Lesson 11: Implicit Differentiation (handout)
-
Upload
matthew-leingang -
Category
Technology
-
view
1.518 -
download
0
description
Transcript of Lesson 11: Implicit Differentiation (handout)
.
.
..
Sec on 2.6Implicit Differen a on
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
February 28, 2011
.
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.
.
Notes
.
Notes
.
Notes
. 1.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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, 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
.
.
Notes
.
Notes
.
Notes
. 2.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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
..
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.
.
Notes
.
Notes
.
Notes
. 3.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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
.
Notes
.
Notes
.
Notes
. 4.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
Yet Another ExampleExample
Find y′ if y5 + x2y3 = 1+ y sin(x2).
Solu on
Differen 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)
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).
.
Notes
.
Notes
.
Notes
. 5.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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]
.
Notes
.
Notes
.
Notes
. 6.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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.
.
Tangents
...
(−2
3 ,2
3√3
)..(
−23 ,−
23√3
)..node
..(−1, 0)
.
Notes
.
Notes
.
Notes
. 7.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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, continued
I 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.
.
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.
.
Notes
.
Notes
.
Notes
. 8.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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.
.
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
.
Notes
.
Notes
.
Notes
. 9.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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.
.
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
.
Notes
.
Notes
.
Notes
. 10.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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
?
.
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).
.
Notes
.
Notes
.
Notes
. 11.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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.
.
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
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
.
Notes
.
Notes
.
Notes
. 12.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011
.
.
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.
.
.
.
Notes
.
Notes
.
Notes
. 13.
. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011