Calculus Notes

October 12, 2015

1

The Binomial Theorem

A binomial is an algebraic statement of the form (a + b)n and the Binomial Theorem relates to expanding this into a polynomial. I’ve compiled a history of the Binomial Theorem largely based on the work of mathematical historian J.L. Coolidge who live in the first half of the 20th century. Historical Development: Pre Newton • From Euclid’s the elements, ”If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle of the segments.” • Diophantus gave example cases for cubing binomials, such as (n-1). • Applied to finding square roots by Hero of Alexandria (although earlier use are possible) and to finding cube roots by Hindus Aryabhata (indian mathematician/Astronomer). • From Wikipedia: Hindu lyricist Pingala (c. 200 B.C.), Persian poet Omar Khayyam (11th century), Arabian mathematician Al-Karaji (11th Century). • Zhu Shijie (1249-1314) who lived during the Yuan Dynasty gave the first recorded version of what we today call pascals triangle, though no proof is given of it relationship to binomial expansions. • Variations on this theme were given by Michael Stifel 1 2 3

3

4

6

5

10

1

(1.1)

10

in 1544 (probably arrived at independently) and Blaise Pascal

1

1

1

1

...

1

2

3

4

...

1

3

6

10 . . .

1 .. .

4 .. .

10 .. .

20 . . . .. .

(1.2)

who also wrote down the general formula  n  n(n − 1)(n − 2) · · · (n − r + 1)  =  r(r − 1)(r − 2) · · · 1 r 

(1.3)

• The first results for fractional powers was James Gregory in 1670 by a rather indirect route involving logarithms. This takes us up to the time of Newton, who was working on a problem originally studied by John Wallis. Wallis had been developing techniques to find the area under curves of the form (1 − x2 )n and had successfully determined formulas for integer values of n by way of the formula Z

xp =

1 xp+1 1+p

(1.4)

Newton extended this to n = 1/2, 3/2, 1/3. Note: these cases may be obtained by the application of things like Hero’s method or by taking Pascal’s formula and assuming it continues to work for non-integer n. 1 (1 − x2 )1/2 = 1 − x2 − 2 3 2 2 3/2 (1 − x ) =1− x − 2 1 2 1/3 (1 − x ) = 1 − x2 − 3

1 4 x − 8 3 4 x − 8 1 4 x − 9

1 6 x ... 16 1 6 x ... 16 5 6 x ... 81

(1.5a) (1.5b) (1.5c) (1.5d)

A general proof of the binomial theorem can be found in a 1742 article by Giovanni Salvemini. This proof breaks the problem into three part, i) natural number power, ii) fractional powers, iii) negative integer powers. A more concise proof is given by Colin Maclaurin in his treatise Fluxions, also in 1742, which relies on taking derivates. However, all these previous proofs (including ones attributed to Colson and Euler), fail to deal with the problem of convergence though the mathematicians were aware of the issue. Neil Hendik 2

Abel (who died at the age of 26 of tuberculosis) gives proof for the convergence of the series for (1 + x)n where x is a complex number. This proof shows the series converge for |x| < 1 and diverge for |x| > 1 and proceeds to go into great depth concerning what happens when |x| = 1

2

The Tangent Line Problem

The word tangent means touching. If we consider a circle, a tangent line is one which intersects the circle at exactly one point. When extending this idea to general curves the earliest definition was ”a right line which touches a curve, but which when produced, does not cut it”. But, this is too restrictive and ignores things like inflection points. By Liebniz’s definition the tangent line is one which passes through two points on the curve which are infinitesimally close, and this is the definition we will work with. Historical Predecessors • Euclid Book III (323283 BC): Tangent found for circle. ”A straight line is said to touch a circle which is meeting the circle and being produced does not cut the circle”. • Apollonius I (15 100 CE): Tangent given for general conic sections. ”If a straight line be drawn through the extremity of the diameter of any conic parallel to the ordinates to that diameter, the straight line will touch the conic, and no other straight line can fall between it and the conic.” Apollonius speaks of his first four books as having been worked out by his predecessors, and it’s likely his theorem was known to Euclid. • Archimedes (?287 BC 212 BC): Studied the tangent line problem for spirals (the curve traced by a point advancing uniformly on a line which turns uniformly about a fixed point). He used the method of exhaustion and considered infinitely short chords, which parallels the treatment of the problem in modern calculus. • Fermat 1629: The method of adequality (Latin: adaequalitas) is developed by Pierre de Fermat to find the maxima and minima of curves. This method is later extended to finding tangent lines. Adequality means ”approximate equality” and in the method Fermat ”adequates” two expressions which differ by a small term ”e”, say P (x) ≈ P (x + e), which is to be ignored after all algebraic reductions are made. • 1637: The method of normals is developed by Rene Descartes to find the normal and tangent lines to a curve. This method begins by constructing a circle tangent to the curve and then using the radius through the point of intersection to find the slope of the normal line which in turn gives you the slope of the tangent line.

