1 Differentiation

Physics 201 and 202 are calculus-based physics courses. This means that calculus will be used on a routine basis, most likely every day in class and on most homework assignments. It is very important that you be able to perform the basic calculus operations;
1. computations of limits and derivatives,
2. expansions of functions,
3. simple integrations.

In this review we will make sure that you are up to speed on all of these basic skills, and we will review trigonometry while we are at it.

There is no substitute for a full year of calculus instruction, and so math 221 is an absolute prerequisite for physics 201, and math 222 is an absolute prerequisite for physics 202. There is no back door.

A smooth function y = f(x) of a variable x is differentiable at x if its derivative

df (x) f (x + dx) - f (x) f ′(x) = --------= lim ------------------------ dx dx →0 dx

exists (is a number) at x. The limit-taking process is very simple; we expand the numerator in ascending powers of dx, perform the division, and take the limit by setting dx = 0 afterwards (in a nutshell).

The geometrical interpretation of the derivative of f(x) at x₀ is that f′(x₀) is the slope of the line tangent to f(x) at x₀;

Therefore a useful formula for translating calculus to geometry is

df (x) tan θ = (--------)x=x dx 0

for the tangent to the curve at x₀.

The limit taking process is most easily handled for polynomial functions such as

N∑ n f (x) = an x n=0

by use of the following simple rule;

-d-- df-(x)-- dg(x)--- (f (x) + g(x)) = + dx dx dx

which is easy to prove from the deﬁnition above. This says that the derivative of a (ﬁnite) sum is the sum of the derivatives of the summands.

Example. Let f(x) = x³, the steps taken in computing the derivative are;
Step 1. Write out the fraction

3 3 f-(x--+--dx)------f(x)-- (x--+--dx)-------x-- = dx dx

(x3 + 3x2 dx + 3x (dx)2 + (dx)3) - x3 = --------------------------------------------------- dx

Step 2. Perform the division;

(x3 + 3x2 dx + 3x (dx)2 + (dx)3) - x3 3x2 dx + 3x (dx)2 + (dx)3 --------------------------------------------------- = ----------------------------------- dx dx

2 2 = 3x + 3x dx + (dx)

Step 3. Perform the limit (set dx = 0);

dx3 -----= lim (3x2 + 3x dx + (dx)2) = 3x2 dx dx→0

Example. Let f(x) = a + bx², in which a and b are constants.
Step 1. Write out the fraction

f (x + dx) - f (x) (a + b(x + dx)2) - (a + bx2) ------------------------ = --------------------------------------- dx dx

(a + b(x2 + 2x dx + (dx)2)) - (a + bx2) = ------------------------------------------------------ dx

Step 2. Perform the division;

(a + b(x2 + 2x dx + (dx)2)) - (a + bx2) 2bx dx + b(dx)2 ------------------------------------------------------= ----------------------= 2bx+b dx dx dx

Step 3. Perform the limit (set dx = 0);

-d-- 2 (a + bx ) = lim (2bx + b dx) = 2bx dx dx→0

which we can see is the sum of the derivatives of the two terms a and bx², the derivative of a constant being zero.

1.1 Problems

1 Compute the derivative of

1 f (t) = x0 + v0t + --a t2 2

with respect to t. x₀, v₀ and a are constants. In other words

df-(t)- f-(t-+--dt)-----f-(t)- = lditm→0 dt dt

2 Compute the derivative of

f (t) = a (t - b)4

with respect to t. a and b are constants.

