The Mother Of All Calculus Quizzes

The Mother of All Calculus Quizzes Louis A. Talman, Ph.D. Department of Mathematical & Computer Sciences Metropolitan State College of Denver February 22, 2008

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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 Question 7: Implicit Functions Question 8: Improper Integrals

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A Derivative I

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 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

Is it true that the function f given by

f (x) =

(

x2 sin(1/x), 0,

when x 6= 0

when x = 0

(1)

is differentiable at x = 0? The differentiation rules give

f ′ (x) = 2x sin(1/x) − cos(1/x),

(2)

and this is undefined when x = 0. What gives?

Question 8: Improper Integrals

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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:

Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

If F (x) = u(x) · v(x), and if u′ (x0 ) and v ′ (x0 ) both exist, then 1. F ′ (x0 ) exists, and 2. is given by

F ′ (x0 ) = u′ (x0 ) · v(x0 ) + u(x0 ) · v ′ (x0 ).

(3)

Question 8: Improper Integrals

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A Derivative III

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

For the function

( x2 sin(1/x), f (x) = 0,

Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

when x 6= 0

when x = 0,

(4)

the difference quotient calculation gives us

h2 sin(1/h) − 0 f (h) − f (0) = lim f (0) = lim h→0 h→0 h h = lim h sin(1/h) = 0.• ′

h→0

(5) (6)

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A Derivative—The Issue

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

The Issue

Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length

When a theorem fails of applicability, that doesn’t necessarily mean that no part of its conclusion can be true.

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue

Question 2: Increasing Functions

Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions I

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue

How can we say that the function f (x) = x3 is increasing on the interval [−1, 1], when f ′ (0) = 0 so that f isn’t increasing at 0?

Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions II

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation

Def: f is increasing on a set A whenever u ∈ A, v ∈ A, and u < v implies f (u) < f (v). For the cubing function, we note that if u, v ∈ [−1, 1] with u < v then u − v < 0, whence

u3 − v 3 = (u − v)(u2 + uv + v 2 )  !2  √  2 v v 3   = (u − v) u + < 0, + 2 2

(7) (8)

so u3 < v 3 .•

Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions III

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation

Theorem: If f is continuous on [a, b] and increasing on a dense subset D of [a, b], then f is increasing on [a, b]. Choose u, v ∈ [a, b], with u < v , and suppose that one or both of u, v do not lie in D . (Otherwise f (u) < f (v) and there is nothing to prove.) Select d0 ∈ (u, v) ∩ D . For each k ∈ N take αk−1 to be the midpoint of (dk−1 , v)and choose dk ∈ (αk−1 , v) ∩ D. Then d0 < d1 < · · · < dk < dk+1 < · · ·, with limk→∞ dk = v ,so f (d0 ) < f (d1 ) < · · · < f (dk ) < f (dk+1 ) < · · ·, with limk→∞ f (dk ) = f (v). It follows now that f (d0 ) < f (v). Similarly, f (u) < f (d0 ), and thus f (u) < f (v).•

Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions IV

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV

It isn’t really clear what “increasing at 0” means. If g is given by

( x/2 + x2 sin(1/x) when x 6= 0, g(x) = 0 when x = 0,

(9)

then g ′ (0) = 1/2. But g isn’t increasing on any interval (−δ, δ).

Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions V

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length

d ′ g (0) = dx

x=0



  1 1 1 x 2 + x sin = +0= , 2 x 2 2

(10)

while x 6= 0 gives

g ′ (x) =

1 + 2x sin(1/x) − cos(1/x). 2

(11)

So every interval (−δ, δ) contains sub-intervals on which f is decreasing—even though f ′ (0) > 0.•

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Increasing Functions VI

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I

0.06

Increasing Functions II Increasing Functions III Increasing Functions IV

0.04

Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity Question 4: Local Minima

0.02

-0.10

0.05

-0.05

0.10

-0.02

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

-0.04 -0.06 13 / 56

Increasing Functions VII

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I

0.06

Increasing Functions II Increasing Functions III Increasing Functions IV

0.04

Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue Question 3: Concavity

y = x  2 + x2

0.02

-0.10

0.05

-0.05

Question 4: Local Minima

0.10

-0.02

Question 5: Polar Arc-Length

-0.04

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

y = x  2 - x2

-0.06 14 / 56

Increasing Functions—The Issue

Question 1: A Derivative Question 2: Increasing Functions Increasing Functions I Increasing Functions II Increasing Functions III Increasing Functions IV Increasing Functions V Increasing Functions VI Increasing Functions VII Increasing Functions—The Issue