3

• Roberval and Torricell: Two rival geometers. Methods based on instantaneous motion. • Baron DeSluse (1652) and Issac Barrow : Derived formula’s for the tangent line of curves f (x, y) = 0 in a method closely related to that of Fermat. Example 2.1. Find the line tangent to the curve y = 3x − x2 at the point (1, 2) Proof. We begin by solving a related problem: Find the secant line through the points (1, 2) and (x0 , y0 ).

Example 2.2. Find the line tangent to the curve y = 1/x at the point (2, 1/2)

3

Limits

The term ”limit” was introduced by Augustin-Louis Cauchy in ”Cours d’Analyse d’Ecole Royal Polytechnique” (1821), though the concept goes all the way back the accent greeks who used the idea to calculate π to several digits. Cauchy’s definition: ”When the values successively attributed to a variable approach indefinitely to a fixed value, in a manner so as to end by differing from it by as little as one wishes, this last is called the limit of all the others.” Along with this Bernard Bolzano (1817) gave the first modern,  − δ, definition for the limit of a function, and Carl Weierstass introduced the modern notation for limits and well as the formal definition used today. Notation: Limit lim f (x) = L

x→a

(3.1)

Read as: The limit of f (x), as x approaches a equals L. It may be give by the alternate shorthand:

f (x) → L as x → a

Read as: f(x) approaches L as x approaches a.

3.1

Types of Limits

• Easy: Example 3.1. Find the limit x2 + x x→3 x + 1 lim

4

(3.2)

• Point Discontinuity: Example 3.2. Find the limit x2 + x x→−1 x + 1 lim

x2 + x x(x + 1) = = x whenever x 6= −1, x+1 x+1

Solution: Here we may reduce the equation by noticing thus f (x) =

  

x

  DNE

x 6= −1 x = −1

And so the limit is −1 since f (x) approaches −1 even though it never obtains this value. Here’s another example of the point discountinuity Example 3.3. Find the limit

√ lim

t→0

t+9−3 t

• Asymptotes: Example 3.4. Find the limit 1 =∞ x→1 (x − 1)2 lim

Solution: Here only the denominator goes to zero as x approaches 1 and f (x) becomes arbitrarily large. In this case we say f (x) goes to infinity as x approaches 1. Example 3.5. Find the limit lim

x→1

1 = DNE (x − 1)

Solution: Here |f (x)| goes to infinity, but f (x) > 0 if x > 1 and f (x) < 0 if x < 1. In this case be must introduce the idea of a left limit and a right limit.

lim−

x→1

1 = −∞ (x − 1)

Read as: The limit as x approaches 1 from the left . . .

lim

x→1+

1 =∞ (x − 1)

Read as: The limit as x approaches 1 from the right . . .

5

3.2

Exact Definition

The exact definition of a limit which we use today is the work of Carl Weierstass (1815-1897). Weiestrass was originally a gymnasium (High School) teacher of arithmetic and calligraphy. In 1854 he publishes a memoir on integrals which launches him into a position at the University of Berlin. He was notorious for teaching while seated at the front of the room (due to sever vertigo) while his students wrote what he said. He also didn’t care much for publishing and many of his results are first published by his pupils. {Add Picture} Definition 3.1. Let f be a function defined on some open interval that contains the number a, except at possibly a itself. We say lim f (x) = L

x→a

(3.3)

if and only if, for every  > 0 there exists a δ > 0, such that,|f (x) − L| <  whenever 0 < |x − a| < δ. Example 3.6. Prove using the  − δ definition that x2 + x − 6 =5 x→2 x−2 lim

(3.4)

