Quiz

The Mother of All Calculus Quizzes

Louis A. Talman, Ph.D.
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver

February 22, 2008

1 / 56

Question 1: A
Derivative
A Derivative I
A Derivative II
A Derivative III
A Derivative—The
Issue
Question 2: Increasing
Functions

Question 3: Concavity
Question 1: A Derivative
Question 4: Local
Minima
Question 5: Polar
Arc-Length

Question 6: Implicit
Differentiation
Functions
Question 8: Improper
Integrals

2 / 56

A Derivative I

Question 1: A
Derivative
A Derivative I
A Derivative II
A Derivative III Is it true that the function f given by
A Derivative—The
Issue
Question 2: Increasing x2 sin(1/x), when x = 0
Functions f (x) = (1)
Question 3: Concavity 0, when x = 0
Question 4: Local
Minima
is differentiable at x = 0? The differentiation rules give
Question 5: Polar
Arc-Length

Question 6: Implicit f ′ (x) = 2x sin(1/x) − cos(1/x), (2)
Differentiation
Functions and this is undeﬁned when x = 0. What gives?
Integrals

3 / 56

A Derivative II

Question 1: A
Derivative
A Derivative I
A Derivative II
A Derivative III
A Derivative—The
Issue Here’s what the Product Rule actually says:
Functions

If F (x) = u(x) · v(x), and if u′ (x0 ) and v ′ (x0 ) both exist, then
Question 4: Local
Minima
Question 5: Polar
1. F ′ (x0 ) exists, and
Arc-Length 2. is given by
Differentiation
Question 7: Implicit F ′ (x0 ) = u′ (x0 ) · v(x0 ) + u(x0 ) · v ′ (x0 ). (3)
Functions
Integrals

4 / 56

A Derivative III

Question 1: A
Derivative
A Derivative I
A Derivative II
A Derivative III For the function
A Derivative—The
Issue
Question 2: Increasing x2 sin(1/x), when x = 0
Functions f (x) = (4)
0, when x = 0,
Question 4: Local
Minima the difference quotient calculation gives us
Question 5: Polar
Arc-Length

′ f (h) − f (0) h2 sin(1/h) − 0
f (0) = lim = lim (5)
Differentiation
h→0 h h→0 h
Functions = lim h sin(1/h) = 0.• (6)
Question 8: Improper h→0
Integrals

5 / 56

A Derivative—The Issue

Question 1: A
Derivative
A Derivative I
A Derivative II
A Derivative III
A Derivative—The
Issue
The Issue
Functions

Question 3: Concavity When a theorem fails of applicability, that doesn’t
Question 4: Local
Minima
necessarily mean that no part of its conclusion can be
Question 5: Polar true.
Arc-Length

Differentiation
Functions
Integrals

6 / 56

Question 1: A
Derivative
Functions
Increasing Functions I
Increasing Functions II
Increasing Functions
III
IV
Increasing Functions V
Increasing Functions Functions
VI
VII
Increasing
Functions—The Issue


Question 4: Local
Minima
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals 7 / 56


Question 1: A
Derivative
Functions
III
Increasing Functions How can we say that the function f (x) = x3 is increasing on the
IV
Increasing Functions V interval [−1, 1], when f ′ (0) = 0 so that f isn’t increasing at 0?
VI
VII
Increasing


Question 4: Local
Minima
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals 8 / 56


Question 1: A
Derivative
Functions
Def: f is increasing on a set A whenever u ∈ A, v ∈ A, and u < v
Increasing Functions II implies f (u) < f (v).
III
Increasing Functions For the cubing function, we note that if u, v ∈ [−1, 1] with u < v then
IV
Increasing Functions V u − v < 0, whence
VI
VII
u3 − v 3 = (u − v)(u2 + uv + v 2 ) (7)
Increasing
√
 
2
 u+ v + v 3
2
Question 3: Concavity = (u − v)  < 0, (8)
Question 4: Local
2 2
Minima
Question 5: Polar
Arc-Length
so u3 < v 3 .•
Differentiation
Functions
Integrals 9 / 56

