1.4. Applications and Examples http://www.ck12.org
1.4 Applications and Examples
Here are some situations where the ideas covered in the chapter are useful.
Question: The lengths of the sides of a cube are doubling each second. At what rate is the volume increasing?
Solution:The cube side length,x, is doubling every second. Therefore after 1 second it becomes 2x. The volume of
the first cube of sidexisx×x×x=x^3. The volume of the second cube of side 2xis 2x× 2 x× 2 x= 8 x^3. The ratio
of the second volume to the first volume is 8x^3 /x^3 =8. Thus the volume is increasing by a factor of 8 every second.
Fermi Questions
The late great physicist Enrico Fermi used to solve problems by making educated guesses. Say you want to
guesstimate the number of cans of soda drunk in San Francisco in one year. You’ll come pretty close if you guess
that there are about 800,000 people in S.F. and that one person drinks on average about 100 cans per year. So, around
80,000,000 cans are consumed every year. Sure, this answer is not exactly right, but it is likely not off by more than
a factor of 10 (i.e., an “order of magnitude”). That is, even though we guessed, we’re going to be in theballparkof
the right answer. This is often the first step in working out a physics problem.
Dimensional Analysis
Question: find (up to a proportionality constant) the period of a pendulum hanging on a string; that is, find how long
it takes such a pendulum to swing through one cycle, knowing that it depends only the acceleration due to gravity
g= 9 .8m/s^2 and its length,l, which is measured in meters.
Solution: since the period is in units of time, the answer needs to have units of time (seconds). The only way to
obtain seconds from the given quantities is to take the square root of the reciprocal ofg(which will have units of
seconds over square root of meters) and multiply it by the square root ofl, which has units of square root of meters
— this will get rid of the meters altogether. In other words, the period will be proportional to the square root ofl
divided by the square root ofg:
T∝
√
l
g