6.2 Exponential Growth and Decay Greg Kelly, Hanford High School, Richland, Washington

In many applications the rate of change of a variable is proportional to the value of the variable itself. Examples include population of living creatures, radioactive material, and money. 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 .

Exponential Change: Where C can be considered as the initial amount at t=0. If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay.

6.2 Exponential Growth and Decay Day 2 HW Day 1: p. 420 #’s 1, 2, 3-13 odd, 17-25 odd

Radioactive Decay The equation for the amount of a radioactive element left after time t is: If we are dealing with decay, what is the expected sign for k ? The half-life is the time required for half the material to decay.

p 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. (It is assumed that the surrounding 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. Show that this is True. p

Newton’s Law of Cooling Example 𝑇− 𝑇 𝑠 =𝐶 𝑒 𝑘𝑡 A pizza pan is removed at 3:00 PM from an oven whose temperature is fixed at 450 F into a room that is a constant 70 F. After 5 minutes, the pizza pan is at 300 F. How long will it take for the pan to cool to 135 F? Find C Find k t @ 135 F 300−70=380 𝑒 𝑘5 135−70=380 𝑒 −0.100418𝑡 230=380 𝑒 𝑘5 65=380 𝑒 −0.100418𝑡 230 380 = 𝑒 𝑘5 65 380 = 𝑒 −0.100418𝑡 𝑙𝑛 23 38 =5𝑘 𝑙𝑛 13 75 =−0.100418𝑡 𝑘= 𝑙𝑛 23 38 5 =−0.100418 𝑡=17.45 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 © 2010 Pearson Education, Inc. All rights reserved

Calculating Half-life, knowing k value Where k is the decay constant.