Increasing Functions III

Question 1: A
Derivative Theorem: If f is continuous on [a, b] and increasing on a dense subset D
Question 2: Increasing of [a, b], then f is increasing on [a, b].
Functions
Increasing Functions II Choose u, v ∈ [a, b], with u < v , and suppose that one or both of u, v do
III not lie in D . (Otherwise f (u) < f (v) and there is nothing to prove.)
IV
Increasing Functions V Select d0 ∈ (u, v) ∩ D . For each k ∈ N take αk−1 to be the midpoint of
VI (dk−1 , v)and choose dk ∈ (αk−1 , v) ∩ D.
VII
Increasing Then d0 < d1 < · · · < dk < dk+1 < · · ·, with limk→∞ dk = v ,so
f (d0 ) < f (d1 ) < · · · < f (dk ) < f (dk+1 ) < · · ·, with

Question 4: Local
limk→∞ f (dk ) = f (v).
Minima
Question 5: Polar It follows now that f (d0 ) < f (v).
Arc-Length

Question 6: Implicit Similarly, f (u) < f (d0 ), and thus f (u) < f (v).•
Differentiation
Functions
Integrals 10 / 56

Increasing Functions IV

Question 1: A
Derivative
It isn’t really clear what “increasing at 0” means. If g is given by
Functions
x/2 + x2 sin(1/x) when x = 0,
g(x) = (9)
0 when x = 0,
III
IV then g ′ (0) = 1/2. But g isn’t increasing on any interval (−δ, δ).
VI
VII
Increasing


Question 4: Local
Minima
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals 11 / 56


Question 1: A
Derivative
Functions
′ d x 2 1 1 1
g (0) = + x sin = +0= , (10)
III
dx x=0 2 x 2 2
IV
Increasing Functions V while x = 0 gives
VI
Increasing Functions 1
VII g ′ (x) = + 2x sin(1/x) − cos(1/x). (11)
Increasing 2

Question 3: Concavity So every interval (−δ, δ) contains sub-intervals on which f is
Question 4: Local
Minima
decreasing—even though f ′ (0) > 0.•
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals 12 / 56

Increasing Functions VI

Question 1: A
Derivative
Functions
Increasing Functions II 0.06
III
IV 0.04
VI 0.02
VII
Increasing

Question 3: Concavity 0.10 0.05 0.05 0.10
Question 4: Local
Minima 0.02
Question 5: Polar
Arc-Length

Question 6: Implicit 0.04
Differentiation
Functions 0.06
Integrals 13 / 56

Increasing Functions VII

Question 1: A
Derivative
Functions
Increasing Functions II 0.06
Increasing Functions y x 2 x2
III
IV 0.04
VI 0.02
VII
Increasing

Question 3: Concavity 0.10 0.05 0.05 0.10
Question 4: Local
Minima 0.02
Question 5: Polar
Arc-Length

Question 6: Implicit 0.04
Differentiation
Question 7: Implicit y x 2 x2
Functions 0.06
Integrals 14 / 56

Increasing Functions—The Issue

Question 1: A
Derivative
Functions
III
The Issue
IV
I The real problem here lies in failure to understand
VI
VII — the relationship between theorems and
Increasing
deﬁnitions, and, ultimately,

Question 4: Local
— ˆ
the role of deﬁnition in mathematics.
Minima
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals 15 / 56

Question 1: A
Derivative
Functions

Concavity I
Concavity II
Concavity—The Moral

Question 4: Local Question 3: Concavity
Minima
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals

16 / 56

Concavity I

Question 1: A
Derivative
Functions

Concavity I
Concavity II
Concavity—The Moral If y = 6x2 − x4 , then y ′′ = 12 − 12x2 , and this is positive exactly
Question 4: Local
Minima
when −1 < x < 1. Where is the curve concave upward?
Question 5: Polar
Arc-Length

Differentiation
Question 7: Implicit Is the answer “(−1, 1)”, or is it “[−1, 1]”?
Functions
Integrals

17 / 56

Concavity II

Question 1: A
Derivative
This is trickier than the last question. f is concave upward on an interval I
Question 2: Increasing provided:
Functions

