11 - Lab 11 Exponential Functions.sagews

##Lab 11 Exponential Functions This lab takes a look at different forms of exponential functions. The basic form of all exponential functions is $A(t)=A_0 b^t$. $A_0$ is called the initial value and $b$ is the growth/decay factor over a unit of time, depending on whether it's less than or greater than 1. For instance, the rate of inflation is currently 2.1% each year. So the price of a gallon of gas next year is $1+.021$ times its current price. A function that models the value of a gallon of gas, which let's say is $2.59 right now is $P(t)=2.59(1+.021)^t=2.59(1.021)^t$. If things don't change, what will a $2.59 gallon cost 10 years from now?

t=var('t')
P(t)=2.59*1.021^t#created the price function
plot(P(t),xmin=-1,xmax=10,ymin=-1, ymax=5,axes_labels=['$t$','$P(t)$'])#plotting
show(P(10).n(15))#evaluating at 10 years
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%md
Create a function that models this situation and find the solution:  A \$45000 pickup truck depreciates at a rate of 15% a year.  When will the truck be worth only half its original value?

Create a function that models this situation and find the solution: A $45000 pickup truck depreciates at a rate of 15% a year. When will the truck be worth only half its original value?

%md
Another common form is **change over some specified unit of time**.  The two most popular are *doubling time* and *half-life*.  These have the form $A(t)=A_0(2)^{\frac{t}{k}}$ and $A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{k}}$.  The k-value is the time it takes to either double or halve.  How does this work?  Let's take a look. . .  Suppose we start with \$100 and are told the amount will double every 25 years.  We'll create a loop to evaluate the amount for every one of the first 25 years.  We'll do this twice, the first time showing exact values and the second time approximations.  You will see that the exponent for base 2 slowly grows to the whole number one.

Another common form is change over some specified unit of time. The two most popular are doubling time and half-life. These have the form $A(t)=A_0(2)^{\frac{t}{k}}$ and $A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{k}}$. The k-value is the time it takes to either double or halve. How does this work? Let's take a look. . . Suppose we start with $100 and are told the amount will double every 25 years. We'll create a loop to evaluate the amount for every one of the first 25 years.

t=var('t')
A(t)=100*2^(t/25)
for i in [0..25]:
    [i,A(i)]
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#now to approximate the same thing
t=var('t')
A(t)=100*2^(t/25)
for i in [0..25]:
    [i,A(i).n(20)]
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%md
Use my code with modifications to handle these situations:
1.  You start with 200 pounds and drop to 3/4ths of this in 6 months.  What's the weight after 3 months?  When will the weight be cut in half?
2.  The polar ice cap has shrunk to 95% of its size in the last 20 years.  How long does it take to lose 1%?(You'll have to do a little extra calculation!.  Also you can just use 1 for $A_0$.)

Use my code with modifications to handle these situations:

1. You start with 200 pounds and drop to 3/4ths of this in 6 months. What's the weight after 3 months? When will the weight be cut in half?

2. The polar ice cap has shrunk to 95% of its size in the last 20 years. How long does it take to lose 1%?(You'll have to do a little extra calculation!. Also you can just use 1 for $A_0$.)

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%md
Finally, let's look at the situation where you don't know the initial value nor do you know the growth/decay factor.  All you know is that there are two points and that exponential change is occurring.  F'rinstance,  at the 6 year mark the value was \$1000 then at the 20 year mark it was \$900.  The change occurred exponentially.  Write a function model for this.
Steps:
1.  Find the ratio $\frac{\mbox{new}}{\mbox{old}}=\frac{900}{1000}$.
2.  Now find the amount of time it took for this fractional change to take place:  $20-6=14$
3.  We can now use the form, $A(t)=A_0(\frac{900}{1000})^{\frac{t}{14}}$
4.  We still don't know what $A_0$, the initial value is.  But we can use either of the points (6,1000) or (20,900) as values of $t$ and $A(t)$, plug in and solve for A_0.
Study my code.   You'll notice that I am letting SAGE do everything!

Finally, let's look at the situation where you don't know the initial value nor do you know the growth/decay factor. All you know is that there are two points and that exponential change is occurring. F'rinstance, at the 6 year mark the value was $1000 then at the 20 year mark it was $900. The change occurred exponentially. Write a function model for this. Steps:

1. Find the ratio $\frac{\mbox{new}}{\mbox{old}}=\frac{900}{1000}$.

2. Now find the amount of time it took for this fractional change to take place: $20-6=14$

3. We can now use the form, $A(t)=A_0(\frac{900}{1000})^{\frac{t}{14}}$

4. We still don't know what $A_0$, the initial value is. But we can use either of the points (6,1000) or (20,900) as values of $t$ and $A(t)$, plug in and solve for A_0. Study my code. You'll notice that I am letting SAGE do everything!

t,A_0=var('t,A_0')
k=20-6
r=900/1000
A(t)=A_0*r^(t/k)
show(A(t))
solve(900==A(20),A_0)
show(solve(900==A(20),A_0))

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#now I'll approximate A_0 two different ways
#first way
1000/9*10^(3/7)*9^(4/7).n(20)
#second way(gets the job done better)
numerical_approx(1000/9*10^(3/7)*9^(4/7))

389.98*10^(3/7) 1046.18949071864

So a function model for the above situation is $A(t)=1046.19(.9)^{\frac{t}{14}}$. Your turn, create a function model for a biological specimen that had 300 cells at the 5 minute mark and then at the 30 minute mark had grown to 8,000,000 cells. How many cells were there when the clock started? Use the model to approximate how long it takes the organism to double in size.

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