The Issue n

The real problem here lies in failure to understand —

Question 3: Concavity Question 4: Local Minima

—

the relationship between theorems and definitions, and, ultimately, ˆ of definition in mathematics. the role

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Concavity I Concavity II Concavity—The Moral Question 4: Local Minima

Question 3: Concavity

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Concavity I

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Concavity I Concavity II Concavity—The Moral Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

If y = 6x2 − x4 , then y ′′ = 12 − 12x2 , and this is positive exactly when −1 < x < 1. Where is the curve concave upward?

Is the answer “(−1, 1)”, or is it “[−1, 1]”?

Question 8: Improper Integrals

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Concavity II

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Concavity I Concavity II Concavity—The Moral Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

This is trickier than the last question. f is concave upward on an interval I provided:

n f ′′ (x) > 0 when x ∈ I . (G. L. Bradley & K. J. Smith, 1999; S. K. Stein, 1977) n f ′ is an increasing function on I . (R. Larson, R. Hostetler & B. H. Edwards, 2007; J. Stewart, 2005)

n The tangent line at each point of the curve lies (locally) below the curve in I . (C. H. Edwards & D. E. Penney, 2008; M. P. Fobes & R. B. Smyth, 1963)

n f [(1 − λ)x1 ] + f (λx2 ) ≤ (1 − λ)f (x1 ) + λf (x2 ) when x1 , x2 ∈ I and

0 < λ < 1. (G. B. Thomas, Jr., 1972)

n {(x, y) : x ∈ I ⇒ y ≥ f (x)} is a convex set. (R. P. Agnew, 1962)

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Concavity—The Moral

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Concavity I Concavity II

The Moral?

Concavity—The Moral Question 4: Local Minima Question 5: Polar Arc-Length

Read your author’s definitions.

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity 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 Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Local Minima I

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Local Minima I Local Minima II Local Minima III Local Minima IV Local Minima—The Issue

If a smooth function f has a local minimum at x = x0 , must there be δ > 0 so that f ′ (x) ≤ 0 on (x0 − δ, x0 ) but f ′ (x) ≥ 0 on (x0 , x0 + δ)?

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Local Minima II

Question 1: A Derivative Question 2: Increasing Functions

The First Derivative Test, of course, says:

Question 3: Concavity Question 4: Local Minima Local Minima I

If there is a δ > 0 such that f ′ is negative on (x0 − δ, x0 ) and positive on (x0 , x0 + δ), then f has a local minimum at x = x0 .

Local Minima II Local Minima III Local Minima IV Local Minima—The Issue Question 5: Polar Arc-Length Question 6: Implicit Differentiation

And here’s a counter example to the converse:

( 4x4 − 3x4 cos(1/x), when x 6= 0; f (x) = 0, when x = 0.

(12)

Question 7: Implicit Functions Question 8: Improper Integrals

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Local Minima III

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

0.0015

Question 4: Local Minima Local Minima I Local Minima II Local Minima III Local Minima IV Local Minima—The Issue

0.0010

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

0.0005

Question 8: Improper Integrals

-0.2

-0.1

0.1

0.2/ 56 23

Local Minima IV

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

0.0015

Question 4: Local Minima

y = 7x4

Local Minima I Local Minima II Local Minima III Local Minima IV Local Minima—The Issue

0.0010

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

0.0005

Question 8: Improper Integrals

y = x4 -0.2

-0.1

0.1

0.2/ 56 24

Local Minima—The Issue

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Local Minima I

The issue?

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

Confusion of a theorem with its converse, among other things.