3 Compute the second derivative of

1 f (t) = x + v t + --a t3 0 0 3

with respect to t. This means ﬁrst compute

df (t) v(t) = ------- dt

and then compute

d2f (t) dv(t) ---------= f ′′(t) = ------- dt2 dt

4 The following problem is very useful in the study of one dimensional motion at constant acceleration. The average velocity of an object over the time interval from t - Δt
2 to t + Δt
2 is

Δt Δt x(t + -2-) - x(t - -2-) ˉv = ----------------------------- Δt

with no limit being taken, Δt can be of any size.
Show that if the average velocity equals the instantaneous velocity at the interval midpoint, namely that if

dx(t) x(t + Δt-) - x(t - Δt-) v(t) = --------= ˉv = ---------2----------------2-- dt Δt

then the acceleration is constant. The acceleration is

d2x(t) a(t) = --------- dt2

2 Derivative rules and formulas; products

Derivatives of a product f(x) g(x) can be easily computed from the deﬁnition,

d f (x + dx)g(x + dx) - f (x)g(x) ----(f (x)g(x)) = lim ------------------------------------------- dx dx→0 dx

by replacing f(x + dx) with f(x + dx) - f(x) + f(x) and rearranging

d (f (x + dx) - f (x) + f (x))g(x + dx) - f (x)g(x) ----(f (x)g(x)) = lim ----------------------------------------------------------------- dx dx →0 dx

(f-(x--+--dx)-----f-(x))g(x----+--dx)---+--f-(x)(g(x---+--dx)------g(x))--- = lim dx →0 dx

f (x + dx) - f (x) g(x + dx) - g(x) = lim (------------------------g(x+dx))+f (x) lim ( -----------------------) dx →0 dx dx→0 dx

If f(x), f′(x), g(x) and g′(x) all exist, then

df (x) dg(x) = --------g(x) + f (x) -------- dx dx

We state this as being the product rule for derivatives.

d df (x) dg(x) ----(f (x)g(x)) = --------g(x) + f (x) -------- dx dx dx

2.1 The binomial theorem

This theorem is of great antiquity, and is extremely useful for both algebraic and calculus applications. It says that

( ) N∑ N (a + b)N = |( |)ambN - m m=0 m

where the number

( ) N N ! |( |) = ----------------- m m!(N - m)!

is a binomial coefficient, and the factorial of an integer N is

N ! = N ⋅ (N - 1) ⋅ (N - 2) ⋅ ⋅ ⋅ 2 ⋅ 1, 1! = 1, 0! = 1

(the last relation is a deﬁnition). For example

2 2 2 (a + b) = a + 2ab + b

3 3 2 2 3 (a + b) = a + 3a b + 3ab + b

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b3

2.2 Problems

5 Use the product rule to show that

-d-- 2 ′ df-(x)-- (f (x)) = 2f (x) f (x) = 2f (x) dx dx

and that in general

d ----(f (x))n = n(f (x))n - 1 f ′(x) dx

2.3 A power tool, series expansion

This last calculation was a little on the tricky side, but there exists a powerful tool for performing most of the operations of calculus in a simple way, the series expansion.

We suppose that the function f(x) exists at the point x₀, and for that matter that it exists near x₀. Let x-x₀ be small, so that x is close to x₀. The idea of the series expansion is that in the neighborhood of x₀, we could replace f(x) with a polynomial

N∑ an-- n Pf (x) = (x - x0) n=0 n!

in which n! = n ⋅ (n - 1) ⋅ (n - 2) ⋅⋅⋅

2 ⋅ 1 is our factorial of the integer n.

The number of terms N that we need to calculate to get P_f depends on what we want to do with it, and is based on the following concept: the function f(x) and polynomial P_f(x) agree at x₀, and have the same derivative at x₀, and the same second derivative, and so on up to the N^th derivative. We would call P_f(x) an N^th order series expansion of f(x) about the point x₀.
Step 1. Both f(x) and P_f(x) agree at x₀;

a2 2 aN N f (x0) = Pf (x0) = a0+a1(x0 - x0)+ --- (x0 - x0) + ⋅ ⋅ ⋅+ ----(x0 - x0) = a0 2! N !

requires that a₀ = f(x₀).
Step 2. Both f′(x) and d _ dxP_f(x) agree at x₀;

