Calculus: The Language of Change

Exercise Set 5.4
Derivatives of Log and Exp-CD

by Wyatt Hadley

Exercise 1: Fixed Percentage Changes

Part 1

Find an exponential function f[x]=(a^x) that doubles every time x increases by 1. Write this English question as a mathematical equation and solve it using properties of exponents.

Solution of part 1

We want the difference of f[x+1] and f[x] to be f[x] so that f[x] doubles whenever x increases by 1 or
		f[x+1] - f[x] = f[x]
so		f[x+1] = 2 f[x]
for f[x] = (a^x)	 (a^(x + 1)) = 2  (a^x)
		 (a^x) (a^1) = 2  (a^x)
so		a = 2 and f[x] =  (2^x)

Part 2

Find an exponential function f[x]= (b^x) that triples every time x increases by 1. Write this English question as a mathematical equation and solve it using properties of exponents.

Solution of part 2

We want the difference of f[x+1] and f[x] to be 2 f[x] so that f[x] triples whenever x increases by 1 or
		f[x+1] - f[x] = 2 f[x]
so		f[x+1] = 3 f[x]
for f[x] = (b^x)	 (b^(x + 1)) = 3  (b^x)
		 (b^x) (b^1) = 3  (b^x)
so		b = 3 and f[x] = (3^x)

Part 3

Find an exponential function f[x]= (c^x) that increases by 50% every time x increases by 1/2 x-unit. Write this English question as a mathematical  equation and solve it using properties of exponents and the logarithm.

Solution of part 3

We want the difference of f[x+1/2] and f[x] to be 0.5 f[x] so that f[x] increases by 50% whenever x increases by 1/2 or
		f[x+1/2] - f[x] = 0.5 f[x]
so		f[x+1/2] = 1.5 f[x]
for f[x] = (c^x)	(c^(x + 1/2)) = 1.5  (c^x)
		(c^x) (c^(1/2)) = 1.5  (c^x)
so		(c^(1/2)) = 1.5 and c = (1.5^2) = 2.25
		f[x] = (2.25^x)

Part 4

Show that the exponential base a from the first part is NOT equal to c from the third part. Should not a function that increases by 50% in 1/2 unit be the same as one that increases by 100% in 1 unit? Why not?

Solution of part 4

The value of a in part one was 2; the value of c in part 3 was 2.25.  We have shown that they are not equal, but why is this the case?  It is important to remember, that if a function increases by 50% in 1/2 unit, the increase of the function for the second half unit will be 50% of the original, plus 50% of the previous increase!  So in this case we have a total increase of 50% plus 75% or 125% compared to 100% in the first part.  This is why compounding interest is greater than simple interest.

Part 5

Let f[x] be an unknown function, h and k unknown constants. Write the  statement, "The change in f[x] as x increases by h equals k times f[x] " as a mathematical equation. In other words, your equation should say, "f[x] increases by k*100% as x increases by h."

Solution of part 5

We want the difference of f[x+h] and f[x] to be k * f[x] so that f[x] increases by k*100%  whenever x increases by h or
		f[x+h] - f[x] = k * f[x]

Continuing as above we have
		f[x+h] = (k+1) * f[x]
for f[x] = (a^x)	(a^(x + h)) = (k+1)  (a^x)
		(a^x) (a^h) = (k+1)  (a^x)
so		a = (((k + 1))^(1/h)) and f[x] = (((k + 1))^(x/h))

Exercise 2 : A doubling Exponential

Suppose algae cells in a warm pond double every 6 hours and at time t=0 (hrs) there is one cell.  How many cells are there in 6 hours?  How many cells are there in 12 hours?  How many in 18 hours? 

    How many 6-hour periods are there in t hours? (A formula.) 

    Give the number of cells n as a function of t, n[t]=2???

   Suppose at time (Subscript[t, 1]) there are a billion cells. How many are there at time (Subscript[t, 1])+6? 

    What is the formula for the rate of growth of algae cells in a 6-hour period beginning at an unknown time t? (Compare your work to Problem CD-28.2 and the program ExpGth of Chapter 28.) 

    The next exercise has you practice using the functional identities for the logarithm. The point of the exercise is that we need only one limit, η → 0, and the functional identity.

part a

Suppose algae cells in a warm pond double every 6 hours and at time t=0 (hrs) there is one cell. How many cells are there in 6 hours? How many cells are there in 12 hours?  How many in 18 hours?

Solution of part a

Quite simply, one can take the number of cells from 6 hrs ago and multiply it by two to get the current number of cells.  So, if there was one cell at time t=0, there would be 2 cells at t=6; 4 cells at t=12; 8 cells at t=18; etc...

part b

How many 6-hour periods are there in t hours? (A formula.)

