Exact Solution to Constant Instantaneous Percentage Growth

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Notebook Overview

This NoteBook looks at various aspects of the important fact that exponentials satisfy the basic growth law that the quantity changes in proportion to the amount.

(dy/dt) ∝  
        y  ⟺  (dy/dt) =  
      k*y   ⟺ (1/y) (dy/dt) =  k,  
    for a constant k

Official Definition of  (e^t):

y[t] = (e^t)

is the unique function satisfying

y[0] = 1

and

(dy/dt)  =  y

Mathematica knows this:

\(Clear[t, y]; 

    DSolve[{y[0]\  == \ 1, \ \(y'\)[t] == \ y[t]}, y[t], t]\)

See the NoteBook ExpEquns for more details.

Exponential Growth Laws in General: y[t]  =  c (e^( k (t -  (Subscript[t, 0]))))

More generally, if  c  and  k are constants and  (Subscript[t, 0]) is a fixed value of  t, then

y[t]  =  c (e^( k (t -  (Subscript[t, 0]))))

is the unique (one and only) solution to the differential equation

(dy/dt)  =  k y  satisfying   y[(Subscript[t, 0])] = c

Mathematica knows this, too:

\(Clear[t, y]; 

    DSolve[{y[2]\  == \ 3, \ \(y'\)[t] == \ 5\ y[t]}, y[t], t]\)

\(Clear[t, y, k, c]; 

    DSolve[{y[t\_0]\  == \ c, \ \(y'\)[t] == \ k\ y[t]}, y[t], t]\)

The General Exponential Growth Law Solution: y[t] = h (e^(k t))

Another way to say the general solution above is:

Every solution to the differential equation

(dy/dt)  =  k y

can be written in the form

y[t] = h (e^(k t))

The FORM of the solution  y[t]  =  c (e^( k (t -  (Subscript[t, 0]))))  above can be re-written as follows:

Begin with  y[t] = c (e^( k (t -  (Subscript[t, 0])))), for  c, k, (Subscript[t, 0]) constants. (Every solution takes a value  c  at some time  (Subscript[t, 0]).)

c (e^( k (t -  (Subscript[t, 0])))) =

= c (e^( k t -  k (Subscript[t, 0])))  , 	expanding the exponent

= c 
        ((e^( k t))*(e^( -k  (Subscript[t, 0]))))
          ,	by the law of exponents  (a^(x + y)) = (a^x) (a^y)

= h (e^( k t)),		combing constants,  h = c (e^( -k  (Subscript[t, 0])))

Exponential Growth with Other Bases: y[t] = h (b^t)

The function  y[t] = h (e^(k t))  can be expressed in a third equivalent way.

(e^(k t))= ((((e^k)))^t),		by the law of exponents ((((a^x)))^y)  =  (a^(x*y))

= (b^t),			taking  b = (e^k)

So,  y[t] = h (e^(k t))= h (b^t) .

Proportional Growth

Another way to say the differential equation

(dy/dt)  =  k y

is:  "The instantaneous rate of growth of the function  y[t]  is a constant PERCENTAGE of the value  y[t]."  The rate of growth is the derivative.  If the derivative is a fixed percentage of y, then  (1/y)(dy/dt)× 100%  is the percentage rate of growth, which the statement says is constant:

(1/y) (dy/dt) = k

The discussion above shows that this expression is equivalent to the statement

y[t] = h (e^(k t))  or equivalently  y[t] = h (b^t), for  k = Log[b]

for some constant h.

Exercises

Exercise 8.2.1

a

Let  y[t] = (b^t) for a fixed, but unknown positive constant.  Use the chain rule to show that  y[t] satisfies:

y[0] = 1(dy/dt) = k y

What is the value of the constant k?

\(Clear[y, b, t]; 
y[t_] := b\^t; 
?? \)

b

Let  y[t]  =  (e^(k t)) for a fixed constant k.  Show that y[t]

y[0] = 1(dy/dt) = k y

\(Clear[y, k, t]; 
y[t_] := E\^\(k*t\); 
?? \)

c

Solve the initial value problem

y[0] = 5(dy/dt) =  y

Do this on paper and then check:

\(Clear[y, t]; 
DSolve[{y[0] == 5, \(y'\)[t] == y[t]}, y[t], t]\)

d

Solve the initial value problem

y[0] = 5(dy/dt) =  k y

where  k = Log[2].  Do this on paper and then check:

\(Clear[y, t]; 

    DSolve[{y[0] == 5, \(y'\)[t] == Log[2] y[t]}, y[t], t]\)

\(Clear[y, t, k]; 

    soln\  = \ DSolve[{y[0] == 5, \(y'\)[t] == k\ y[t]}, y[t], t]; 

    y[t_]\  = \ y[t] /. soln; 
y[t]\)k  =  Log[2]y[t]

Problem 8.1 The Canary's Postmortem

Let  T[t]  =  (Subscript[T, 0]) (e^(-k*t)) and show that  T[t]  satisfies:

T[0]  =  (Subscript[T, 0])

and

(dT/dt) =  - k T

If  T[0] = 75, solve with paper and pencil for the value of  k that satisfies  T[10] = 60.

