6.4 Exponential Growth and Decay, p. 350 AP Calculus AB/BC 6.4 Exponential Growth and Decay, p. 350

Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. Example 1: Multiply both sides by dx and divide both sides by y2 to separate the variables. (Assume y2 is never zero.)

Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. Example 1: Combined constants of integration

Now, use the initial condition y = 2 when x = 1.

Example 2: Separable differential equation Combined constants of integration

Example 2 (continued): We now have y as an implicit function of x. We can find y as an explicit function of x by taking the tangent of both sides. Notice that we can not factor out the constant C, because the distributive property does not work with tangent.

The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by:

Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides.

Integrate both sides. Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication.

Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication. Since is a constant, let .

Since is a constant, let . At , . This is the solution to our original initial value problem.

Exponential Change: If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay. Note: This lesson will talk about exponential change formulas and where they come from. The problems in this section of the book mostly involve using those formulas. There are good examples in the book, which will not be repeated here.

Continuously Compounded Interest If money is invested in a fixed-interest account where the interest is added to the account k times per year, the amount present after t years is: If the money is added back more frequently, you will make a little more money. The best you can do is if the interest is added continuously.

Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine. We could calculate: but we won’t learn how to find this limit until chapter 8. (The TI-89 can do it now if you would like to try it.) Since the interest is proportional to the amount present, the equation becomes: You may also use: which is the same thing. Continuously Compounded Interest: p

Radioactive Decay The equation for the amount of a radioactive element left after time t is: This allows the decay constant, k, to be positive. The half-life is the time required for half the material to decay.

Half-life Half-life:

Newton’s Law of Cooling Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air. (It is assumed that the air temperature is constant.) If we solve the differential equation: we get: Newton’s Law of Cooling where is the temperature of the surrounding medium, which is a constant.

Example 3: Suppose that a cup of soup cooled from 90°C to 60°C in 10 minutes in a room whose temperature was 20°C. Use Newton’s Law of Cooling to find how much longer would it take the soup to cool to 35°C. T0 = 90°C, TS = 20°C, T = 60°C, when t = 10 min; find t, T = 35°C

T0 = 90°C, TS = 20°C, T = 60°C, when t = 10 min; find t, T = 35°C p