I f ′′ (x) > 0 when x ∈ I . (G. L. Bradley & K. J. Smith, 1999; S. K. Stein, 1977)
Concavity I
Concavity II I f ′ is an increasing function on I . (R. Larson, R. Hostetler & B. H. Edwards,
Concavity—The Moral 2007; J. Stewart, 2005)
Question 4: Local
Minima
I The tangent line at each point of the curve lies (locally) below the curve in I .
Question 5: Polar
Arc-Length (C. H. Edwards & D. E. Penney, 2008; M. P. Fobes & R. B. Smyth, 1963)
Differentiation I f [(1 − λ)x1 ] + f (λx2 ) ≤ (1 − λ)f (x1 ) + λf (x2 ) when x1 , x2 ∈ I and
Functions
0 < λ < 1. (G. B. Thomas, Jr., 1972)
Integrals I {(x, y) : x ∈ I ⇒ y ≥ f (x)} is a convex set. (R. P. Agnew, 1962)

18 / 56


Question 1: A
Derivative
Functions

Concavity I
Concavity II The Moral?

Question 4: Local
Minima
Question 5: Polar
Arc-Length
Read your author’s deﬁnitions.
Differentiation
Functions
Integrals

19 / 56

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Local Minima I
Local Minima II Question 4: Local Minima
Local Minima III
Local Minima IV
Local Minima—The
Issue
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals

20 / 56

Local Minima I

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Local Minima I
Local Minima II
If a smooth function f has a local minimum at x = x0 , must there be
Local Minima III δ > 0 so that f ′ (x) ≤ 0 on (x0 − δ, x0 ) but f ′ (x) ≥ 0 on
Local Minima IV
Local Minima—The (x0 , x0 + δ)?
Issue
Question 5: Polar
Arc-Length

Differentiation
Functions
Integrals

21 / 56

Local Minima II

Question 1: A
Derivative
Functions
The First Derivative Test, of course, says:

Question 4: Local If there is a δ > 0 such that f ′ is negative on (x0 − δ, x0 ) and positive on
Minima
Local Minima I (x0 , x0 + δ), then f has a local minimum at x = x0 .
Local Minima II
Local Minima III
Local Minima IV
Local Minima—The And here’s a counter example to the converse:
Issue
Question 5: Polar
Arc-Length 4x4 − 3x4 cos(1/x), when x = 0;
f (x) = (12)
Differentiation 0, when x = 0.
Functions
Integrals

22 / 56

Local Minima III

Question 1: A
Derivative
Functions

0.0015
Question 4: Local
Minima
Local Minima I
Local Minima II
Local Minima III
Local Minima IV
Local Minima—The 0.0010
Issue
Question 5: Polar
Arc-Length

Differentiation
Functions
0.0005
Integrals

0.2 0.1 0.1 0.2
23 / 56

Local Minima IV

Question 1: A
Derivative
Functions

0.0015
Question 4: Local
Minima y 7x4
Local Minima I
Local Minima II
Local Minima III
Local Minima IV
Local Minima—The 0.0010
Issue
Question 5: Polar
Arc-Length

Differentiation
Functions
0.0005
Integrals

y x4

0.2 0.1 0.1 0.2
24 / 56

Local Minima—The Issue

Question 1: A
Derivative
Functions


The issue?
Question 4: Local
Minima
Local Minima I
Local Minima II
Local Minima III
Local Minima IV
Local Minima—The
Issue
Confusion of a theorem with its converse,
Question 5: Polar
Arc-Length among other things.
Differentiation
Functions
Integrals

25 / 56

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar
Arc-Length Question 5: Polar Arc-Length
Polar Arc-Length I
Polar Arc-Length II
Polar Arc-length III
Polar Arc-length IV

Differentiation
Functions
Integrals

26 / 56

Polar Arc-Length I

Question 1: A
Derivative
Functions
Why don’t we approach arc-length in polar coordinates the way we
do in cartesian coordinates?
Question 4: Local
Minima
Question 5: Polar
Arc-Length
Polar Arc-Length I In cartesian coordinates:
Polar Arc-Length II
Polar Arc-length IV s = lim (xk − xk−1 )2 + [f (xk ) − f (xk−1 )]2 (13)
Differentiation = lim (xk − xk−1 )2 + [f ′ (ξk )]2 (xk − xk−1 )2 (by MVT) (14)
Functions
= lim 1 + [f ′ (ξk )]2 (xk − xk−1 ) (15)
Integrals
b
= 1 + [f ′ (x)]2 dx (16)
a

