In this episode the FBI investigates a bizarre string of sniper attacks which seem to have little in common. To determine the location of the sniper in each shooting, Charlie uses ballistic trajectory modelling. Exponential growth and regression to the mean are also briefly mentioned, and the first of these we explore in depth below.

A bullet, like any other object flying through the air, is subject to the forces of gravity, air resistance, and wind. One way to closely approximate the actual trajectory is to ignore the effects of drag and wind, instead looking only at gravity.

Consider the figure to the right. A bullet leaves the barrel of a gun
inclined at a 30 deg angle and flies a horizontal distance of d before
reaching the starting elevation. The force of gravity acts on the
bullet, creating a downward acceleration of g=9.8 m/sec^{2} and
so influencing the vertical component of the velocity vector (see
diagram below) over time.
Since we disregard drag, the horizontal component of the velocity
does not change.

The next activity involves figuring out the equations describing the speed and position of an object in freefall. These derivations make some use of calculus. Try to follow them and do the exercises, but if you can't, just use the equations mentioned in order to do activity 2.

Technically, the term *velocity* means the vector pointing in the
direction of motion with magnitude equal to the *speed*
of the object. However, in everyday usage and even in many
physics textbooks the term velocity is used to denote both the
vector and its magnitude, the speed. It is usually easy to
figure out which is being meant from the context: just ask
yourself, is the sentence talking about a vector or a scalar?

Let us first consider only the vertical direction of motion. For the sake of brevity, I'll just write v(t) below instead of v

- Recall that the acceleration of an object is equal to the instantaneous change in velocity, i.e. a(t)=v'(t). Apply the fundamental theorem of calculus to this equality to deduce that , where g is the acceleration due to gravity.
- We can do even better by applying the same trick to velocity. Namely, we know that v(t)=y'(t), where y(t) is the vertical position of the object at time t. Apply the fundamental theorem of calculus again to show that
- Now write down an equation for the
*horizontal*position x(t) of the bullet in terms of the initial horizontal velocity v_{horizontal}(0). (Hint: remember, we disregard drag and wind so the only acting force is gravity.)

Suppose the bullet is fired at an angle of 30 deg as in the first picture, with a speed of 900m/s from an initial point x(0)=0 and y(0)=0 (i.e. from the origin) at time t=0. Use the equations from activity 1 to answer the following questions.

- What is the maximum height achieved by the bullet? At what time is this height achieved? (Hint: what is the vertical velocity of the bullet when it's at a peak height?)
- What is the horizontal distance of the bullet from the origin at the time of peak height? What is the distance to the point at which the bullet is again at the height from which it was fired, that is, y=0?
- Show that the trajectory of the bullet is a parabola.

Analysing the general situation, in which both wind and drag affect the path of a bullet, is in fact very complicated. You can get a taste of the difficulties involved by reading the wikipedia article on external ballistics. Furthermore, mathematically recreating the path of a bullet after it has hit a target, thus only knowing its angle of entry, is much harder.

Here are a few recent uses of the term exponential growth in the news media:

The company has had a spectacular two years, riding the exponential growth in oil prices that helped to increase profits by a fifth in 2006 to £28.5 million. (

Business Big Shot: Alasdair Locke, The Times, Dec 20, 2007)

After years of exponential growth, there has recently been a slow down in the Northern Ireland property market. (

Well-known property firms merge, BBC News, Dec 7, 2007)

While the above excerpts describe growth in entirely different areas, the one thing they have in common is the use of the termKessler himself came under university scrutiny for alleged financial irregularities. In January 2005, an anonymous source contended he "spent or formally committed all of the reserves of the dean's office and has also incurred substantial long-term debt in the form of lavish salary increases and exponential growth in new, highly compensated faculty and staff directly reporting to him." (

UCSF dean is fired, cites whistle-blowing, Los Angeles Times, Dec 15, 2007)

- Suppose yesterday you heard that annual inflation was 3% in the last year. If x is the price of a representative basket of goods, and t is measured in years, what is the corresponding proportionality constant k in the exponential growth equation that models the price increase? (Hint: note that in this case dt=1 year.)
- What if t is measured in days instead?

The reason why such growth is called exponential is that when the
time variable t is continuous, we can
solve the differential equation . By separating variables we get ,
integrating we arrive at ,
where C is some constant, and exponentiating both sides, we finally get
x=De^{kt}, where D
is a constant. We can solve for D by plugging in t=0, the starting
time, to arrive at the general solution . Exponential growth is much
faster than polynomial, as the example below illustrates in case of
e^{t} versus t^{3}.

- Find a constant r so that 2
^{t}=e^{rt}. - Show that 2
^{t}becomes larger than any polynomial in t, for sufficiently large t. (Hint: suppose p(t)=t^{n}for some positive integer n. For which t is 2^{t}> p(t)?) - Can you think of a function f(t) which grows faster than an exponential function, in the sense of part 2 above?

In practice, when talking about compound interest two quantities are
important. One is the annual interest rate, sometimes called the annual
percentage rate (APR). The other is the number of compounding periods per
year: how many times per year is the interest added to the principal
amount. For instance, say you have $100 credit card debt with an APR of
20%. Usually credit cards compound monthly, so there are 12 compounding
periods per year. Thus if you make no payments (and incur no additional
penalties or expenses) for a whole year, your debt will *not* simply
be 100+100*0.2=120, which it would if the interest was compounded
*only once per year*. Instead, after the first month, you'll owe
100+100*(0.2/12)=101.67 dollars. After the second month, you'll owe
101.67+101.67*(0.2/12)=103.36 dollars, and so on. At the end of the
year, with such monthly compounding, you'll owe $121.94. Might not seem
like a huge difference from the once a year compounding sum of $120, but
over longer periods of time, the difference becomes substantial.

- You open a savings account which earns 2% interest with a deposit of $1000. Would you rather the interest compound daily or monthly? Write down the formula for the amount of money in the account after a year in both cases. (Hint: write down the expression for the amount of money after one period of compounding, now after two periods (don't simplify!), then three... See the pattern?)
- Suppose we decide to compound not once a month or a day, but
once every split second. In fact, we can let the number of
compounding periods go to infinity, thus letting the length of
each period approach zero. Use the fact that , for any real number y, to
show that when the number of compounding intervals goes to
infinity, then after t years,
your account will have 1000*e
^{0.02*t}dollars. This is continuous compounding.

In popular usage, the expression "exponential growth" is often
used as a synonym for "very fast growth". There's no good
reason to describe faculty hiring practices, as the third quote
in the beginning of this section does, in terms of exponential
growth. While an exceptional number of faculty might have been
added during Kessler's tenure as dean, there's no sense in
which "faculty makes more faculty" proportionally to existing
numbers. At other times, "exponential growth" can be more
accurately described as sigmoidal
(remember that strange function used in logistical
regression?). While similar in the low range to the
exponential function, sigmoidal growth reflects the fact that
at some point growth must slow down due to lack of resources.