′ a2- a3- 2 aN-- N - 1 f (x0) = 0+a1+2 (x0 - x0)+3 (x0 - x0) + ⋅ ⋅ ⋅+N (x0 - x0) = a1 2! 3! N !

requires that a₁ = f′(x₀) = df dx_x=x₀.
Step 3. Both f′′(x) and d _ dxP_f(x) agree at x₀;

′′ a2- a3- aN-- N - 2 f (x0) = 0+0+2 +3 ⋅2 (x0 - x0)+ ⋅ ⋅ ⋅+N ⋅(N - 1) (x0 - x0) = a2 2! 3! N !

requires that a₂ = f′′(x₀) = d²f dx² _x=x₀.

For ninety percent of all of the calculus applications in our physics text, this is enough; the polynomial P_f(x) that agrees with f(x) up to two derivatives in the neighborhood of x₀ is

1 Pf (x) = f (x0) + f ′(x0) (x - x0) + --f ′′(x0) (x - x0)2 + ⋅ ⋅ ⋅ 2

This is called the Euler-Maclaurin or Taylor series for f(x) near x₀, and it may be substituted in place of f(x) in the neighborhood of x₀.

What do we use it for? For starters it can be used to get formulas for derivatives of products and quotients. In most applications you only need to keep one or two terms in a Taylor series. For example, let x = t + dt and x₀ = t, then

1 P (t + dt) = f (t) + f ′(t) dt + --f ′′(t) (dt)2 + ⋅ ⋅ ⋅ f 2

and you can replace any occurrence of f(t + dt) in a formula that involves taking the limit dt → 0 with this expression.

Example

d (f (t + dt)g(t + dt) - f (t)g(t)) --- (f (t)g(t)) = lim ----------------------------------------- dt dt→0 dt

′ ′ (f-(t)--+--f-(t)-dt--+--⋅-⋅ ⋅ )(g(t)-+--g-(t)-dt--+--⋅ ⋅-⋅ )---f-(t)g(t)- = lim dt→0 dt

g(t) f′(t) dt + g ′(t) f (t) dt + ⋅ ⋅ ⋅ = lim -------------------------------------------= g(t) f ′(t) + g ′(t) f (t) dt→0 dt

and we are done quickly and cleanly, all of the terms in ( ⋅⋅⋅

) contain at least two factors of dt, and so in the limit become zero.

Example; l’Hospitals rule is a formula for computing the limit of the ratio of two functions that both vanish at x₀,

lim f (x) = 0 = lim g(x) x→x0 x→x0

The limit of the ratio is then

′ 1- ′′ 2 f-(x)-- f-(x0)--+--f--(x0)--(x-----x0)--+--2f---(x0)(x------x0)---+--⋅-⋅ ⋅- xl→ixm = xli→mx ′ 1- ′′ 2 0 g(x) 0 g(x0) + g (x0) (x - x0) + 2g (x0)(x - x0) + ⋅ ⋅ ⋅

but both f(x₀) = 0 and g(x₀) = 0, so

f (x) f ′(x ) (x - x ) + 1f ′′(x )(x - x )2 + ⋅ ⋅ ⋅ lim -------= lim ------0----------0------2------0-----------0----------- x→x0 g(x) x →x0 g ′(x ) (x - x ) + 1g ′′(x )(x - x )2 + ⋅ ⋅ ⋅ 0 0 2 0 0

divide out the factor (x - x₀) from numerator and denominator:

f ′(x ) + 1f ′′(x )(x - x ) + ⋅ ⋅ ⋅ = lim ------0------2------0----------0---------- x→x0 g′(x0) + 1g ′′(x0)(x - x0) + ⋅ ⋅ ⋅ 2

and in the limit x → x₀, (x - x₀) → 0,

′ 1- ′′ ′ f--(x0)--+---2f--(x0)(x-------x0)--+--⋅ ⋅-⋅ f--(x0)-- xli→mx0 ′ 1- ′′ = ′ g (x0) + 2g (x0)(x - x0) + ⋅ ⋅ ⋅ g (x0)