Definition 3.2. Let f be a function defined on some open interval that contains the number a, except at possibly a itself. We say lim f (x) = ∞

x→a

(3.5)

if and only if, for every M > 0 there exists a δ > 0, such that, f (x) > M whenever 0 < |x − a| < δ. Example 3.7. Prove using the M − δ definition that

lim

x→1

3.3

1 =∞ (x − 1)2

Limit Laws

Suppose that the following limits exist

lim f (x) = L

x→a

lim g(x) = K

x→a

• Linearity: 6

(3.6)

h i h i – lim [f (x) + g(x)] = lim f (x) + lim g(x) x→a x→a x→a h i – lim [cf (x)] = c lim f (x) x→a

x→a

• Products: h ih i – lim [f (x)g(x)] = lim f (x) lim g(x) x→a x→a x→a h in – lim [f (x)n ] = lim f (x) x→a

x→a

• Inverses: i−1   h , whenever k 6= 0. – lim f (x)−1 = lim f (x) x→a

x→a

• Similar Functions: (Super Important!) – If f(x) = g(x) when x 6= a, then lim f (x) = lim g(x) x→a

x→a

Theorem 3.1. The Squeeze Theorem If f (x) ≤ g(x) ≤ h(x) when x is near (except possibly at) a and

lim f (x) = lim h(x) = L,

(3.7)

lim g(x) = L

(3.8)

x→a

x→a

then x→a

A proof of this follow directly from the lemma 3.2 bellow. By way of example, we will provide a full proof for this lemma using the formal definition of limits. Similar proofs may be produced for the all the limit laws. Lemma 3.2. If f (x) ≤ g(x) when x is near (except possibly at) a and the limits of f and g both exist at a, then lim f (x) ≤ lim g(x).

x→a

x→a

(3.9)

Proof. We prove this by contradiction, which is to say we assume that the lemma is false to begin with and then show this assumption leads to a contradiction. If the lemma is false then there exist functions f (x) and g(x) with f (x) ≤ g(x) near a and both with limits which exist as a such that limx→a [f (x) − g(x)] = M for some M > 0. By the definition of a limit this means for all  there exists a δ such that |f (x) − g(x) − M | <  whenever 0 < |x − a| < δ, or in shorthand

∀ ∃δ, s.t. if 0 < |x − a| < δ, then |f (x) − g(x) − M | <  7

If we pick  = M there exists a δM such that |f (x) − g(x) − M | < M whenever 0 < |x − a| < δM . Which, if we take away the absolute values, means (f (x) − g(x) − M ) < M and −(f (x) − g(x) − M ) < M . From the second of these two statements we have g(x) < f (x) whenever 0 < |x − a| < δM . But this contradicts the original statement that f (x) ≤ g(x) near a. This is a contradiction and thus the lemma must be true.

Lemma 3.2 implies that Example 3.8. Use the Squeeze theorem to show that lim

θ→0

sin θ . θ

Proof. For a clear illustration of what follows see Figure 2. Take the unit circle with a radius drawn. Let the angle the radius makes with the x axis be θ. Since θ is given in radians the length of the associated arch (from the x axis to the point the radius intersects the circle) is |θ|. If we draw a line perpendicular to the x axis through the point where the radius intersects the circle, then this forms a right triangle whose hypotenuse is the radius and whose height is | sin θ|. If we extend the radius beyond the circle and draw a line perpendicular to the x axis and through the point (1, 0), then this forms a triangle whose base is the radius and whose height is | tan θ|. lim θ →0

sinθ θ

1

y

θ tanθ sinθ θ 0 0

1

x Figure 1: Illustration of the relationship, sin θ < θ < tan θ. Now, the smaller triangle has height less than the length of the arc and the larger triangle has height greater than the length of the arc. Or, expressed as an inequality.

| sin θ| < |θ| < | tan θ|

8

If we take the reciprocal of each term, this becomes 1 1 1 > > sin θ θ tan θ Or, after multiplying by | sin θ| sin θ > | cos θ|. 1 > θ If we now take the limit as θ → 0 we have

lim 1 ≥ lim

θ→0

θ→0

sin θ ≥ lim cos θ θ→0 θ

where the absolute values have been dropped since all terms are positive for θ near 0. And, since the first and last limits are easy, plugging in 0 for θ works and we are left with sin θ ≥1 θ→0 θ

1 ≥ lim

Or, lim

θ→0

sin θ =1 θ

as we set out to prove.