\(Clear[T, t, k]; 
T[t_]\  := \ 75\ Exp[\(-k\)*t]; 

    Solve[T[10]\  == \ 60, k]\)NSolve[T[10]  ==  60, k]

\(k\  = \ 
      \(?? ;
Plot[T[t], {t, 0, 600}, PlotStyle -> {Blue}, 
            AxesLabel -> {"time", "Temp"}]; 
T[10]
T[600.0]\)\)

Problem 8.5 Bugs Bunny's Law of Gravity

A correct statement is that falling objects go faster the farther they fall.  If  s  is the distance fallen since release and  t  is the time elapsed since release, the speed is the derivative:

(ds/dt)

The correct comment only says  (ds/dt)  increases  as  s  increases.

The expression   k*s  also increases as  s  increases  when  k  is constant, so it is plausible to postulate that objects fall according to th law:

(ds/dt)= k s, 	 for a constant k.

While this IS plausible, your job is to show that it is wrong. (Notice that  the expression  k (Sqrt[ s])  also increases as  s increases, so the law   (ds/dt)= k (Sqrt[ s]) is also plausible - and RIGHT, but "ill-posed.")

Solve the problem that says we release an object from rest and it falls according to the linear law above.  Show that your solution has the property that it doesn't fall,  s[t] = 0, for all t.

\(Clear[k, s, t]; 

    DSolve[{s[0]\  == \ 0, \(s'\)[t] == k*s[t]}, s[t], t]\)

Good cartoon characters don't fall when they run off a cliff, so we call the differential equation  (ds/dt)= k s  Bugs Bunny's Law of Gravity.  You have just shown that it only works in the cartoons.  See Chapter 10 for Galileo's Law of Gravity that works in vacuum or the Bungee Diver project for a law of gravity that includes air resistance and a bungee cord.

Exercise 21.1.2

Show that the function  x[t] = 7 (e^(3 t)) satisfies

x[0]  =  7(dx/(dt )) =  3 x

\(Clear[x, t]; 
x[t_]\  := \ \(?? 
x[0]
\(x'\)[t]/x[t]\)\)

In other words, x[t] satisfies:

\(Clear[x, t]; 
DSolve[{x[0] == 7, \(x'\)[t] == 3  x[t]}, x[t], t]\)

Exercise 21.1.3

Solve the initial value problem

x[0]  =  57(dx/dt) =  -23 x

\(Clear[x, t]; 

    DSolve[{x[0]\  == \ 57, \(x'\)[t] == \(-23\)\ x[t]}, x[t], t]\)

Exercise 21.1.4

Solve the initial value problem

x[0]  =  (Subscript[x, 0])(dx/dt) =  r x

\(Clear[x, t, x0, r]; 

    DSolve[{x[0]\  == \ x0, \(x'\)[t] == r\ x[t]}, x[t], t]\)

Exercise 21.1.5

Show that the algae growth model

x[0]  =  57(dx/dt) =   x

predicts that  x = 1000000  when  t = Log[((10^6)/57)]

Solve[57 (E^t) == (10^6), t]NSolve[57 (E^t) == (10^6), t]

Exercise 21.1.6

Show that the algae growth model

x[0]  =  (Subscript[x, 0])(dx/dt) =   r x

predicts that  x = 1000000  when  t = Log[((10^6)/(Subscript[x, 0]))]/r

Solve[x0 (E^(r t)) == (10^6), t]

Exercise 21.1.7

a

A function  x[t]  satisfies the differential equation  (dx/dt) = 3 x .  Find the time  (Subscript[t, 2]) where  x[(Subscript[t, 2])] = 2 x[0].

\(Clear[x, t]; 

    x[t_] := \(?? \ 
Solve[x[t]==2\ x[0]]
NSolve[x[t] == 2\ x[0]]\)\)

b

Prove that the function also satisfies  x[t + (Subscript[t, 2])] = 2 x[t]

\(t2\  = \ \(?? ;
Expand[x[t + t2]/x[t]]\)\)

Exercise 21.1.8

Radioactive substances decay at a constant percentage of the amount left.  Show that the function  x[t]  satisfying  x[0] = (Subscript[x, 0]) and  (dx/dt) = -r x has

x[(Subscript[t, 1/2])] = x[0]/2  when  (Subscript[t, 1/2])= Log[2]/r

\(Clear[x, t, r]; 
x[t_]\  := \ \(?? 
Solve[x[t]==x[0]/2,t]\)\)

\(t2\  = \ \(?? ;
Expand[x[t + t2]/x[t]]\)\)

Exercise 21.1.9

The mouse population in my neighbor's field doubles every five days.  What is the instantaneous rate of growth?  What is the instantaneous percentage rate of growth?