We restate this as l’Hospitals rule;

f (x) f ′(x0) lim -------= --------- x →x0 g(x) g′(x0)

provided g′(x₀) is non-zero. If g′(x₀) and f′(x₀) are in fact both zero, we simply repeat the process noting that in

f ′(x ) + 1f ′′(x )(x - x ) + ⋅ ⋅ ⋅ lim -----0------2-------0----------0---------- x→x0 g ′(x ) + 1g ′′(x )(x - x ) + ⋅ ⋅ ⋅ 0 2 0 0

the ﬁrst term in both numerator and denominator are zero and we can divide out another factor of (x - x₀);

1- ′′ ′′ 0-+--2-f--(x0)--+--⋅-⋅ ⋅ f--(x0)-- = xl→imx0 1- ′′ = ′′ 0 + 2 g (x0) + ⋅ ⋅ ⋅ g (x0)

3 Non-polynomial functions

The most complicated derivatives that you will need to perform are of functions such as

1 √ -- f (x) = ---, f(x) = n x, f (x) = sin ax xn

which are not polynomials. These can be very simply differentiated by using the product rule alone, resulting in differentiation rules for radicals and quotients.

3.1 Rational functions

Consider a function that is the ratio of two functions, both of which you can differentiate;

f-(x)-- h(x) = g(x)

To compute the derivative of h(x) we take these algebraic steps, ﬁrst

h(x) g(x) = f (x)

now apply the product rule

d d ----(h(x) g(x)) = ----f (x) = f ′(x), h′(x) g(x)+h(x) g′(x) = f (x) dx dx

and rearrange

′ ′ ′ f(x)- ′ ′ ′ ′ f--(x)-----f-(x)--g-(x)-- f--(x)-----g(x)-g--(x)- f--(x)--g(x)------f-(x)-g--(x)- h (x) = = = 2 g(x) g(x) g (x)

which we will call the quotient rule.

3.2 Radicals

Consider the radical

√n -- f(x) = x

To compute its derivative, ﬁrst raise both sides to the n^th power

n n- 1 f (x) = f (x)f (x) = x

Now differentiate and apply the product rule repeatedly to the left side

d n d n- 1 ′ ----f (x) = ----x = 1, n f (x)f (x) = 1 dx dx

solve for f′(x);

′ ------1------ ----1---- -1 1n- 1 f (x) = n - 1 = n--1 = x nf (x) n x n n

We have shown that

d √ -- d 1 1 1 ----n x = ---x n- = --x n-- 1 dx dx n

and therefore for any power a, integral, rational or otherwise

d ---xa = a xa - 1 dx

which we call the power rule for differentiation.

Example Find a series expansion for f(x) = √ --
x valid near x₀ = 4.
The ﬁrst step is to compute a few derivatives, using the power rule with a = 1
2 ,

d √ -- 1 1 d2 √ -- d 1 1 1 1 ---- x = --√----, ----- x = ------√---- = - ---√------ dx 2 x dx2 dx 2 x 22 x3

and so inserting this all into Eq. 5 we ﬁnd that

2 √ --------- √ -- 1 1 1 1 2 dx (dx) 4 + dx = 4+ -√----dx+ --(- ---√-----) (dx) + ⋅ ⋅ ⋅ = 2+ ----- -------+ ⋅ ⋅ ⋅ 2 4 2 22 43 4 64

This should be written using x = 4 + dx, as

√ -- 2 (x-----4)- (x-----4)-- x = 2 + - + ⋅ ⋅ ⋅ 4 64

and in any formula involving √ --
x

that will be used for x near 4, this is a valid replacement.
In particular, this can be used to calculate square roots of numbers close to 4, such as 5 for which

√ -- 1 1 5 ≈ 2 + --- ----+ ⋅ ⋅ ⋅ = 2.234375... 4 64

which gives quite good accuracy (we are off in the third decimal place) with only these three terms in the series.

