Saturday, July 21, 2012

Calculus: Rates of Change in Other Sciences

There are many uses for derivatives in the natural and social sciences, including physics, chemistry, biology, and economics, just to name a few. You will encounter a lot of word problems dealing with these subjects and how they use derivatives. Most people become weary right after reading the problem because the physics/chemistry/biology/economics wording can make the question and information difficult to understand. Don't let that scare you off, though! You don't need to know these other subjects in order to work these word problems, and I'll show you why.

First, let's talk about the typical format. You will (almost always) be given an equation with variables that represent a numerical value in that field of study. Then, you will be asked a specific rate of change question, either to find a formula for the rate of change, or to find the rate of change at a particular variable value. Sometimes you will be asked to explain what the rate of change means in the particular situation.

That's it! Doesn't sound too hard, does it? Let's look at some examples. First, a physics problem.

I know this is a loaded problem, so we'll take it one step at a time! First, notice the format. You don't need to know anything about physics to work this problem, but you do need to understand derivatives. We are given an equation that represents the particle's position to the time. We know this because the variable x is given in meters, which is a position unit.

In the first question, we are asked for the velocity at time t. This means we need to find a formula for the rate of change of position with respect to time (see your textbook for the definition of velocity). We simply need to find the derivative of the position function with respect to time.

I'm not going to show you part (b) because it is simple. To find the velocity at a particular time (any time, but in this case, 3 seconds), simply plug 3 in place of t in the velocity equation you just found. That's simple algebra.

Part (c) might be a challenge if you've never been faced with something like it before. The term "at rest" means just what it sounds like--"not moving." We are asked to find when the particle is not moving. If a particle is at rest, then its position is not changing. Therefore, the change in s is 0, but time is still passing, so the change in t is a value other than 0. So then, it's velocity is 0 divided by some number--which is 0. Thus, when a particle is at rest, its velocity is zero. All we need to do here, then, is find when the velocity of this particle is zero.
(See image above.) Anytime you are solving for a variable that appears in a quadratic equation, you have to use the quadratic formula to find the value. In some cases, you can factor the quadratic equation, but more often than not, it's faster to use the formula than to try and guess the factors. Also remember that when using the formula, you will obtain two answers--make sure both answers make sense! In this case, both times are positive, so we should report both answers, but if one were negative, we could ignore that answer.

Part (d) might also be a challenge for you. Again, you don't need to know physics to solve this problem, you just need to use your brain! Let's think of an example situation. Envision a particle on top of a ruler at the 0 meters mark. That particle moves forward 1 meter in 1 second. The particle's average velocity is the (second position minus the first position) divided by the total time,which comes out to (1-0)/1, or 1 m/s, and the value is positive. Now the particle moves backward 1 meter in 1 second. The average velocity now is (0-1)/1, or -1 m/s. It is negative because the particle moved from the 1 meter mark to the 0 meter mark, instead of the other way.

I hope you can see now that when a particle moves forward, it has a positive velocity, and when it moves backward, it has a negative velocity. In this way, velocity is different than speed.

For this question then, we need to find at what time interval the particle's velocity is positive. To do so, we can look at the graph of the velocity function, or we can find the roots of the velocity function and then determine where the velocity is positive. Notice that we already found the roots of the velocity function in part (c).
(See imagine above). I like to use a chart/table for this. In the first row, divide your time into segments using the roots you found in part (c). Then, in the next row, input the sign that the velocity has in those time intervals. For example, to find the sign for the first interval (t < 2), I input t = 0 into the velocity equation and get v = 36, which is positive. For the second interval, I chose t = 3 and got v = -9. You can choose any number greater than 6 to find the last sign. Then, since we want to know when the particle is moving forward, we answer with the intervals in which the velocity is positive.
In part (e), we are asked when the particle is speeding up and slowing down. This question is asking about the acceleration of the particle, which is the rate of change of the velocity of the particle with respect to time. Similarly to velocity, speeding up occurs when the acceleration is positive, and slowing down when the acceleration is negative. So, these answers can be found in a similar way to part (d), with the only difference being the equation we use. We first have to find the acceleration function (which is the derivative of the velocity function), and then find the roots of the acceleration function before finally using a chart to find the intervals desired.
That wasn't too hard, right? Now let's look at a problem from economics.
First, a few words about the cost function. The cost function describes how much it will cost to manufacture a number of goods. Notice that as x increases, that is, if the company produces a large number of jeans, the cost increases less and less. We know this to be true--we know that if a company produces 250 light bulbs, it will be cheaper per bulb than if they only produced 50. This is the relationship that the cost function describes. In this little scenario (using the function from the problem), it costs \$2200 to produce 50 lightbulbs, which comes out to \$44/bulb (I know, it's not realistic). It costs \$6,500 to produce 650 bulbs, which comes out to \$26/bulb, which is less than the cost per bulb to produce only 50 bulbs.

Now onto this problem. We are first asked to find the marginal cost function, which is simply the derivative of the cost function given to us.
In part (b), we are told to find the marginal cost at x = 100 and to explain its meaning. We get a number of 11 for this number, but what are its units? The derivative dC/dx is said "the rate of change of cost with respect to x". In this case, x is the number of pairs of jeans. So, we have cost (\$) over jeans (or pairs). This number then represents the cost per pair of jeans, but it is an instantaneous rate of change that is occurring at the exact moment the 100th pair of jeans is produced. That is, once the 101st pair of jeans is produced, the cost will change yet again, and before the 100th pair of jeans was produced, the marginal cost was not \$11/pair! Therefore, we can conclude that this number predicts the cost of producing the 101st (next) pair of jeans.
In part (c), we are to compare the marginal cost at the 100th pair of jeans with the cost of producing the 101st pair of jeans. We found the marginal cost at x = 100 to be \$11/pair. What about the cost of producing the 101st pair of jeans? Well, C(100) is the cost to produce 100 pairs of jeans and C(101) is the cost to produce 101 pairs of jeans--thus, the cost of just that last pair of jeans is simply the difference C(101) - C(100)!
That's all we have time for now. Other problems may come from the fields of chemistry or biology and might deal with reaction rates or growth rates. These problems, in all honesty, are similar to the ones presented here, only in a different context. If you understand the concept of a derivative and use formulas given in your textbook, you should be able to answer questions.

If you have a word problem type not covered here that you need help with, leave a comment below. If you have any other questions or comments or suggestions for future topics, let me know!