Example 3.9. Use the Limit Laws to show that lim

θ→0

lim θ →0

cos θ − 1 . θ

1 − cosθ θ

y

θ sinθ 0

θ 1-cosθ

0

1

x Figure 2: Illustration of the relationship between 1 − cos θ and θ.

9

Proof. We begin by ”multiplying by one” and then reducing things.    1 − cos θ 1 − cos θ 1 + cos θ = lim θ→0 θ→0 θ θ 1 + cos θ 2 1 − cos θ = lim θ→0 θ(1 + cos θ) lim

sin2 θ θ→0 θ(1 + cos θ)

= lim

From here we use the product law for limits to break this into two simpler limits that we are can already compute.    sin2 θ sin θ sin θ lim = lim lim θ→0 θ(1 + cos θ) θ→0 θ θ→0 1 + cos θ =1·0 =0

Thus, lim

θ→0

4

cos θ − 1 =0 θ

Continuity

In Cauchy’s ”Cours d’Analyse” he defines continuity as: ”an infinitely small increment α of the independent variable x always produces and infinitely small change f (x + α) − f (x) of the dependent variable y.” Which, in layman’s terms, means that the curve is a single connected line. Definition 4.1. A function is continuous at a point a if and only if

lim f (x) = f (a)

x→a

(4.1)

Clearly, if f is continuous then it’s limit exists at that point. If not f is said to be discontinuous or to have a discontinuity at a. Furthermore, the connection to limits implies that all of the limit laws are also

10

continuity laws. This means if f (x) and g(x) then so are the functions

f (x) + g(x),

4.1

cf (x),

[f (x)]−1 if f (x) 6= 0

f (x)g(x),

Types of Discontinuities

• Jump Discontinuity: This consists of functions whose left and right limits exist, but the left limit is unequal to the right limit. For examples we mostly look at function defined piecewise. Example 4.1. Find all values of c such that    x2 − cx f (x) = x   c2 + 2

x a we know from the IVT that there must be some x which satisfies this equation. These two facts together give us all rational powers, but what about irrational powers? Here we must essentially fill in the blanks, i.e. we wish for f (x) = ax 21

to be a continuous function. So, we define for x irrational

f (x) = lim pq →x ax where we write

p q

to mean the converges of the limit over the rational numbers.

The Number e Let us begin with the limit definition of a derivative as applied to our exponential function f (x + h) − f (x) h ax+h − ax = lim h→0 h ax ah − ax = lim h→0 h  h  a −1 = lim ax h→0 h

f 0 (x) = lim

h→0

ah − 1 h→0 h

= ax lim

Now, define e to be the number such that limh→0

eh −1 h

= 1 and let ln a be the number such that eln a = a

Now
h eln a − 1 ah − 1 = lim lim h→0 h→0 h h h ln a e −1 = lim h→0 h eh ln a − 1 ln a = lim h→0 h ln a eh − 1 e e h→0 h e

= ln a lim = ln a

Thus, we come to our next rule for derivatives Theorem 5.6. Exponential Rule: If f (x) = ax , then f 0 (x) = (ln a) ax .

22

where e h = h ln a

5.6

Product Rule

Consider a rectangle with height h and width w, so the area is given by A = hw. We wish to see how a small change in the height or width effects the area.

A + ∆A = (h + ∆h)(w + ∆w)

(5.22)

A + ∆A = hw + ∆h w + h∆w + ∆h∆w

(5.23)

Now, using the fact that A = hw we can cancel these terms out and arrive at

∆A = ∆h w + h∆w + ∆h∆w

(5.24)

Since the changes in the height and width are assumed to be small, the product of the small changes is on a smaller order ∆h∆w  ∆h ∼ ∆w and we drop the ∆h∆w term.

