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Problem Solving
As often as possible, we will attempt to use variable names consisting of single letters which abbreviate the name of the quantity involved. In fact, we have already run into trouble this way, when we looked for an equation for power in terms of pressure. We will address this logistical problem in one of several ways. We will capitalize some variables while leaving others lower case. We will use Greek letters on occasion. Sometimes we will use an apostrophe (read "prime") to distinguish between similar quantities which are associated with different objects, or with different parts of a process. And sometimes we will use subscripts. When all else fails, we will simply use more than one letter (usually the first syllable) of the quantity as a variable name. In all cases, we will attempt to be consistent with general usage in the physics community. This will cause some problems when comparing notes with biologists and chemists, because common usage in one discipline is not always the same as common usage in another. We will attempt to clarify as many of these contradictions as possible, but when in doubt, check the units.
The single most confusing aspect of physics for the beginning student is making use of equations that are filled with variables. Most students have rarely treated equations with more than two variables. It is not unusual in physics to begin with the "general form" of an equation which has no numbers at all! Consider the equations for the propagated error above. Excepting the exponents, there were no numbers in those formulas. In order to use them, we had to match the general formula with a specific example in order to compute the propagated error for the example (for u v 2, A = 1, B = 1 and C = 2). The key to working with such equations is to realize that every variable in an equation represents a number. Some of them will be given in the statement of the problem, and some will be the unknowns for which we are to solve. General forms of equations are good when asking yourself the question, "how does this variable change if I change this other quantity?". When solving problems for numerical answers, the general form of the equation simplifies greatly as soon as all of the known quantities are "plugged in". All that you need is practice!
In a very real sense, a college physics course is a year-long exercise in solving word problems. Since we are examining the universe around us, we must describe things as we perceive them. For this we require words. But to make numerical predictions about what will happen under different circumstances, we must use mathematics. It therefore behooves us to have a model for solving word problems which we can apply in any case. In the model presented here, there are nine basic steps to solving a word problem:
- Read the problem carefully.
- Ignore all unnecessary verbiage (pronouns, prepositions and the like, which do not alter the description of the problem). Think in terms of the model, not in terms of the realistic details which are in the problem to make it relevent but are not part of the simplified model. Also, don't bring outside knowledge that you have to the problem: the details will almost certainly confuse the simplifications which we need to make in order to treat the problem mathematically.
- Identify the information that is given.
- Identify what the problem asks you to compute.
- If appropriate, draw a picture, graph or movie (series of pictures) to clarify coordinate systems and steps of the problem.
- Select the equations which relate the unknown(s) to the information given. Here is your chance to use your knowledge of units!
Remember when choosing the equations you will use to solve a problem that the number of independent equations must equal the number of unknowns for a unique solution to exist.
Also remember that it is possible to pose a problem which does not contain enough information to find a solution, or which contains too much or contradictory information.
Always try to use the minimal set of equations which contain the givens and the unknowns as variables: keep it simple!
- Solve the equations for the unknown(s). This is the easy part: simple algebraic manipulation. Some hints may be helpful:
Do not substitute numbers into the equation until all of the algebra is done. This will help you avoid mistakes, due both to transcription or keypad error, and to intermediate rounding (rounding of numbers much less than one in the denominator can have disasterous results for the final answer!). It also will indicate situations where the answer to the problem is independent of one or more of the givens: sometimes variables will subtract or divide out of the equation.
Be sure to use units when you do the numerical computation, and make sure the units are correct in the result.
When converting squared or cubed units, remember to square or cube the coefficient as well. For example, to convert square centimeters to square meters, use:
(1 m / 100 cm) 2 = 1 m 2 / 10,000 cm 2 . - To check your work, substitute your answer(s) into the original, or an alternate, statement of the problem (because you need to check your selection of equations as well as your computation).
- Ask yourself if the results are physically sensible for the problem (this often allows you to catch mistakes in scale).
