6 Differential Equations Copyright © Cengage Learning. All rights reserved.

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6.2 Differential Equations: Growth and Decay Copyright © Cengage Learning. All rights reserved.

Objectives Use separation of variables to solve a simple differential equation. Use exponential functions to model growth and decay in applied problems.

Warm-Up Rewrite the natural log equation in exponential form: Simplify the exponential expression:

Differential Equations

Differential Equations Analytically, you have learned to solve only two types of differential equations—those of the forms y' = f(x) and y'' = f(x). In this section, you will learn how to solve a more general type of differential equation. The strategy is to rewrite the equation so that each variable occurs on only one side of the equation. This strategy is called separation of variables.

Example 1 – Solving a Differential Equation Apply separation of variables. Integrate each side. So, the general solution is given by: y2 – 2x2 = C.

Slope Fields In the last lesson we created a slope field to help us visualize solutions for the differential equation: Of course, we can also solve this differential equation analytically by separating variables:

Slope Fields For the given slope field, sketch two approximate solutions , one of which passes through the given point: Now, let’s solve the differential equation passing through the point (4, 2) analytically: Solution:

Growth and Decay Models

Proportionality 2 quantities are proportional if one is a multiple of the other. In each example, the rate of change of y is proportional to the value of y.

Growth and Decay Models In many applications, the rate of change of a variable y is proportional to the value of y. If y is a function of time t, the proportion can be written as follows.

So, all solutions of y’ = ky are in the form of y = Cekt Solve the differential equation by separating the variables: So, all solutions of y’ = ky are in the form of y = Cekt

Growth and Decay Models The general solution of this differential equation is given in the following theorem. So, if y’ = ky, then y = Cekt

Growth and Decay Models

Example 2 – Using an Exponential Growth Model The rate of change of y is proportional to y. When t = 0, y = 2, and when t = 2, y = 4. What is the value of y when t = 3? Solution: Because y' = ky, you know that y and t are related by the equation y = Cekt. You can find the values of the constants C and k by applying the initial conditions.

Example 2 – Solution So, the model is y ≈ 2e0.3466t . cont'd So, the model is y ≈ 2e0.3466t . When t = 3, the value of y is 2e0.3466(3) ≈ 5.657. Figure 6.8

Growth and Decay Models Notice that we did not actually have to solve the differential equation y' = ky, we just used its solution The next example demonstrates a problem whose solution involves the separation of variables technique. The example concerns Newton's Law of Cooling, which states that: the rate of change in the temperature of an object is proportional to the difference between the object’s temperature and the temperature of the surrounding medium.

Example 6 – Newton's Law of Cooling The rate of change in the temperature of an object is proportional to the difference between the object’s temperature and the temperature of the surrounding medium. Let y represent the temperature (in ºF) of an object in a room whose temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10 minutes, how much longer will it take for its temperature to decrease to 80º? Solution: From Newton's Law of Cooling, you know that the rate of change in y is proportional to the difference between y and 60. This can be written as y' = k(y – 60), 80 ≤ y ≤ 100.

cont'd Let y represent the temperature (in ºF) of an object in a room whose temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10 minutes, how much longer will it take for its temperature to decrease to 80º? To solve this differential equation, use separation of variables, as follows.

The object cools from 100º to 90º in 10 minutes. cont'd Because y > 60, |y – 60| = y – 60, and you can omit the absolute value signs. Rewriting the solution in exponential form, you have Using y = 100 when t = 0, you obtain: 100 = 60 + Cek(0) so, 100 = 60 + C, which implies that C = 40 and Because y = 90 when t = 10, 90 = 60 + 40ek(10) 30 = 40e10k

So, the model is: y = 60 + 40e–0.02877t Cooling model cont'd Let y represent the temperature (in ºF) of an object in a room whose temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10 minutes, how much longer will it take for its temperature to decrease to 80º? So, the model is: y = 60 + 40e–0.02877t Cooling model and finally, when y = 80, you obtain So, it will require about 14.09 more minutes for the object to cool to a temperature of 80º (see Figure 6.11). Figure 6.11

Extra Example– Newton's Law of Cooling When an object is removed from a furnace and placed in a room with a constant temperature of 80 , its temperature is 1500 One hour after it is removed its temperature is 1120 Find its temperature 5 hours after the object it is removed from the furnace. Solution: From Newton's Law of Cooling, you know that the rate of change in y is proportional to the difference between y and 80. This can be written as y' = k(y – 80),

Solution con’t. y' = k(y – 80)

6.2 Differential Equations: Growth & Decay Uranium (U238) 4,510,000,000 years Plutonium (Pu239) 24,100 years Carbon (C14) 5730 years Radium (Ra226) 1620 years Einsteinium (Es254) 270 days Nobelium (No257) 23 seconds

6.2 Differential Equations: Growth & Decay A sample contains 1 gram of radium. How much radium will remain after 1000 years? (Use a half-life of 1620 years.) We know that y= C = 1 grams at time t = 0. First, find k. Take the ln of both sides.

Suppose that 10 grams of the plutonium isotope Pu-239 was released in the Chernobyl nuclear accident. How long will it take for the 10 grams to decay to 1 gram? Pu-239 has a half life of 24,100 years We know that C = 10 grams at time t = 0. First, find k. y = Cekt 5 = 10ek(24,100) So, the model is: y = 10e-.000028761t To find the time it would take for 10 grams to decay to 1 gram, solve for t in: 1 = 10e-.000028761t t = 80,059 years

6.2 Differential Equations: Growth & Decay No Initial Population??

Extra Exponential Examples: Money is deposited in an account for which interest is compounded continuously. If the balance doubles in 6 years, what is the annual percentage rate? A = Pert 2P = Pert 2 = e6r ln 2 = 6r

Homework 6.2: Pg. 418 1-27 odds, 41, 43, 49, 53, 62, 71.