27 / 56

Polar Arc-Length II

Question 1: A
Derivative
When r = f (θ) in polar coordinates (so that f (θk ) = rk ), the Law of Cosines
Question 2: Increasing gives:
Functions

Question 3: Concavity 2 2
s = lim rk + rk−1 − 2rk rk−1 cos(θk − θk−1 ) (17)
Question 4: Local
Minima
Question 5: Polar
= lim (rk − rk−1 )2 + 2(1 − cos ∆θk )rk rk−1 (18)
Arc-Length
Polar Arc-Length I 1 − cos ∆θk
Polar Arc-Length II = lim [f ′ (ξk )]2 + 2f (θk )f (θk−1 ) 2
∆θk . (19)
Polar Arc-length III (∆θk )
Polar Arc-length IV

Question 6: Implicit This is not a Riemann sum. . . and I see no way to fudge it into one.
Differentiation
The fact that
Functions
1 − cos t 1
lim = (20)
Integrals
t→0+ t2 2
is very suggestive—though not particularly helpful.

28 / 56


Question 1: A
Derivative Duhamel’s Theorem (Standard Model)
Functions

Question 3: Concavity Theorem1 : Let f be a continuous function of three variables on
Question 4: Local
Minima
[a, b] × [a, b] × [a, b]. If P = {x0 , x1 , . . . , xn }, where
Question 5: Polar a = x0 < x1 < x2 < · · · < xn−1 < xn = b, is a partition of [a, b], with
Arc-Length
xk−1 ≤ ξk , ηk , ζk ≤ xk for k = 1, 2, . . . , n, then for every ǫ > 0 there is a
Polar Arc-Length I
Polar Arc-Length II δ > 0 such that whenever P < δ it follows that
Polar Arc-length IV n b
Question 6: Implicit f (ξk , ηk , ζk )(xk − xk−1 ) − f (t, t, t) dt < ǫ. (21)
Differentiation
k=1 a
Functions
Integrals

1
Adapted from Advanced Calculus, David V. Widder, Second Edition, Prentice-Hall, 1961,
and reprinted by Dover, 1989; p 174.
29 / 56

Polar Arc-length IV

Question 1: A
Derivative Duhamel’s Theorem (Deluxe Model)
Functions

Theorem: Let η > 0, and suppose that F is a continuous function from
Question 4: Local [a, b] × [a, b] × [a, b] × [0, η] to R. To each partition
Minima
Question 5: Polar
P = {x0 , x1 , . . . , xn }, where a = x0 < x1 < · · · < xn = b, and to
Arc-Length each choice of triples of numbers ξk , ηk , ζk ∈ [xk−1 , xk ], k = 1, . . . , n,
Polar Arc-Length I
Polar Arc-Length II we associate the sum
Polar Arc-length IV n
Question 6: Implicit S(F, [a, b], P, {(ξk , ηk , ζk )}n ) =
k=1 F (ξk , ηk , ζk , ∆xk ) ∆xk .
Differentiation
k=1
Functions
Question 8: Improper If ǫ > 0, there is a δ > 0 such that P < δ implies
Integrals

b
S(F, [a, b], P, {(ξk , ηk , ζk )}n )
k=1 − F (t, t, t, 0) dt < ǫ.
a

30 / 56

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar Question 6: Implicit
Arc-Length

Question 6: Implicit Differentiation
Differentiation
Implicit Differentiation I
Implicit Differentiation
II
III
IV
V
VI
VII
Implicit
Differentiation—The
Issue
Functions 31 / 56


Question 1: A
Derivative y 2 (2 − x)
Given the problem “Find y ′ when 2 2+1
= 1,”
Functions
x +y
Question 3: Concavity I Æthelbert differentiated both sides (correctly), solved (correctly),
Question 4: Local
′ y(x2 − y 2 − 4x − 1)
Minima and got yÆ = 3 − 2x2 + x − 2)
.
Question 5: Polar
2(x
Arc-Length

