Transcript Calculus FDWK Section 6.4

Imagine this
much bacteria
in a Petri dish
Now this
amount of the
same bacteria
Assuming that each bacterium would reproduce at the
same rate, which dish will have a larger rate of growth?
Answer: The second one simply because there are more
of them.
dy
So if we presume that the rate of growth is given by:
dt
y = amount of bacteria, t = time
And that the population produces at a rate proportional to
itself with the proportion represented by the constant k
dy
So if we presume that the rate of growth is given by:
dt
y = amount of bacteria, t = time
And that the population produces at a rate proportional to
itself with the proportion represented by the constant k
dy
 ?k  y
Then
dt
Use your newfound skills for solving differential equations
to solve for y here:
dy
 ky
dt
1
dy  k dt
y
1
 y dy   k dt
ln y  kt  C
ln y
e
e
kt C
y  eC  e kt
y  Ae
kt
Since the initial amount is at t = 0
y  Ae
0
A  y0
In this case, y0 is the initial amount
So our equation for this type of
growth would be…
Exponential Change:
y  y0e
kt
Remember too that we’ve just shown that
y  Ae
kt
is the solution to the differential equation
dy
 ky
dt
If the constant k is positive then the equation represents
growth. If k is negative then the equation represents
decay.
There is a similar growth equation used in finance that you
may remember from pre-calc…and we’ll talk about that soon
One straight-forward application of this is
1
 kt
y0  y0 e
2
1
 kt
e
2
1
 kt
ln    ln e
2
 
ln1  ln 2  kt
ln 2  kt
ln 2
half-life 
k
Half-life is the period of time it
takes for a substance undergoing
decay to decrease by half.
In this case, think of y0 as the initial
amount of a substance undergoing
decay. To find its half life…
ln 2
t
k
Compounded Interest
If money is invested in a fixed-interest account where the
total interest r (which is a % written as a decimal) is
broken into k equal portions and added to the account k
times per year, the amount present after t years is:
 r
A  t   A0 1  
 k
Initial investment
kt
# times per year over t years
% added each time
100% (initial investment)
If the interest is broken down more and added back more
frequently (k is larger), you will make a little more money.
We can add as many times as we want which means we
can make k as large as…

The larger k gets, the more times per year we compound
the interest. If we can theoretically compound an infinite
number of times, we say that the interest is compounded
continuously
 r
We could calculate: lim A0 1  
k 
 k
kt
k
 1
lim 1   
k 
k
Using an old limit from pre-calc
k
 r
lim1    er
k 
 k
e
The larger k gets, the more times per year we compound
the interest. If we can theoretically compound an infinite
number of times, we say that the interest is compounded
continuously
kt
 r
We could calculate: lim A0 1   which turns out to be:
k 
 k
Continuously Compounded Interest:
A0e
rt
Just like the exponential growth model we just saw, the
interest is directly proportional to the amount present.
You may also use:
A  Pert
Remember PERT?
Same equation, different letters
Newton’s Law of Cooling
Coffee left in a cup will cool to the temperature of the
surrounding air. The rate of cooling is proportional to the
difference in temperature between the liquid and the air.
(The colder the air, the faster the coffee cools)
This would give us the differential equation:
dT
 k T  Ts 
dt
where Ts is the temperature
of the surrounding medium,
which is a constant.
Newton’s Law of Cooling
Coffee left in a cup will cool to the temperature of the
surrounding air. The rate of cooling is proportional to the
difference in temperature between the liquid and the air.
(The colder the air, the faster the coffee cools)
If we solve the differential
equation, we get:
T  Ts  T0  Ts  e
 kt
Newton’s Law of Cooling
Don’t be afraid of the size of this
equation. It really is not that different
from the first exponential growth/decay
equation. Don’t forget also that TS and
T0 are constants. Just look at this
comparison…
Newton’s Law of Cooling
Coffee left in a cup will cool to the temperature of the
surrounding air. The rate of cooling is proportional to the
difference in temperature between the liquid and the air.
(The colder the air, the faster the coffee cools)
If we solve the differential
equation, we get:
T  Ts  T0  Ts  e
 kt
Newton’s Law of Cooling
T  Ts  T0  Ts  e
 kt
y  Ae
kt
It’s just a matter of sorting through the constants
The End