Solution of part b

As stated in the problem, 't' is expressed in units of hrs.  At any given 't', there are 't'/6 6-hour periods. ( i.e. at t=18, there are 18/6=3 6-hour periods.)

part c

Give the number of cells n as a function of t, n[t]=2???

Solution of part c

Let (Subscript[n, 0]) be the number of cells at time 0.  Then, every 6 hours, this value will double.  This can be expressed as n[t] = (Subscript[n, 0])(2^(t/6)) .   Notice that, n[0]=(Subscript[n, 0]), for the exponent becomes 0 and (2^0) = 1.

part d

Suppose at time (Subscript[t, 1]) there are a billion cells. How many are there at time (Subscript[t, 1])+6?

Solution of part d

The number of cells doubles every 6 hours, so there will be twice as many, or two billion cells.  This can be shown mathematically using the solution to part c),   by assigning (Subscript[t, 1])=0 and (Subscript[n, 0])=(10^9).  Put into the function for n[t], this becomes:

\(n[t_] = n\_0*2\^\(t/6\); 
n\_0 = 10\^9; 
t\_1 = 0; 
n[t\_1 + 6]\)

Exercise 3 : The Derivative of Natural Log

1)

Given that

(( δ Log[x])/δ)x = (1/x),

\(TextForm
    \`\(write\ the\ increment\ approximation\ for\ \ Log\ at\ x = 
      1\ to\ show\ that\ \)\)

Log[1 + δx] = δx + ϵ*δx

with ϵ≈0 when δx ≈0.

Solution of part 1

First, take the increment approximation equation:

f[x+δx]-f[x]=f'[x]*δx+ϵ*δx

Then set f[x]=Log[x]:

Log[x + δx] - Log[x] = 
      (1/x) δx + ϵ*δx

Now set x=1...

\(TextForm\`\(	\ \ \ \ \ \ \ \ \ \ \ \ \ \ 
    Log[1 + δx] - Log[1] = 
      \(1\/1\) δx + ϵ*δx\)\)

Simplify..

Log[1+δx] - 0= 1*δx+ϵ*δ

And viola!

Log[1+δx] = δx+ϵ*δx    with ϵ≈0 when δx ≈0.

2)

Use properties of logs to show that

\(TextForm
    \`\(Log[x + δx] - Log[x]\)\/δx = 
      Log[1 + δx\/x]\/δx\)

Solution of part 2

Remember:  Log[a]-Log[b]=Log[a/b].  So,

\(TextForm
    \`\(\(Log[x + δx] - Log[x]\)\/δx = 
      \(Log[\(x + δx\)\/x]\/δx = 
        Log[1 + δx\/x]\/δx\)
\)\)

3)

Suppose you are given that

Log[1+η]=η+ϵ*η

with ϵ≈0 when η≈0. Use these two facts to prove that for all positive x, bounded away from 0,

Log[x + δx] = Log[x] + δ(x/x) + ϵ*δx

with ϵ≈0 when η≈0.

Solution of part 3

We are given:

Log[1+η]=η+ϵ*η

\(TextForm\`\(
Then, \ we\ set\ η = \ δx\/x\)\)

\(TextForm
    \`\(Log[1 + δx\/x] = δx\/x + ϵ*δx\/x

    \)\)

\(TextForm
    \`Log[\(x + δx\)\/x] = δx\/x + ϵ*δx\/x\)

\(TextForm
    \`Log[x + δx] - Log[x] = 
      δx\/x + ϵ*δx\/x\)

\(TextForm
    \`Log[x + δ] = Log[x] + δx\/x + ϵ*δx\/x\)

Now, use the increment approximation equation:

f[x+δx]-f[x]=f'[x]*δx+ϵ*δx

Move f[x] to the other side:

f[x+δx]=f[x]+f'[x]*δx+ϵ*δx

Set f[x]=Log[x]:

Log[x + δx] = Log[x] + (1/x)*δx + ϵ*δx

And finally, simplify:

Log[x + δx] = Log[x] + δ(x/x) + ϵ*δx

4)

Use the above to show that

\(\(
δx  →\&lim 0 
        \( Log[x + δx] - Log[x]\)\/δx = 1\/x\)\)

Solution of part 4

From above we have:
	   
      Log[x + δx] = 
        Log[x] + δ(x/x) + ϵ*δx
so	   
      Log[x + δx]  -  Log[x]  =  
        δ(x/x) + ϵ*δx
	((Log[x + δx] - Log[x])/δ)x = (1/x) + ϵ
	δx
           (Overscript[→ , lim])0 
            (( Log[x + δx] - Log[x])/δ)x = (1/x)

Exercise 4: What is wrong with the following computation?

