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6.4 Exponential Growth and Decay. What you’ll learn about Separable Differential Equations Law of Exponential Change Continuously Compounded Interest.

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Presentation on theme: "6.4 Exponential Growth and Decay. What you’ll learn about Separable Differential Equations Law of Exponential Change Continuously Compounded Interest."— Presentation transcript:

1 6.4 Exponential Growth and Decay

2 What you’ll learn about Separable Differential Equations Law of Exponential Change Continuously Compounded Interest Modeling Growth with Other Bases Newton’s Law of Cooling … and why Understanding the differential equation gives us new insight into exponential growth and decay.

3 Separable Differential Equation

4 Example Solving by Separation of Variables

5 The Law of Exponential Change

6 Continuously Compounded Interest

7 Example Compounding Interest Continuously

8 Example Finding Half-Life

9 Half-life

10 Newton’s Law of Cooling

11 Example Using Newton’s Law of Cooling A temperature probe is removed from a cup of coffee and placed in water that has a temperature of T = 4.5 C. Temperature readings T, as recorded in the table below, are taken after 2 sec, 5 sec, and every 5 sec thereafter. Estimate (a)the coffee's temperature at the time the temperature probe was removed. (b)the time when the temperature probe reading will be 8 C. o S o

12 Example Using Newton’s Law of Cooling

13 is a model for the  t, T – T  = ( t,T  4.5) data. Thus,  4.5  61.66 0.9277 Example Using Newton’s Law of Cooling  Use exponential regression to find that According to Newton's Law of Cooling, T  T =  T – T  e  kt S O S where T = 4.5 and T is the temperature of the coffee at t  0. S O t S T  T  4.5 + 61.66 0.9277 is a model of the  t,T , data.  t (b) The figure below shows the graphs of y  8 and y  T  4.5 + 61.66  0.9277  t (a)At time t  0 the temperature was T  4.5 + 61.66  0.9277   66.16 C 


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