∆A = ∆h w + h∆w

(5.25)

This the core result of the product rule, it now remains to clean things up a little bit. By letting A, h and w all vary with respect to some common parameter t, we can divide by ∆t to get ∆A ∆h ∆w = w+h ∆t ∆t ∆t

(5.26)

In the limit ∆t → 0   ∆A ∆h ∆w = lim w+h ∆t→0 ∆t ∆t→0 ∆t ∆t     ∆h ∆A ∆w = lim lim w+h lim ∆t→0 ∆t ∆t→0 ∆t ∆t→0 ∆t dA dh dw = w+h dt dt dt lim

(5.27) (5.28) (5.29)

This becomes the product rule for taking derivatives. Rewriting this in terms of the product of two functions f (x) and g(x) Theorem 5.7. Let f (x) and g(x) be two differentiable functions, then d d d [f (x)g(x)] = [f (x)] g(x) + f (x) [g(x)] dx dx dx

23

(5.30)

or if u and v are two curves 0

(uv) = u0 v + u v 0

5.7

(5.31)

Quotient Rule

Here we present two way of getting at the quotient rule. If we think in terms of small changes to a function f (x) =

u(x) v(x) .

∆ (f (x)) = f (x + ∆x) − f (x) u + ∆u u − v + ∆v v (u + ∆u) v − u (v + ∆v) = v (v + ∆v) v∆u − u∆v = v (v + ∆v) =

Thus, ∆(u/v) d u = lim ∆x→0 dx v ∆x ∆u ∆v v −u ∆x ∆x = lim ∆x→0 v (v + ∆v) vu0 − uv 0 = v2 Now instead, we can get the quotient rule from the product rule since u(x) = f (x)v(x).

u = fv u0 = f 0 v + f v 0 u0 − f v 0 v u 0 u − v0 0 v f = v vu0 − uv 0 f0 = v2 f0 =

Examples: • f (x) = • y

=

ex x

x3

x+1 +x−2

24

• g(t) =

1 − tet t + et

Example: An alternative proof for finding the derivative of y = xn . Example: Extending the power rule to negative integer powers.

5.8

Trigonometric Derivatives

We now find the derivative of sin by d sin(x + h) − sin(x) [sin(x)] = lim h→0 dx h sin(x) cos(h) + sin(h) cos(x) − sin(x) = lim h→0 h   cos(h) − 1 sin(h) = lim sin(x) + cos(x) h→0 h h sin(h) cos(h) − 1 + cos(x) lim = sin(x) lim h→0 h→0 h h

(5.32b) (5.32c) (5.32d)

= sin(x) · 0 + cos(x) · 1

(5.32e)

= cos(x)

(5.32f)

d cos(x + h) − cos(x) [cos(x)] = lim h→0 dx h cos(x) cos(h) − sin(h) sin(x) − cos(x) = lim h→0 h   cos(h) − 1 sin(h) = lim cos(x) − sin(x) h→0 h h cos(h) − 1 sin(h) = cos(x) lim − sin(x) lim h→0 h→0 h h

(5.33a) (5.33b) (5.33c) (5.33d)

= cos(x) · 0 − sin(x) · 1

(5.33e)

= − sin(x)

(5.33f)

Examples: • f (x) = tan(x) • y=

(5.32a)

1 − sec(x) tan(x)

• A(t) = tet cos(t)

25

5.9

The Chain Rule

Consider the composite of two functions z = f (g(x))

(5.34)

If we take y = g(x) as the intermediate variable, we have

∆y ≈ g 0 (x)∆x + O(∆2 )

(5.35a)

∆z ≈ f 0 (y)∆y + O(∆2 )

(5.35b) (5.35c)

Substituting in for y and ∆y, leaves us with

∆z ≈ f 0 (g(x))g 0 (x)∆x + O(∆2 )

(5.36)

∆z ≈ f 0 (g(x))g 0 (x) + O(∆) ∆x

(5.37)

Or, after diving be ∆x

Now by taking the limit, we arrive at the chain rule

lim

∆x→0

∆z = lim f 0 (g(x))g 0 (x) + O(∆) ∆x ∆x→0 dz = f 0 (g(x))g 0 (x) dx

(5.38) (5.39)

Theorem 5.8. The Chain Rule: Let f and g be differentiable functions, then d [f (g(x))] = f 0 (g(x))g 0 (x) dx

(5.40)

We may also think about the chain rule in terms of the fractional interpretation of the derivative ∆z ∆z ∆y = · ∆x ∆y ∆x

26

(5.41)

Taking the limit we now arrive at ∆z ∆z ∆y = lim · ∆→0 ∆x ∆→0 ∆y ∆x ∆z ∆z ∆y lim = lim · lim ∆→0 ∆x ∆→0 ∆y ∆→0 ∆x dz dy dz = · dx dy dx lim