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity 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 Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

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Polar Arc-Length I

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

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 III Polar Arc-length IV Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

s = lim = lim

Xp

Xp

Xp

(xk − xk−1 )2 + [f (xk ) − f (xk−1 )]2

(13)

(xk − xk−1 )2 + [f ′ (ξk )]2 (xk − xk−1 )2 (by MVT) (14)

1 + [f ′ (ξk )]2 (xk − xk−1 ) = lim Z bp 1 + [f ′ (x)]2 dx =

(15) (16)

a

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Polar Arc-Length II

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Polar Arc-Length I Polar Arc-Length II Polar Arc-length III Polar Arc-length IV Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

When r = f (θ) in polar coordinates (so that f (θk ) = rk ), the Law of Cosines gives:

Xq 2 s = lim − 2rk rk−1 cos(θk − θk−1 ) rk2 + rk−1 Xp (rk − rk−1 )2 + 2(1 − cos ∆θk )rk rk−1 = lim s X 1 − cos ∆θk = lim [f ′ (ξk )]2 + 2f (θk )f (θk−1 ) ∆θk . 2 (∆θk )

(17) (18) (19)

This is not a Riemann sum. . . and I see no way to fudge it into one. The fact that

1 1 − cos t = lim t2 2 t→0+

(20)

is very suggestive—though not particularly helpful.

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Polar Arc-length III

Duhamel’s Theorem (Standard Model)

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Polar Arc-Length I Polar Arc-Length II Polar Arc-length III

Theorem1 : Let f be a continuous function of three variables on [a, b] × [a, b] × [a, b]. If P = {x0 , x1 , . . . , xn }, where a = x0 < x1 < x2 < · · · < xn−1 < xn = b, is a partition of [a, b], with xk−1 ≤ ξk , ηk , ζk ≤ xk for k = 1, 2, . . . , n, then for every ǫ > 0 there is a δ > 0 such that whenever kPk < δ it follows that

n Z b X f (ξk , ηk , ζk )(xk − xk−1 ) − f (t, t, t) dt < ǫ. a

Polar Arc-length IV Question 6: Implicit Differentiation

(21)

k=1

Question 7: Implicit Functions Question 8: Improper 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 Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Polar Arc-Length I Polar Arc-Length II

Duhamel’s Theorem (Deluxe Model) Theorem: Let η > 0, and suppose that F is a continuous function from [a, b] × [a, b] × [a, b] × [0, η] to R. To each partition P = {x0 , x1 , . . . , xn }, where a = x0 < x1 < · · · < xn = b, and to each choice of triples of numbers ξk , ηk , ζk ∈ [xk−1 , xk ], k = 1, . . . , n, we associate the sum

Polar Arc-length III Polar Arc-length IV Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

S(F, [a, b], P, {(ξk , ηk , ζk )}nk=1 ) =

n X

F (ξk , ηk , ζk , ∆xk ) ∆xk .

k=1

If ǫ > 0, there is a δ > 0 such that kPk < δ implies

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

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation

Question 6: Implicit Differentiation

Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

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Implicit Differentiation I

Question 2: Increasing Functions

2 (2 − x) y Given the problem “Find y ′ when 2 = 1,” 2 x +y +1

Question 3: Concavity

n

Æthelbert differentiated both sides (correctly), solved (correctly),

n

Brunhilde ¨ multiplied through by x2 + y 2 + 1 (correctly) before she differentiated (correctly), and when she solved (correctly),

Question 1: A Derivative

Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

2 − y 2 − 4x − 1) y(x ′ = . and got yÆ 3 2 2(x − 2x + x − 2)

2 2x + y she got yB′ = . 2y(1 − x)

Who was wrong?

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Implicit Differentiation II

Question 1: A Derivative

Here are the slope fields2 :

Question 2: Increasing Functions

2

2

1

1

Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

-2

1

-1

2

-2

1

-1

-1

-1

-2

-2

′ yÆ

2

yB′

2

My thanks to Prof. Diane Davis for the ideas that underlie the Mathematica code I used to generate these slope fields. 33 / 56

Implicit Differentiation III

Question 1: A Derivative

Pick a point on the curve—say (−1, 1):

Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

y 2 (2

− x) n x2 + y 2 + 1

(−1,1)

′ n yÆ

(−1,1)

′ n yB

(−1,1)

y(x2

12 (3) = 1, so (−1, 1) is on the curve. = 3 y2

− − 4x − 1) 1 · (3) 1 = = =− . 3 2 2(x − 2x + x − 2) (−1,1) 2 · (−6) 4 y2

2x + = 2y(1 − x)

= (−1,1)

−2 + 1 1 =− . 2·1·2 4

The issue goes away.