Differentiation I Brunhilde multiplied through by x2 + y 2 + 1 (correctly) before
¨
she differentiated (correctly), and when she solved (correctly),
II
′ 2x + y 2
she got yB = .
III
2y(1 − x)
IV
V
VI
Who was wrong?
VII
Implicit
Issue
Functions 32 / 56

Implicit Differentiation II

Here are the slope ﬁelds2 :
Question 1: A
Derivative
Functions
2 2

Question 4: Local
Minima 1 1
Question 5: Polar
Arc-Length

Differentiation 2 1 1 2 2 1 1 2
II 1 1
III
IV 2 2
V
Implicit Differentiation ′ ′
VI yÆ yB
VII
Implicit
2
Issue My thanks to Prof. Diane Davis for the ideas that underlie the Mathematica code I used
Question 7: Implicit to generate these slope ﬁelds.
Functions 33 / 56

Implicit Differentiation III

Question 1: A
Derivative Pick a point on the curve—say (−1, 1):
Functions

Question 3: Concavity y 2 (2 − x) 12 (3)
I = = 1, so (−1, 1) is on the curve.
Question 4: Local
Minima
x2 + y 2 + 1 (−1,1) 3
Question 5: Polar
Arc-Length

Question 6: Implicit ′ y(x2 − y 2 − 4x − 1) 1 · (3) 1
Differentiation I yÆ = = =− .
Implicit Differentiation I (−1,1) 2(x3 − 2x2 + x − 2) (−1,1) 2 · (−6) 4
II
III
′ 2x + y 2 −2 + 1 1
Implicit Differentiation I yB = = =− .
IV
Implicit Differentiation (−1,1) 2y(1 − x) (−1,1) 2·1·2 4
V
VI
VII
The issue goes away.
Implicit
Issue
Functions 34 / 56

Implicit Differentiation IV

Question 1: A
Derivative
Functions

Question 3: Concavity Let y be deﬁned implicitly as a function of x by
Question 4: Local
Minima
F (x, y)
Question 5: Polar = H(x, y). (22)
Arc-Length G(x, y)
Differentiation
Then
II
′ Fx G − F Gx − G2 Hx
yÆ = − , and (23)
III
Fy G − F Gy − G2 Hy
IV
Implicit Differentiation ′ Fx − Gx H − GHx
V yB = − . (24)
Implicit Differentiation Fy − Gy H − GHy
VI
VII
Implicit
It is easy to use (22) to reduce(23) to (24).
Issue
Functions 35 / 56

Implicit Differentiation V

Question 1: A
Derivative
Exercise: Assume that
Functions y 2 (2 − x)
2 + y2 + 1
= 1, (25)
x
Question 4: Local
Minima
and show—without using the analysis just given—how to reduce
Question 5: Polar
Arc-Length

′y(x2 − y 2 − 4x − 1)
Differentiation yÆ = (26)
Implicit Differentiation I 2(x3 − 2x2 + x − 2)
II
III
to
IV
′ 2x + y 2
V yB = . (27)
VI
2y(1 − x)
VII
Implicit
Issue
Functions 36 / 56

Implicit Differentiation VI

Question 1: A
Derivative 2
Functions


Question 4: Local
Minima
1
Question 5: Polar
Arc-Length

Differentiation
II 2 1 1 2
III
IV
V
Implicit Differentiation 1
VI
VII
Implicit
Issue
Question 7: Implicit 2
Functions 37 / 56

Implicit Differentiation VII

Question 1: A
Derivative
The implicit differentiation technique is justiﬁed by the
Functions
Implicit Function Theorem: Let f be a smooth real-valued function

Question 4: Local
deﬁned on an open subset D of R, and let (x0 , y0 ) be a solution of the
Minima
equation f (x, y) = 0. If fy (x0 , y0 ) = 0, there are positive numbers, ǫ
Question 5: Polar
Arc-Length and δ , and a smooth function ϕ : (x0 − δ, x0 + δ) → (y0 − ǫ, y0 + ǫ)
Differentiation
such that for each x ∈ (x0 − δ, x0 + δ), y = ϕ(x) is the only solution of
Implicit Differentiation I f (x, y) = 0 lying in (y0 − ǫ, y0 + ǫ). Moreover, for each
II x ∈ (x0 − δ, x0 + δ),
III
′ fx [x, ϕ(x)]
IV
ϕ (x) = − . (28)
V fy [x, ϕ(x)]
VI
VII
Implicit
Issue
Functions 38 / 56