3.3 Problems

6 Compute

√ --------- -d-- 2 x + 1 dx

7 Compute

√ --------- -d--3 2 x + x dx

8 Compute

d 1 ---( ---------) dx x3 + x

9 Compute

2 -d-( 1----x---) dx x3 + x

10 Find a series expansion for

1 f (x) = ---- x2

valid near x = 1 that agrees with f(x) up through the third derivative at x = 1.

11 Find a series expansion for

1 f (x) = ----------- (1 - x)2

valid near x = 0 that agrees with f(x) up through the third derivative at x = 0.

12 There is a certain function with the truly unique property that at any point x, all of its derivatives are the same;

′ ′′ ′′′ f(x) = f (x) = f (x) = f (x) = ⋅ ⋅ ⋅

If we deﬁne f(0) = 1, ﬁnd the Taylor series expansion for x near zero for this function.

13 Show that

∑N 1 - xN +1 xn = 1 + x + x2 + ⋅ ⋅ ⋅ + xN = ------------- 1 - x n=0

This can be done by purely elementary means.
Use l’Hopital’s rule to show that

d N∑ ∑N N (N + 1) ---- xn ∣ = n = -------------- dx n=0 x=1 n=0 2

14 Use l’Hopital’s rule to compute

N - N x-------x----- xlim→1 - 1 x - x

15 Suppose that you can ﬁnd two functions f(x) and g(x) such that

-d-- -d-- f (x) = g(x), g(x) = - f (x) dx dx

for any point x. Prove that

f 2(x) + g2(x) is a constant for any x

Can you think of two functions for which this is true?

4 Integration

Integration is the anti-derivative; this is how we will deﬁne it. The process of integration must undo the process of differentiation, and so we deﬁne

∫ b df (x) --------dx = f (b) - f (a) a dx

A Riemann sum will do the trick; consider

∫ b N∑ (b----a)-- (b----a)-- a f (x) dx = lim f (a + n) N → ∞ n=0 N N

which is the area under the curve f(x) from a to b, being made up of little strips of width (b-a) N of height f(a + n(b-a) N ).

How does this undo the derivative? Let

(b - a) Δ = ---------- N

and write the Riemann sum as

∫ b N∑ a f (x) dx = lim f (a + n Δ) Δ N → ∞ n=0

and insert into this a differentiated function f(x) = dg(x) dx = lim _Δ→0g(x+Δ)-g(x) Δ ;

∫ b a f (x) dx = lim Δ(f (a)+f (a+ Δ)+ ⋅ ⋅ ⋅+f (a+(N - 1) Δ)+f (a+N Δ)) N → ∞

g(a + Δ) - g(a) g(a + 2Δ) - g(a + Δ) = lim Δ( ---------------------- + ------------------------------+ ⋅ ⋅ ⋅ N → ∞ Δ Δ

g(a + (N - 1) Δ) - g(a + (N - 2)Δ) + -------------------------------------------------- Δ

g(a + N Δ) - g(a + (N - 1) Δ) + ------------------------------------------) Δ

You can see that consecutive terms partially cancel, and so does the Δ in the denominator, leaving

(b-----a)- = lim (g(a+N Δ) - g(a)) = lim (g(a+N )- g(a)) = g(b) - g(a) N → ∞ N → ∞ N

Integration is a harder problem than differentiation, since the only procedure for performing integrals is to either do the Riemann sum directly, or ﬁnd a function whose derivative is the integrand, or to perform variable changes that put the integrand into a more readily recognized form.