(5.42a) (5.42b) (5.42c)

Theorem 5.9. An alternative statement of the Chain Rule: Let z, y and x be three related variables dz dy dz = · dx dy dx

(5.43)

Examples: • y = sec(x) • y = e2x+1 • y = e3x + 1


2

√ • y = sin x2 + 1 • y=

p

1 − sin2 x

5.10

Implicit Derivatives

5.11

Log Derivatives

Fancy Techniques: Logarithmic Differentiation We have considered exponentiation in the case where the exponent is constant, i.e. polynomials, and wher the base is constant, i.e. exponential functions. However, there is no reason why both can’t be functions of the independent variable. Consider the curve

y = xsin x

(5.44)

Since, the exponent varies in x this function is not covered by the power rule, and since the base varies in x it is also not covered by the exponent rule. Here we put forward two alternative methods for finding the derivative

dy dx ,

both of which take advantage

of implicit differentiation. The first is logarithmic differentiation; here we take the natural log of both sides

27

of equation 5.44 and use the log identities to eliminate the exponential expression.

ln y = ln xsin x




= sin x ln x

We may now take the derivative of both sides and then solve for

dy dx

dy 1 sin x = cos x ln x + dx y x dy sin x = y cos x ln x + y dx x dy sin x =x cos x ln x + (sin x) xsin x−1 dx

For an alternative treatment we introduce a dummy variable z = sin x, so that equation 5.44 becomes

y = xz

from our understanding of implicit differentiation, when we take the derivative of both sides. dxz dz dxz dy = + dx dz dx dx dz = (ln x)xz + zxz−1 dx = (ln x)xsin x cos x + (sin x) xsin x−1

6 6.1

Applications of the Derivative Example: Damped Springs

Consider an object with mass m attached to a spring. Examples include actual springs, pendulums, beams which are fixed at one end. Let A(t), the amplitude, be the deviation of the objects center of mass from it’s resting position. Hooke’s Law says that the restoring force from the spring is proportional to the amplitude, FSpring = −kA. From Newton’s Second Law of Motion, F = ma, which states that the sum of all external forces, F , on an object balances against the mass, m, times the acceleration, a, we have

m

d2 A = −kA. dt2

28

(6.1)

This is the equation of a basic mass spring system, which has general solution r A(t) = c1 sin

k t m

!

r + c2 cos

k t m

!

While we won’t get into the derivation here, we can check the validity of the solution by taking the first two derivatives of A. ! ! r r k k k t + c2 sin t A (t) = c1 m m m ! ! r r k k k k 00 t − c2 cos t A (t) = −c1 sin m m m m ! !! r r k k k =− c1 sin t + c2 cos t m m m r

0

=−

k cos m

r

k A(t) m

Taking this a step further we consider the case of a damped spring. In this case the force due to friction is proportional to the velocity, FFriction = −βv. So the sum of all external forces is F = FSpring + FFriction and the force balance equation is m

d2 A dA = −kA − β . 2 dt dt

(6.2)

Rather than deal with a general solution, here we look at specific numbers. Say, m = 0.5 Kg, k = 5 β=1

Kg s .

So, the equation for the spring is 1 00 A + A0 + 5A = 0 2

Then a particular solution is A(t) = e−t cos (3t) which we check by substituting this solution back into the damped spring equation

A0 (t) = −(cos(3t) + 3 sin(3t))e−t A00 (t) = (−8 cos(3t) + 6 sin(3t))e−t

29

Kg s2 ,

1 00 A + A0 + 5A = (−4 cos(3t) + 3 sin(3t))e−t − (cos(3t) + 3 sin(3t))e−t + 5 cos (3t) e−t 2

6.2

(6.3)

= ((−4 − 1 + 5) cos (3t) + (3 − 3) sin (3t)) e−t

(6.4)

=0

(6.5)

Example: Supply and Demand

As supreme leader I want to erect statues to commemorate my greatness. The cost of producing x statues is given by the function C(x) = 1000 + 24x + x2

(6.6)

The tourist revenue per statue drops off the more that are built since they start to compete with each other, this demand is given in dollars by the equation

D(x) = 600 − 15x

(6.7)

From here we seek to find the ideal number of statues to erect, in order to maximize the profit from our endeavor. First we find the revenue stream given by the money generated by the demand per statue

R(x) = xD(x) 600x − 15x2

and then the profit is given by the incoming revenue minus the cost.

