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Differential equations, believe it or not, show up everywhere.
There are math departments studying how differential equations relate to biology using differential equations for the study of diseases, books relating differential equations and economics and whole journals devoted to differential equations and their applications. Calculus classes usually introduce differential equations, but don't do a whole lot with them. To get more, there are usually differential-equations-only classes offered after calculus.
We're used to functions that eat numbers and spit out other numbers. The function
f (x) = x2
eats a number and spits out the square of that number. For example,
f (4) = 16.
There are also vector functions that may eat and/or spit out multiple numbers. The function
f (u, v) = + v, u – v>
eats two numbers and spits out a vector. For example,
f (2, 3) = <2 + 3, 2 – 3) = <5, -1>.
Now, for extra fun, there are functions that eat functions and spit out other functions! The act of taking a derivative is one example.
We can think of as a function that eats a function and spits out the derivative of that function (which is itself a function):
For example, if we feed the function f (x) = x2 to the function we get
The function ate the function f (x) = x2 and spit out its derivative function, f '(x) = 2x.
There are three steps to solving a math problem.
Model the following situation with a differential equation: A bucket starts out with 5 gallons of water. A diluted mixture of red dye (40 percent red dye; 60 percent water) is dripping into the bucket at a rate of 1.5 gallons per minute. The dye mixture swirls around and mixes evenly with the liquid already in the bucket, then drips out a hole in the bottom of the bucket at a rate of 1.5 gallons per minute. How can we describe how the amount of red dye in the bucket is changing over time?
• Figure out what the problem is asking.
Let's figure out the general situation.
We have a bucket. Red dye is dripping into the bucket, mixing, then dripping out again.
We're interested in the quantity of red dye in the water. We're given numbers for how fast the liquid is flowing and how much red dye is in the incoming liquid, and we need to translate this into math.
• Solve the problem.
Let R stand for the amount of red dye in the water, measured in gallons, and let t stand for time (measured in minutes). We know that
Liquid is dripping into the bucket at a rate of 1.5 gallons per minute, but only 40 percent of this liquid is red dye. Every minute, the amount of red dye that enters the bucket is
0.4(1.5) = 0.6 gallons.
Liquid is dripping out of the bucket at a rate of 1.5 gallons per minute. We need to know what percent of this liquid is red dye. The amount of red dye in the bucket is given by R. Since there are 5 gallons of liquid in the bucket, the percentage of the liquid in the bucket that is red dye is
If 1.5 gallons of liquid flow out of the bucket every minute, and of those gallons are red dye, the rate at which red dye flows out of the bucket is
Putting this information together, we have
• Check the answer.
This particular question doesn't really have an answer to check. After you learn how to solve differential equations, you'll get asked questions like "how much red dye is in the bucket after five minutes" or something like that. We'll give you ways to check answers to those sorts of problems when we get there.