Example. It is not hard to show that

∑N N (N + 1) n = -------------- n=1 2

To do it lay out all numbers 1 through N in a row

1 + 2 + 3 + 4 + 5 + 6 + ⋅ ⋅ ⋅ + (N - 2) + (N - 1) + N

and below it all numbers in reverse order

N + (N - 1) + (N - 2) + ⋅ ⋅ ⋅ + 6 + 5 + 4 + 3 + 2 + 1

and add the two rows

(N +1)+(N +1)+(N +1)+ ⋅ ⋅ ⋅+(N +1)+(N +1)+(N +1) = N (N +1)

but this is each number counted twice.
We can use this to integrate

∫ b N∑ N∑ x dx = lim (a + n Δ) Δ = lim ((N + 1)a Δ + Δ2 n) a N → ∞ n=0 N → ∞ n=0

with Δ = b-a N . Put this in;

∫ b a(b-----a)-- 1- b----a-- 2 a x dx = lim ((N + 1) + N (N + 1)( ) ) N → ∞ N 2 N

1 2 1 2 1 2 = a(b - a) + --(b - a) = --b - -a 2 2 2

By considering more difficult Riemann sums we can establish

∫ b n ---1---- n+1 n+1 a x dx = (b - a ) n + 1

This formula is valid for any n

- 1, even irrational values.

Like differentiation, integration is a linear operation

∫ ∫ ∫ b(f (x) + g(x)) dx = b f (x) dx + b g(x) dx a a a

and another very useful property inherited from the Riemann sum deﬁnition is

∫ c ∫ b ∫ c f (x) dx = f (x) dx + f (x) dx a a b

What is integration used for in 201? Suppose that you know the value of a function, such as the position of a body, at time t₀, and the velocity at all times. Integration is used to recover the position at any time t from

dx ∫ t v(t) = ----, x(t) = x(t0) + v(t) dt dt t0

4.1 Problems

These problems are similar to any integration exercises that you may encounter in the homework for the ﬁrst few chapters of the book.

16 Evaluate

∫ 3 (2.0 + 4.0 x) dx 1

17 Evaluate

∫ 1 3(1.0 x + 1.0 ----) dx 1 x2

18 The distance traveled, or your odometer reading, depends on your speed, not your velocity, the odometer does not care which way you are going. Speed is s(t) = ∣v(t)∣ = ∣ dx(t)
dt ∣. Suppose that your position versus time is

2 x(t) = 10.0 + 4.0 t - t

Find your velocity v(t), and the distance traveled from t = 0 to t = 3 seconds;

∫ 3 ∣v(t) ∣ dx 0

Hint Find the time at which your velocity switches from positive to negative, and break the integral up into two parts.

5 Trigonometric functions

We will use the radian as the measure of an angle in 201 and 202. The angle θ in radians equals the ratio of the arc-length subtended by the angle θ on a circle of radius R, to the radius;

s-- θ (rad) = R

as in the ﬁgure.

This means that technically, the radian is a dimensionless quantity, being the ratio of two lengths, and there are 2π rad in a full circle. The conversion factor is then

o 360 = 2π rad

Trig functions are deﬁned as the base and height of right-triangles inscribed within the circle, with R cos(θ) the base and R sin(θ) the height. The Pythagorean theorem then states that

2 2 2 2 2 R = (R cos( θ)) + (R sin( θ)) , 1 = cos (θ) + sin ( θ)

Virtually all trig identities can be derived from this picture by stacking triangles. Consider the ﬁgure below, for which;

a + b = cos( φ), c = cos( θ + φ), c + d = cos( θ)

e + f = sin( θ + φ), f = a sin( θ), b = e sin( θ), c = a cos( θ)

Now just apply the Pythagorean theorem to the two little triangles

┌│ ---------------- 2 2 2 2 2 2 ││ cos2( θ + φ) cos( θ + φ) a = c +f = cos (θ+ φ)+a sin (θ), a = │∘ ----------2---- = -------------- 1 - sin (θ) cos( θ)

and

┌│ --------------- ││ sin2( φ) sin( φ) e2 = b2+sin2( φ) = e2 sin2( θ)+sin2( φ), e = │∘ --------------- = -------- 1 - sin2( θ) cos( θ)

and from these we get

cos( θ + φ) sin( φ) f = a sin( θ) = -------------- sin( θ), b = e sin( θ) = -------- sin( θ) cos( θ) cos( θ)

