIf y =
√x, then
y2 = x,
so2y
dydx
= 1 =⇒ dydx
=12y
=1
2√x.
The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.
The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differen ate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.Now, differen ate implicitly:
qyq−1dydx
= pxp−1 =⇒ dydx
=pq· x
p−1
yq−1
The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =
pqxp/q−1.
Proof.Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so
xp−1
yq−1 =xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
Summary
I Using implicit differen a on we can treat rela ons which arenot quite func ons like they were func ons.
I In par cular, we can find the slopes of lines tangent to curveswhich are not graphs of func ons.