Let's apply this model to a sample problem:
An uncoiled molecule of DNA can be modeled as a cylinder with a radius of 7.9 x 10 - 4 microns (a micron is 10 - 6 meters). A typical cell nucleus is approximately spherical with a radius of 1.7 microns. If a molecule of my DNA is 1.5 m long, what fraction of the volume of a cell nucleus is DNA?
After steps 1 through 4, we have:
- DNA is cylindrical
- DNA radius is 7.9 x 10 - 4 microns
- DNA length is 1.5 m
- nucleus is spherical
- nucleus radius is 1.7 microns
- What is DNA volume / nucleus volume ?
We will skip step 5 since we have already drawn pictures of cylinders and spheres before, and we do not need to fix a coordinate system since we are not computing distances. To do step 6, we need to remember the formulas for the volume of a cylinder and the volume of a sphere:
Note the use of "different" variables ("r" and "R") for the radii of the cylinder and the sphere. We will often need to do this sort of thing when we have two quantities which have similar meanings.
In order to use the information in the problem, we need to think of the length of the DNA molecule as the height of a cylinder, and we need to put the length in the same units as the rest of the data:
Step 7 is then a simple series of simplifications and substitutions:
Note that we simplified the algebra before substituting numbers for the variables, and that we checked the units before using our calculators: the units of both the numerator and denominator are cubic microns, and so the answer is unitless. Also, the value of p was irrelevent to the problem; this was a result of the volume formulas, and not much an indication that the answer was independent of some physical quantity.
Note also that we rounded the result to a convenient number of significant figures. We will often do this when we do not have any indication of the precision of our data. For example, the value of p is known to (for us) arbitrary precision, and the 4/3 is exact until we enter it into our calculator. The other numbers are averages or estimates, but we have no idea of their absolute or relative error. In those cases, it is not possible to compute the error propagation and we arbitrarily retain three significant digits.
You really don't need a calculator to get a fairly accurate idea of the answer to most problems; all you need is some familiarity with fractions, algebra and the willingness to approximate. For example, in the computation above, we need to square 7.9. Using the binomial expansion (a+b) 2 = a 2 + 2ab + b 2, we can express 7.9 as 8-.1, and so 7.9 2 = 8 2 + 2 x 8 x (-.1) + (-.1) 2 which is 62.41. A similar expansion is available for cubes, but it is a little more tedious to use, so we want to approximate 1.7 as 1 2/3, or 5/3. Then 1.7 3 becomes (approximately) 125/27, which is very close to 5! We now have (ignoring the powers of ten for a moment) 3 x 66 x 1.5 / 4 x 5; since 66 x 1.5 is 99, we can approximate this as (3/4) x (100/5), which is 15. Including our powers of ten: (10 -4) 2 is 10 -8, times 10 6 gives us a net 10 -2; so that our final answer is 15 x 10 -2 or .15. Not bad! In fact, the numbers we will use in our computations are so often approximations themselves, that usually this type of approximate "back-of-the-envelope" calculation will be perfectly adequate for our needs.
We will rarely do a problem in so many steps. We will assume whenever it is reasonable that the algebraic manipulations are not a problem for you. But we will always explicitly state the givens, the unknowns and the equations, as well as the numerical answer, to the examples in the text.
Step 8 may seem somewhat problematic in this example: the problem seems so easy that it is only possible to check our answer by repeating the calculation. This has merit, since most of us use calculators, which are notoriously subject to pilot error! But we can also approach the problem in a different way, and thereby check our work. By computing each volume separately, and then computing the ratio, we avoid some of the algebra which produced our first answer. By doing the problem in two ways, we have a verification of both our technique and our computation.
For step 9, we must ask ourselves if it is reasonable to expect that 14 % of the cell nucleus is DNA. Standard texts on cell biology indicate that the DNA is a major component of the nucleus, and our answer is verified qualitatively by examining the chromatin (which contains the DNA) content of the nucleus in electron micrographs. Ultimately, the physicist must consult the biologist for confirmation of such a calculation.
Problems
The next chapter is about mechanics.
If you have stumbled on this page, and the equations look funny (or you just want to know where you are!), see the College Physics for Students of Biology and Chemistry .
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