Implicit Differentiation—The Issue

Question 1: A
Derivative
Functions
The Issue

Question 4: Local
Minima
Question 5: Polar
Our textbook problems encourage students (and
Arc-Length

teachers) to think about these problems
Differentiation
II
I globally instead of locally, and
III
IV
I without considering the hypotheses needed to justify
V
Implicit Differentiation what they are doing.
VI
VII
Implicit
Issue
Functions 39 / 56

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar
Arc-Length Question 7: Implicit Functions
Differentiation
Functions
Implicit Functions I
Implicit Functions II
Implicit Functions III
Implicit Functions IV
Implicit Functions V
Implicit Functions VI
Implicit
Functions—The
Issues
Integrals

40 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima When I apply implicit differentiation to the equation
Question 5: Polar
Arc-Length
(x2 + y 2 )2 = x2 − y 2 to ﬁnd y ′ , I get
Differentiation
′ x(2y 2 + 2x2 − 1)
Question 7: Implicit y =− 2 + 2x2 + 1)
, (29)
Functions y(2y
which gives the indeterminate form 0/0 at the origin. Can I use limits
Implicit Functions IV to ﬁnd the slope of the line tangent to this curve at the origin? How?
Implicit
Functions—The
Issues
Integrals

41 / 56


Question 1: A y
Derivative
Functions 1.0


Question 4: Local
Minima
Question 5: Polar 0.5
Arc-Length

Differentiation
Functions 0.0 x
Implicit Functions V 0.5
Implicit
Functions—The
Issues
Question 8: Improper 1.0
Integrals

1.0 0.5 0.0 0.5 1.0
42 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar The surplus of tangent lines at (0, 0) results from the fact that there is no
Arc-Length
open rectangle, centered at (0, 0), whose intersection with the curve is the
Differentiation graph of a function.
Functions
But the conclusion of the Implicit Function Theorem asserts that there is
Implicit Functions II such a rectangle. Because the conclusion is false, the IFT must not apply
Implicit Functions IV to this function at the origin.
Implicit
Functions—The
Issues
Integrals

43 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar Actually, we should have known that the IFT doesn’t apply:
Arc-Length

Question 6: Implicit Putting F (x, y) = (x2 + y 2 )2 − x2 + y 2 , we have
Differentiation
Functions
Implicit Functions I Fy (0, 0) = 4(x2 + y 2 )y + 2y = 0, (30)
Implicit Functions II (0,0)
so that one of the hypotheses of the IFT fails.
Implicit
Functions—The
Issues
Integrals

44 / 56


Question 1: A
Derivative
Functions

Question 3: Concavity We could ﬁnd the slope of either branch of the curve by using the implicit
Question 4: Local
Minima derivative if we were to solve, algebraically, for y in terms of x and then
Question 5: Polar replace y with the solution throughout the implicit differentiation expression
Arc-Length

for y ′ —and take the limit as we approach the origin.
Differentiation
Question 7: Implicit That’s nice. . . except that the whole point of implicit differentiation is to
Functions
Implicit Functions I cirvumvent the necessity of solving for y in terms of x. . .
Implicit
Functions—The
Issues
Integrals

45 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar If we absolutely must have the slope of a branch of the curve as it passes
Arc-Length
through the origin, the best option is probably to re-parametrize. In this
Differentiation case, polar coordinates work nicely. They give us the equation
Functions r2 = cos 2θ for our curve, and it’s easy to see from this that the slopes of
the two tangent lines are ±1.
Implicit
Functions—The
Issues
Integrals

46 / 56

Implicit Functions—The Issues

Question 1: A
Derivative
Functions

Question 3: Concavity Two Issues:
Question 4: Local
Minima
Question 5: Polar
Arc-Length
I Hypotheses, hypotheses, hypotheses!
Differentiation
Functions I In this case, 0/0 isn’t an “indeterminate form”— it’s
Implicit Functions II undeﬁned!
Implicit
Functions—The
Issues
Integrals

