Home - Syllabus - Problem Sets - Seminar Section

U4735 Environmental Science for Decision Makers

Basic Mathematical Tools for Environmental Science.

Below are some import analytical tools that we will be using during the semester. Most of the semester will be spent applying these tools to environmental science questions, but for the first several weeks we will spend some time becoming familiar with these tools.

Fractions-A Quick Review

Exponents

a0 = 1

a1 = a

a-1 = 1/a

a1/2 =

a-¥ = 0

ab ´ ac = ab + c

 

Scientific Notation

10-6

0.000001

10-3

0.001

100

1

106

1,000,000

So, 4.7x109 is just 4.7x1,000,000,000 or 4,700,000,000.

    1. It is a compact way of writing large numbers. 4,700,000,000 is cumbersome to write, and 4.7x109 is much shorter. Along the same lines, it is easier to work with numbers in scientific notation. For example, to multiply 4.7x109 by 3.0x103, just multiply 4.7 by 3.0 (14) and add the exponents together (9+3=12). The final answer is 1.4x1013. See below why it isn't 1.41x1013. Also, note that 14 is converted to 1.4x101, and the 1 is added to 12 to get 13.
    2. Scientific notation exactly specifies the number of significant figures (see below).

Significant Figures

Units

 

Exponential Growth

where T is the time in years it takes for the population to double, t is the time in years, N(0) is the initial population (i.e., at t = 0), and N(t) is the population at time t. In the case above, 1000x24/2=4,000. Alternatively, we could define R=1/T, which is the rate of doubles in a year (0.5 in our case). The equation is then:

Again, for our numbers, 1000x20.5x4=4,000.

In this case, r is a percentage growth per year (for example, 5% or 0.05) and t is the time in years. As before, N(0) is the population at t=0.

On your calculator, type in 0.5 and hit the ln key to get 1.65. (Note: significant figures become a bit tricky here: don't worry about it too much.)

Natural logs

Divide each side by N(0) and then take the natural log of both sides to get (remember that ln (ex) = x):

To find ln(1.3), just type 1.3 into your calculator, and then hit the ln key. Finally, r=0.019, so t=13 years. With t=0 in 1983, the answer is the year 1996.

ln e = 1

ln bc = ln b + ln c

ln 1 = 0

e ln a = a

ln e a = a

ln 0 = - ¥

ln a b = b ln a

 

 

Base 10 logs and other bases

Base 10 logs are very similar to natural logs. Just as ln x is the inverse of e x, log10 x (usually the 10 is omitted: log x) is the inverse of 10 x. For example, to solve the problem 10 x = 3.54, take the log of both sides to get x = log 3.54. Using your calculator, you'll find that x = 0.549. In general:

where n is the base of both the log and the exponent.

 

Steady State Systems (Box Models)

Spherical Cow provides a good explanation of box models on page 23. Here is a quick summary of their terminology and assumptions:

If the box is a "first in, first out" situation, the residence time is the amount of time a "particle" spends in the box. For example, if the box is a tube of 10 marbles, and the marbles are being put in and taken out at the rate of one marble per minute, then the residence time is 10 minutes and is equal to the time that a marble spends in the tube. On the other hand, if the marbles are in a bowl and the marbles are removed at random, then the actual time spent in the bowl will vary. But if you can't tell the difference between the marbles, then this doesn't actually matter (e.g., this is the case with pure water).

Residence time and pollution

Residence time is often used to figure out how long a pollutant will stay in the lake after it is no longer being introduced. For example, say that a lake has 10 mg of mercury per cubic meter. The residence time of water in the lake is 10 years. If no more mercury is being introduced into the lake, how long will the mercury take to go away? How much mercury will be left after 10 years?

The mercury will never go away completely. Even though the volume of water in the lake has been replace after 10 years, the mercury has been diluted, but some still remains because the new water that comes into the lake is contaminated with the old mercury. In fact, this is an exponential decay problem (just like population growth, except with a negative sign):

Where C denotes concentration, t is time, and T is the residence time.

Why does this work? Like population growth, there is an analogy of e with 2. You might guess that half the mercury has been removed after 10 years, and this is almost correct. Again, think about marbles. If there are 10 marbles in a bowl, and one is added and one (randomly selected) is removed once per minute, how long does the average marble stay in the bowl? The answer is the residence time, 10 minutes. But what about individual marbles? The chance is one in ten that a marble will be removed after one minute, and there is approximately a 35% percent chance that it will stay in longer than 10 minutes. It is not far from wrong to say that half the marbles will stay in longer than the expected time and that half will be removed quicker. Instead of using 2 as the base, e gives the exact solution. (The reason is the same as with population growth-compound interest, or in this case compound decay.)

Multiple inputs and/or outputs