P (x) = R(x) − C(x) = 600x − 15x2 − (1000 + 24x + x2 ) = −1000 + 576x − 16x2

The profit’s stop growing when P 0 (x) = 0, so

576 − 32x = 0 x = 18

30

6.3

Example: Exponential Growth and Decay

We consider processes where the rate of increase is proportional to the amount already there, like population growth, radioactive decay, bank interest, or heat transfers. dy = ry dt This equation has solution y(t) = Cert for some constant c = y(0). Note:

(6.8) y0 = r is constant and so r is the y

percent increase. One example comes from Newton’s Law of Cooling dT = k(T − T0 ) dt

(6.9)

where T is the temperature of some object and T0 is the temperature of the surrounding area which we assume is sufficiently large so the transfer of heat doesn’t change the background temperature.

6.4

Related Rates

Example 6.1. An observer on the ground watches a rocket taking off from D meters away from the launch π π site. She measures the angle of inclination of the rocket from her position to be θ(t) = − . What 2 t+2 was the velocity of the rocket at t = 0? t = 4? If θ follows the same function of time forever what’s limiting velocity of the rocket?

Example 6.2. Two paths diverge in a wood. Guildenstern takes the path heading North walking at 2 mph while Rosencrantz takes the path heading North East walking at 3 mph. How fast is the distance between them changing after 15 minutes.

Example 6.3. Professor Ratigan sits at the tip of minute hand of Big Ben, while Basil of Baker Street sits at the end of the hour hand. For Basils clever, but violent ”harpooning” solution to the problem he must know have fast they moving away from each other. The minute hand on Big Ben is 12 meters longs while the hour hand is 6 meters long.

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Example 6.4. The relationship between the pressure and volume of air in a balloon is given by the equation

P V 1.4 = C

where C is a constant. Assume V = 400cm3 , P = 80K Pa, and

dV dP = 10K Pa/s. What is ? dt dt

Example 6.5. The top of a ladder slides down a vertical wall at a rate of 0.15 meters per second. The base of the ladder is 3 meters from the wall and sliding away at 0.2 meters per second. How long is the ladder?

Example 6.6. A lamp post 18 ft tall casts the shadow of Alec Baldwin dressed as the Shadow on the ground. The Shadow’s shadow is moving at 60 ft/s, how fast is the Shadow moving? (Hint: Alec Baldwin is 6 ft tall.)

6.5

Curve Sketching

Goal: Draw the graph of y = f (x) using y 0 , y 00 and all tools from previous classes. INSERT TABLE Example 6.7. Sketch the curve f (x) = 3x − x3 . Definition 6.1. When f (x0 ) = 0 we call x = x0 a critical point. We use the critical points as anchors in the graph. We also anchor points like the zeroes of f (x). And, finally we consider the behavior as x → ±∞. Example 6.8. Sketch the curve f (x) =

x+1 . x+2

General Strategy • Looks for Discontinuities. • Plot anchor points: Critical points, zeros, inflection points • Check signs for f 0 and f 00 in-between critical point and inflection points respectively. • Find the asymptotic behavior (x → ±∞). Example 6.9. Sketch the curve f (x) =

x . ln x 32

Example 6.10. Sketch the curve y =

ln x x .

Example 6.11. Sketch the curve y = x3 − 3x2 + 3x. Example 6.12. Sketch the curve y = (x2 − 1)4/3 . Example 6.13. Sketch the curve y = ex sin x. Example 6.14. Sketch the curve y =

x2 −1 x3 .

Example 6.15. Sketch the curve y =

x x2 +1 .

6.6

Optimization Promblems

General Strategy • Draw diagram and label variables. • Write Equations • Use constraint to remove a variable. • Take the derivative to find max or min. Example 6.16. Cut a length of wire into two pieces each of which will be bent into squares. How should the wires be cut to maxims the total area covered? Example 6.17. What is the most efficient can shape for a volume of 355 cm3 ? Is the most efficient glass taller or shorter? Example 6.18. What are the most efficient proportions of a box with no top? Example 6.19. Where does a free weight hang on a string?