These give us the sum formula;

cos(-θ-+--φ)-- sin(φ)-- a + b = cos( φ) = + sin( θ) cos( θ) cos( θ)

rearranging;

cos( θ + φ) = cos( θ) cos( φ) - sin( θ) sin( φ)

In a similar way we get the sum formula for sine;

sin( θ + φ) = sin( θ) cos( φ) + cos( θ) sin( φ)

5.1 The calculus of trig functions

We can see from the triangle deﬁnition that

cos(0) = 1, sin(0) = 0

In addition, since the cosine function decreases from its peak value of 1 as the angle θ moves away from 0, we see that

d ----cos( θ) ∣ = 0 d θ θ=0

We now establish the most important limit regarding trig functions, consider

sin(2 θ) = 2 sin( θ) cos( θ)

and apply this to

θ θ θ θ θ sin( θ) 2 sin( 2) cos( 2) sin( 2-) cos( 2-) θ sin( 2) -------- = -------------------- = --------θ-------- = cos( --)----θ--- θ θ 2- 2 2-

and again...

sin( θ) θ sin( θ) θ θ sin( θ-) -------- = cos( --) -----2--= cos( --) cos( --) -----4-- θ 2 θ- 2 4 -θ 2 4

and keep doing this forever...

sin(-θ)- θ- θ- θ- -θ-- ∞∏ θ-- = cos( ) cos( ) cos( ) cos( ) ⋅ ⋅ ⋅ = cos( n ) θ 2 4 8 16 n=1 2

(the notation is for an inﬁnite product). Now let θ → 0, every term on the right-hand side is 1;

sin(θ) ∞∏ 0 lim --------= cos( ---) = 1 θ→0 θ n=1 2n

We have established that for small θ

2 sin(θ) = θ + 𝒪( θ )

This is the small-angle approximation, it holds in radians only, and you will use it extensively in the course. Differentiate it

d--- 1 sin( θ) = 1 + 𝒪( θ ) dθ

and let θ → 0;

-d-- d θ sin( θ) ∣θ=0 = 1

Return now to

2 2 sin (θ) + cos (θ) = 1

and differentiate;

d d sin( θ) ----sin( θ) + cos( θ) ----cos( θ) = 0 d θ dθ

This has solution

-d-- -d-- sin( θ) = α cos( θ), cos( θ) = - α sin(θ) d θ d θ

Take each of these relations and let θ → 0;

d ----sin( θ) ∣ = 1 = α cos(0) = α, α = 1 d θ θ=0

and we arrive at the derivative formulas

d d ----sin(θ) = cos( θ), ----cos( θ) = - sin( θ) dθ d θ

All other trigonometric derivative formulas can be gotten rom these.

5.2 Problems

19 For a particle with position vector

1 2 1 2 ⃗x = (r0 cos( ωt + --αt ), r0 sin( ωt + --αt ), 0) 2 2

with α,ω and r₀ constant, ﬁnd the velocity

= d dt

and acceleration

vectors.