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Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar
Arc-Length Question 8: Improper Integrals
Differentiation
Functions
Integrals
Improper Integrals I
Improper Integrals II
Improper Integrals III
Improper Integrals IV
Improper Integrals V
Improper Integrals VI
Improper
Integrals—The Moral

48 / 56


Question 1: A
Derivative
Why don’t we use
1 t
Functions 2x 2x
Question 3: Concavity 2
dx = lim 2
dx (31)
−1 1−x t→1− −t 1−x
Question 4: Local
Minima t
Question 5: Polar = lim ln(1 − x2 ) (32)
Arc-Length t→1− −t
Differentiation = lim {ln(1 − t2 ) − ln[1 − (−t)2 )]} = 0 (33)
Question 7: Implicit t→1−
Functions
Question 8: Improper as the elementary-calculus deﬁnition for that improper integral?
Integrals
It would make freshman life so much easier.
Improper

49 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima The calculation we have just examined gives something called the “Cauchy
Question 5: Polar Principal Value” (CPV) of the improper integral. The CPV is written
Arc-Length

1
Differentiation 2x
Question 7: Implicit PV 2
dx.
Functions −1 1−x
Integrals
Improper

50 / 56


Question 1: A
Derivative
Functions


Question 4: Local Why not use the CPV in elementary Calculus?
Minima
Question 5: Polar
Arc-Length The short answer:
Differentiation
Using Cauchy Principal Values would break the equation
Functions
1 ξ 1
Question 8: Improper 2x 2x 2x
Integrals
2
dx = 2
dx + 2
dx. (34)
−1 1−x −1 1−x ξ 1−x
Improper

51 / 56


Question 1: A
Derivative
The long answer:
Functions Choose B , with |B| > 1. Let P be the polynomial function given by

Question 4: Local P (u) = (u − 1)(u + 1)(u + 2B − 1)(u + 2B + 1), (35)
Minima
Question 5: Polar
Arc-Length and put
Differentiation
P ′ (u)
f (u) = − . (36)
Functions
P (u)
Integrals
Improper Integrals I Then f is continuous in (−1, 1), and a tedious calculation shows that
1
Improper Integrals IV B−1
Improper Integrals V PV f (u) du = ln . (37)
Improper Integrals VI −1 B+1
Improper

52 / 56


Question 1: A
Derivative
Functions
Now put g(u) = (u2 + 2Bu − 1)/(2B), and note that g(−1) = −1

Question 4: Local
while g(1) = 1. Putting x = g(u), we ought therefore to be able to write
Minima
Question 5: Polar 1 1
Arc-Length
PV F (x) dx = PV F [g(u)]g ′ (u) du, (38)
Differentiation
−1 −1
Functions where F (x) = 2x/(1 − x2 ).
Integrals
However, the CPV on the left side of (38) is zero, as we have seen; the
Improper Integrals II integrand on the right side of (38) turns out to be the integrand of (37),
B−1
Improper Integrals IV above, and so that CPV is ln . As it happens, the single value this
Improper Integrals V B+1
Improper Integrals VI latter quantity cannot assume is zero.
Improper

53 / 56


Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar
Arc-Length
A very good reason for not using Cauchy Principal Values in
Question 6: Implicit elementary calculus is that they break the Substitution Theorem for
Differentiation
Deﬁnite Integrals—which is too valuable a theorem to give up. . . in
Functions elementary calculus.
Integrals
Improper

54 / 56

Improper Integrals—The Moral

Question 1: A
Derivative
Functions


Question 4: Local Moral
Minima
Question 5: Polar
Arc-Length
The deﬁnitions we adopt condition the theorems we can
Differentiation prove. We have chosen (not necessarily consciously or
Functions
with full knowledge) the deﬁnitions we use in elementary
Question 8: Improper calculus to support the tools we want students to be able
Integrals
to use.
Improper

55 / 56

Question 1: A
Derivative
Functions


Question 4: Local
Minima
Question 5: Polar
Arc-Length

Differentiation
The End
Functions
Integrals
Improper

56 / 56

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