6.7

Linear Approximations f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 )

(6.10)

Example 6.20. f (x) = ln x. We will now show that equation (6.10) is the best possible linear approximation. To do this we assume that there’s a better approximation L(x) = mx + b

33

(6.11)

Since, any approximation must go through the point (x0 , f (x) ) can reduce this to

L(x) = m(x − x0 ) + f (x0 )

(6.12)

so that only m is unknown. If it’s a better approximation than there exists some range of x values x ∈ (x0 − δ, x0 + δ) − {x0 } such that 0 ≤ |f (x) − L(x)| ≤ |f (x) − (f (x0 ) + f 0 (x0 )(x − x0 )) | Now, by dividing all terms by |x − x0 | we get f (x) − f (x0 ) f (x) − f (x0 ) 0 0≤ − m ≤ − f (x0 ) x − x0 x − x0 But, recall that ∆f ∆x→0 ∆x

f 0 (x0 ) = lim By the squeeze theorem

0 ≤ |f 0 (x0 ) − m| ≤ 0 and thus, f 0 (x0 ) = m and L(x) is the same linear approximation. Example 6.21. Find the linear approximation for sin x, cos x, ex . Example 6.22. Find the linear approximation for ln(1 + x), (1 + x)r . e−3x Example 6.23. Find the linear approximation for √ 1+x Better Approximations Guess for small x

f (x) ≈ a0 + a1 x + a2 x2 + a3 x3 + . . .

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Now, we see that

f (0) ≈ a0 f 0 (x) ≈ a1 + 2a2 x + 3a3 x2 + . . . f 0 (0) ≈ a1 f 00 (x) ≈ 2a2 + 6a3 x + . . . f 00 (0) ≈ 2a2

So, the quadratic approximation should be

f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ) +

6.8

f 00 (x0 ) (x − x0 ) 2

Mean Value Theorem

Theorem 6.1. If f (x) is differentiable on a < x < b,and continuous on a ≤ x ≤ b, then there exists c ∈ (a, b) such that f 0 (c) =

f (b) − f (a) . b−a

After a little algebra we see that there’s a Error for Linear Approximation

6.9

Newton’s Method

35

(6.13)

7

Integration

7.1

The Inverse of a Derivative

We’ve already touched upon briefly how the derivative can behave like a function that takes in a function F and spits out its derivative F 0 . Dx [F (x)] = F 0 (x)

(7.1)

using the notation of Euler. And, just like any other function, we can also consider the inverse operation. That is to say defining the inverse derivative that takes in F 0 and spits out the original function F .

Dx−1 [F 0 (x)] = F (x)

(7.2)

Or, as we’ll come to write it in the future Z f (x)dx = F (x)

where f (x) = F 0 (x) = D[F (x)]. In the future sections we’ll get more in depth as to what the

(7.3)

R

symbol

means, but for now we just take it to mean the inverse derivative function. By way of example, the following list shows some very simple inverse derivatives Z • 1 dx = x + C Z •

x dx = Z

•

1 2 x +C 2

xn dx =

1 n x +C n

Z •

sin(x) dx = − cos(x) + C Z

•

cos(x) dx = sin(x) + C Z

• Z •

ex dx = ex + C 1 dx = ln |x| + C x

Notice how all these equation have a C tacked on to the end. This is because the inverse derivative function is indefinite, i.e., there’s more than one function with the same derivative. For instance

    Dx (x + 2)2 = Dx x2 + 4x = 2x + 4 36

(7.4)

We can, however, show that the ambiguity is limited to a constant C difference. Theorem 7.1. If two functions F (x) and G(x) have the same derivative F 0 (x) = G0 (x) = f (x), then the difference between the functions in a constant F (x) − G(x) = C. Proof. Consider the new function H(x) = F (x) − G(x). When we take the derivative 0

H 0 (x) = (F (x) − G(x)) = F 0 (x) − G0 (x) = f (x) − f (x) =0

Thus, we have H 0 (x) = 0. In this reduced form, we readily see that the only functions with derivative zero are constant functions H(x) = C or F (x) − G(x) = C. All the derivative rules have corresponding inverse derivative (integral) rules.

7.2

The Inverse of the Chain Rule

•

R

2x(x2 + 1) dx

•

R

2x(x2 + 1)1/2 dx

•

R

sin x cos x dx

•

R

e5x dx

•

R

x x2 +1

•

R

tan x dx

•

R

tan x sec2 x dx

•

R (1 +

dx

1 √

x)4

dx

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