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. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Notes
Section 2.6
Implicit Differentiation
V63.0121.021, Calculus I
New York University
October 11, 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
Announcements
Notes
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 11, 2010 2 / 34
Objectives
Notes
Use implicit differentation to
find the derivative of a
function defined implicitly.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 3 / 34
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2. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Outline
Notes
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 4 / 34
Motivating Example
y Notes
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: y 2 = 1 − x 2 =⇒ y = − 1 − x 2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1 − x2 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 11, 2010 5 / 34
Motivating Example, another way
Notes
We know that x 2 + y 2 = 1 does not define y as a function of x, but
suppose it did.
Suppose we had y = f (x), so that
x 2 + (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 11, 2010 6 / 34
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3. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Yes, we can!
Notes
The beautiful fact (i.e., deep theorem) is that this works!
y
“Near” most points on the
curve x 2 + y 2 = 1, the curve
resembles the graph of a
function. looks like a function
So f (x) is defined “locally”,
x
almost everywhere and is
differentiable does not look like a
The chain rule then applies function, but that’s
for this local choice. OK—there are only
two points like this looks like a function
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 7 / 34
Motivating Example, again, with Leibniz notation
Notes
Problem
Find the slope of the line which is tangent to the curve x 2 + y 2 = 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 11, 2010 8 / 34
Summary
Notes
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 relation x
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 11, 2010 9 / 34
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4. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Outline
Notes
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 10 / 34
Another Example
Notes
Example
Find y along the curve y 3 + 4xy = x 2 + 3.
Solution
Implicitly differentiating, we have
3y 2 y + 4(1 · y + x · y ) = 2x
Solving for y gives
3y 2 y + 4xy = 2x − 4y
(3y 2 + 4x)y = 2x − 4y
2x − 4y
=⇒ y = 2
3y + 4x
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 11 / 34
Yet Another Example
Notes
Example
Find y if y 5 + x 2 y 3 = 1 + y sin(x 2 ).
Solution
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 12 / 34
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5. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Finding tangent lines with implicit differentitiation
Notes
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
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 13 / 34
Recall: Line equation forms
Notes
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 11, 2010 14 / 34
Horizontal Tangent Lines
Notes
Example
Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2
Solution
We have to solve these two equations:
3x 2 + 2x
= 0
y2 = x3 + x2 2y
1 [(x, y ) is on the curve] 2 [tangent line
is horizontal]
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 15 / 34
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6. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Solution, continued
Notes
Solving the second equation gives
3x 2 + 2x
= 0 =⇒ 3x 2 + 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
y 2 = 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
3 2
2 2 4 2
y2 = − + − = =⇒ y = ± √ ,
3 3 27 3 3
so there are two horizontal tangents.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 16 / 34
Tangents
Notes
− 2 , 3√3
3
2
(−1, 0)
− 2 , − 3√3
3
2 node
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 17 / 34
Example
Find the vertical tangent lines to the same curve: y 2 = x 3 + x 2 Notes
Solution
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 18 / 34
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7. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Solution, continued
Notes
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 19 / 34
Examples
Notes
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.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 21 / 34
Orthogonal Families of Curves
Notes
y
xy
xy
=
3
=
xy
2
=
x2 − y2 = 3
x −y =2
x2 − y2 = 1
1
xy = c
2
x2 − y2 = k x
1
−
2
− 2
=
x 2 − y 2 = −1
−
3
xy
=
x 2 − y 2 = −2
xy
=
x 2 − y 2 = −3
xy
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 22 / 34
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8. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Examples
Notes
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.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 23 / 34
Ideal gases
Notes
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
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 25 / 34
Compressibility
Notes
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 11, 2010 26 / 34
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9. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Compressibility of an ideal gas
Notes
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 11, 2010 27 / 34
Nonideal gasses
Not that there’s anything wrong with that Notes
Example
The van der Waals equation H
makes fewer simplifications:
Oxygen H
n2
P +a 2 (V − nb) = nRT ,
V H
Oxygen Hydrogen bonds
where P is the pressure, V the H
volume, T the temperature, n
the number of moles of the gas, Oxygen H
R a constant, a is a measure of
attraction between particles of H
the gas, and b a measure of
particle size.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 28 / 34
Compressibility of a van der Waals gas
Notes
Differentiating the van der Waals equation by treating V as a function of
P gives
an2 dV 2an2 dV
P+ + (V − bn) 1 − = 0,
V2 dP V 3 dP
so
1 dV V 2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV 3
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 11, 2010 29 / 34
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10. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
Nasty derivatives
Notes
dβ (2abn3 − an2 V + PV 3 )(nV 2 ) − (nbV 2 − V 3 )(2an3 )
=−
db (2abn3 − an2 V + PV 3 )2
nV 3 an2 + PV 2
=− <0
(PV 3 + an2 (2bn − V ))2
dβ n2 (bn − V )(2bn − V )V 2
= >0
da (PV 3 + an2 (2bn − V ))2
(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 11, 2010 30 / 34
Outline
Notes
The big idea, by example
Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 31 / 34
Using implicit differentiation to find derivatives
Notes
Example
dy √
Find if y = x.
dx
Solution
√
If y = x, then
y 2 = 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 11, 2010 32 / 34
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11. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010
The power rule for rational powers
Notes
Theorem
p
If y = x p/q , where p and q are integers, then y = x p/q−1 .
q
Proof.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 33 / 34
Summary
Notes
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 11, 2010 34 / 34
Notes
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