What is wrong with the following computation?

\(TextForm\`\(

    \(Log[x + δx] - Log[x]\)\/δx = 
      \(Log[x + δx - x]\/δx = Log[δx]\/δx\)\)\)

Solution

Log[x + δx] - Log[x] ≠ Log[x + δx - x]

Remember, the sum of two Log's is NOT the Log of the sums. Refer back to part 2 of exercise three in this notebook.

Exercise 5: Percent Growth and the Natural Base

Part 1

We know that exponentials grow at a constant PERCENTAGE  rate for fixed steps. (See the program PercentGth.) Let f[x]=(e^rx) and substitute this into the expression

\(TextForm\`\(\(1\/f[x]\) \(f[x + δx] - f[x]\)\/δx = k, \ 
    k\ constant
\)\)

Solution of part 1

\(TextForm\`\(

    \(1\/e\^rx\) \(e\^\(r \((x + δx)\)\) - e\^rx\)\/δx = k
      \)\)

Part 2

Suppose you know that Log[1+η]=η+ϵ*η with ϵ≈0 when η≈0.  Show that  r≈k when δx≈0.

Solution of part 2

From above:

\(TextForm\`\(

    \(1\/e\^rx\) \(e\^\(r \((x + δx)\)\) - e\^rx\)\/δx = k
      \)\)

\(TextForm\`\(

    \(1\/e\^rx\) \(e\^\((rx + rδx)\) - e\^rx\)\/δx = k\)\)

\(TextForm\`\(

    \(1\/e\^rx\) \(\(e\^rx\) e\^rδx - e\^rx\)\/δx = k\)\)

\(TextForm\`\(

    \(1\/e\^rx\) \(\(e\^\(rx (\)\) e\^rδx - 1)\)\/δx = k\)\)

\(TextForm\`
\((e\^rδx - 1)\)\/δx = k\)\(TextForm\`\((e\^rδx - 1)\)\  = \ δx\ *\ k
\)(e^r)δx  =   δx * k  +  1\(TextForm
    \`We\ now\ take\ the\ Log\ of\ both\ 
        \(sides : 
τ\ \ δx\)\  = \ 
      Log \((\ δx\ *\ k\  + \ 1)\)\)Letting  η  =   δx * k  gives usτ  η/k  =  Log (1 + η)\(TextForm
    \`We\ are\ given\ that\ Log[1 + η] = 
      η + ϵ*η\ with\ ϵ ≈ 
        0\ when\ η ≈ 0\ so\)τ  η/k  =   η + ϵ*η\(TextForm
    \`With\ \ ϵ ≈ 0\ we\ have\ τ\ \ η/k\  
      ≈ \ \ η\ \ or\) 
    τ  η  ≈   η  k  so  τ  
      ≈  k

Exercise 7: Natural Increments

Part 1

Use the formula for the derivative of the natural exponential to write the increment approximation for y=(e^x),

\(TextForm\`\(				
    e\^\(x + δx\) = 
      e\^x + \([\(?]*δx+ϵ*δx\)\)\)\)

Solution of part 1

\(TextForm\`\(	
      e\^\(x + δx\) = 
        e\^x + \(e\^x\) δx + ϵδx\)\)

Part 2

Use the formula for the derivative of the natural logarithm to estimate Log[2.8]=log[e+(2.8-2.71828...)].

Solution of part 2

The basic microscope approximation  is

f[x+dx]-f[x] = f'[x] dx + ϵ dx

where  ϵ  tends to zero as  dx  tends to zero.  Since the derivative  of natural logarithm is  1/x,

Log[x + dx] - Log[x] = (1/x) dx + ϵ dx

If we take  x = e  and  dx = 0.0817182, then  x + dx = 2.8, note:

E + 0.0817182

Subsitituting these specific values in the microscope approximation,

Log[x + dx] - Log[x] = (1/x) dx + ϵ dx

Log[2.8] - Log[e] = (1/e) (0.0817182) + ϵ (0.0817182) ≈ (0.0817182/2.71828), because  ϵ ≈ 0.

(0.0817182/N)[E]

Log[2.8] - Log[e] ≈ 0.0300624

So Log[2.8] ≈ 1 + 0.0300624 = 1.0300624
	Now let's check:

Log[2.8]

We are accurate to only 3 significant digits.  We would have better accuracy if dx were smaller.