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Implicit Differentiation IV

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

Let y be defined implicitly as a function of x by

Question 4: Local Minima

F (x, y) = H(x, y). G(x, y)

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

(22)

Then

Fx G − F Gx − G2 Hx yÆ = − , and Fy G − F Gy − G2 Hy Fx − Gx H − GHx yB′ = − . Fy − Gy H − GHy ′

(23) (24)

It is easy to use (22) to reduce(23) to (24).

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Implicit Differentiation V

Question 1: A Derivative

Exercise: Assume that

Question 2: Increasing Functions

y 2 (2 − x) = 1, 2 2 x +y +1

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

and show—without using the analysis just given—how to reduce

Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

(25)

y(x2 − y 2 − 4x − 1) yÆ = 2(x3 − 2x2 + x − 2)

(26)

2x + y 2 . yB = 2y(1 − x)

(27)

′

to ′

36 / 56

Implicit Differentiation VI

Question 1: A Derivative

2

Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima

1

Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

-2

1

-1

2

-1

-2

37 / 56

Implicit Differentiation VII

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

The implicit differentiation technique is justified by the Implicit Function Theorem: Let f be a smooth real-valued function defined on an open subset D of R, and let (x0 , y0 ) be a solution of the equation f (x, y) = 0. If fy (x0 , y0 ) 6= 0, there are positive numbers, ǫ and δ , and a smooth function ϕ : (x0 − δ, x0 + δ) → (y0 − ǫ, y0 + ǫ) such that for each x ∈ (x0 − δ, x0 + δ), y = ϕ(x) is the only solution of f (x, y) = 0 lying in (y0 − ǫ, y0 + ǫ). Moreover, for each x ∈ (x0 − δ, x0 + δ),

fx [x, ϕ(x)] ϕ (x) = − . fy [x, ϕ(x)] ′

(28)

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Implicit Differentiation—The Issue

Question 1: A Derivative Question 2: Increasing Functions

The Issue

Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Implicit Differentiation I Implicit Differentiation II Implicit Differentiation III Implicit Differentiation IV Implicit Differentiation V Implicit Differentiation VI Implicit Differentiation VII Implicit Differentiation—The Issue Question 7: Implicit Functions Question 8: Improper Integrals

Our textbook problems encourage students (and teachers) to think about these problems n

globally instead of locally, and

n

without considering the hypotheses needed to justify what they are doing.

39 / 56

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length

Question 7: Implicit Functions

Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I Implicit Functions II Implicit Functions III Implicit Functions IV Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

40 / 56

Implicit Functions I

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

When I apply implicit differentiation to the equation (x2 + y 2 )2 = x2 − y 2 to find y ′ , I get

x(2y 2 + 2x2 − 1) y =− , 2 2 y(2y + 2x + 1) ′

(29)

Implicit Functions I Implicit Functions II Implicit Functions III Implicit Functions IV

which gives the indeterminate form 0/0 at the origin. Can I use limits to find the slope of the line tangent to this curve at the origin? How?

Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Implicit Functions II

Question 1: A Derivative Question 2: Increasing Functions

y 1.0

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

0.5

Question 6: Implicit Differentiation Question 7: Implicit Functions

0.0

x

Implicit Functions I Implicit Functions II Implicit Functions III Implicit Functions IV Implicit Functions V

-0.5

Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

-1.0 -1.0

-0.5

0.0

0.5

1.0

42 / 56

Implicit Functions III

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I Implicit Functions II Implicit Functions III Implicit Functions IV

The surplus of tangent lines at (0, 0) results from the fact that there is no open rectangle, centered at (0, 0), whose intersection with the curve is the graph of a function. But the conclusion of the Implicit Function Theorem asserts that there is such a rectangle. Because the conclusion is false, the IFT must not apply to this function at the origin.

Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Implicit Functions IV

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I Implicit Functions II Implicit Functions III Implicit Functions IV Implicit Functions V

Actually, we should have known that the IFT doesn’t apply: Putting F (x, y) = (x2 + y 2 )2 − x2 + y 2 , we have

 2  2 Fy (0, 0) = 4(x + y )y + 2y

= 0,

(30)

(0,0)

so that one of the hypotheses of the IFT fails.

Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Implicit Functions V

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I

We could find the slope of either branch of the curve by using the implicit derivative if we were to solve, algebraically, for y in terms of x and then replace y with the solution throughout the implicit differentiation expression for y ′ —and take the limit as we approach the origin. That’s nice. . . except that the whole point of implicit differentiation is to cirvumvent the necessity of solving for y in terms of x. . .

Implicit Functions II Implicit Functions III Implicit Functions IV Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Implicit Functions VI

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I Implicit Functions II

If we absolutely must have the slope of a branch of the curve as it passes through the origin, the best option is probably to re-parametrize. In this case, polar coordinates work nicely. They give us the equation 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 III Implicit Functions IV Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Implicit Functions—The Issues

Question 1: A Derivative Question 2: Increasing Functions

Two Issues:

Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Implicit Functions I Implicit Functions II Implicit Functions III

n

n

Hypotheses, hypotheses, hypotheses! In this case, 0/0 isn’t an “indeterminate form”— it’s undefined!

Implicit Functions IV Implicit Functions V Implicit Functions VI Implicit Functions—The Issues Question 8: Improper Integrals

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length

Question 8: Improper Integrals

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Improper Integrals I

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

Why don’t we use

Z

1

−1

Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

2x dx = lim 2 1−x t→1−

t

2x dx 2 −t 1 − x t 2 = lim ln(1 − x ) t→1− Z

(31)

(32)

−t

= lim {ln(1 − t2 ) − ln[1 − (−t)2 )]} = 0

(33)

t→1−

as the elementary-calculus definition for that improper integral?

Improper Integrals I Improper Integrals II Improper Integrals III

It would make freshman life so much easier.

Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Improper Integrals II

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals

The calculation we have just examined gives something called the “Cauchy Principal Value” (CPV) of the improper integral. The CPV is written PV

Z

1

−1

2x dx. 2 1−x

Improper Integrals I Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Improper Integrals III

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I Improper Integrals II

Why not use the CPV in elementary Calculus? The short answer: Using Cauchy Principal Values would break the equation

Z

1

−1

2x dx = 2 1−x

Z

ξ

−1

2x dx + 2 1−x

Z

ξ

1

2x dx. 2 1−x

(34)

Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Improper Integrals IV

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity

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

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

P (u) = (u − 1)(u + 1)(u + 2B − 1)(u + 2B + 1),

(35)

P ′ (u) f (u) = − . P (u)

(36)

and put

Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I

Then f is continuous in (−1, 1), and a tedious calculation shows that

Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

1

B−1 . f (u) du = ln PV B+1 −1 Z

(37)

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Improper Integrals V

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

Now put g(u) = (u2 + 2Bu − 1)/(2B), and note that g(−1) = −1 while g(1) = 1. Putting x = g(u), we ought therefore to be able to write PV

Z

1

F (x) dx = PV

−1

Z

1

F [g(u)]g ′ (u) du,

(38)

−1

where F (x) = 2x/(1 − x2 ). However, the CPV on the left side of (38) is zero, as we have seen; the integrand on the right side of (38) turns out to be the integrand of (37),

B−1 above, and so that CPV is ln . As it happens, the single value this B+1 latter quantity cannot assume is zero.

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Improper Integrals VI

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions

A very good reason for not using Cauchy Principal Values in elementary calculus is that they break the Substitution Theorem for Definite Integrals—which is too valuable a theorem to give up. . . in elementary calculus.

Question 8: Improper Integrals Improper Integrals I Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Improper Integrals—The Moral

Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I

Moral The definitions we adopt condition the theorems we can prove. We have chosen (not necessarily consciously or with full knowledge) the definitions we use in elementary calculus to support the tools we want students to be able to use.

Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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Question 1: A Derivative Question 2: Increasing Functions Question 3: Concavity Question 4: Local Minima Question 5: Polar Arc-Length Question 6: Implicit Differentiation

The End

Question 7: Implicit Functions Question 8: Improper Integrals Improper Integrals I Improper Integrals II Improper Integrals III Improper Integrals IV Improper Integrals V Improper Integrals VI Improper Integrals—The Moral

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