20 For a particle with position vector

⃗x = (r0 cos( ωt), r0 sin( ωt), 0)

with ω and r₀ constant, ﬁnd the velocity

= d dt

and acceleration

vectors.
Call

⃗Ω = (0, 0, ω) = ω ˆk

and show that

2 ⃗v = ⃗Ω × ⃗x, ⃗a = - ω ⃗x

6 Lines

To ﬁnd the equation y = ax + b of the line passing through points (x₁,y₁) and (x₂,y₂) we notice that

dy ---- = a dx

is constant (independent of x) for a straight line, and so can be computed without having to take a limit,

rise-- y2-----y1-- a = = run x2 - x1

and we then solve

y2 - y1 y1 = ax1 + b = -----------x1 + b x2 - x1

for the intercept

-y2----y1-- x2y1-----y2x1--- b = y1 - x1 = x2 - x1 x2 - x1

7 e^x and ln x

The function f(x) all of whose derivatives agree at x

f (x) = f ′(x) = f ′′(x) = ⋅ ⋅ ⋅

is special. Consider its Taylor expansion around x;

′ 1 2 ′′ 1 3 ′′′ f (2x) = f (x + x) = f (x) + xf (x) + --x f (x) + ---x f (x) + ⋅ ⋅ ⋅ 2 3!

1 1 = f (x) (1 + x + -x2 + ---x3 + ⋅ ⋅ ⋅) 2 3!

and around 0;

′ 1 2 ′′ 1 3 ′′′ f (x) = f (0 + x) = f(0) + xf (0) + -x f (0) + --x f (0) + ⋅ ⋅ ⋅ 2 3!

1 2 1 3 = f (0) (1 + x + --x + ---x + ⋅ ⋅ ⋅) 2 3!

Combining these two relations we get

---1-- 2 f (2x) = f (x) f (0)

and if we let f(0) = 1, this means that the function obeys the characteristic identity of an exponential function

f(2x) = f 2(x)

In other words there is some number q such that

x f (x) = q

since

2x x 2 q = (q )

We can easily calculate q;

1 1- 2 -1- 3 f (1) = q = q = f (0) (1 + 1 + 1 + 1 + ⋅ ⋅ ⋅) = 2.71828... 2 3!

which is simply called e. We have constructed the classic exponential function as the unique function equal to its own derivative at any point

x ∞∑ 1-- n -d-- x x a+b a b e = x , e = e , e = e e n=0 n! dx

The integral

∫ a dx ---- = g(a) 1 dx

turns out to be the inverse function of the exponential. We will show that

g(ab) = g(a) + g(b)

∫ ab dx ∫ a dx ∫ ab dx g(ab) = ---- = ----+ ---- 1 x 1 x a x

in the last integral let x = ay, when y = 1, x = a, when y = b, x = ab;

∫ ab dx ∫ a dx ∫ b dy ----= ----+ ----= g(a) + g(b) 1 x 1 x 1 y

Any function with this property is an inverse-exponential;

g(a)+g(b) g(a) g(b) g(ab) e = e e = e

and

g(eaeb) = g(ea+b) = g(ea) + g(eb)

These have solution

x g(e ) = x

The function is called the natural logarithm

∫ x dy-- ln x x ln x = 1 , e = x, ln(e ) = x x

x- n ln xy = ln x + ln y, ln = ln x - ln y, ln x = n ln x y

7.1 Problems

21 Compute the derivatives (with respect to x) of

ax+b ax2 2x 2 f1(x) = e , f2(x) = e , f3(x) = (ae + b)

22 Compute the following integrals

∫ ∞ ∫ 1 dx I1 = e- ax dx, I2 = ---------, a, b > 0 0 0 a + bx

23 Consider the simple biological problem of computing the population of a culture of organisms that reproduce by mitosis (cell division). Each organism splits into two complete organisms once it reaches maturity.
Suppose that you begin at time t = 0 with N₀ organisms. If a fraction of them α ΔtN₀, 0 < α < 1 will reach maturity in time Δt, then the population at time 0 + Δt will be N₀ + α ΔtN₀ (the longer you wait, the more mature).
Show that the rate of population N(t) increase follows

dN (t) ---------= αN (t) dt

and demonstrate (using the ideas of this section) that the population at time t will be

N (t) = N eαt 0

8 Partial derivatives

A function of several variables can have rates of change with respect to any of its variables. The partial derivative of a function f(x,y) with respect to x is the rate of change with respect to x;