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6.4

6.4. Exponential Growth and Decay. Quick Review. Quick Review Solutions. 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

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6.4

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  1. 6.4 Exponential Growth and Decay

  2. Quick Review

  3. Quick Review Solutions

  4. 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.

  5. Separable Differential Equation

  6. Example Solving by Separation of Variables

  7. The Law of Exponential Change

  8. Continuously Compounded Interest

  9. Example Compounding Interest Continuously

  10. Example Finding Half-Life

  11. Half-life

  12. Newton’s Law of Cooling

  13. 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 • the coffee's temperature at the time • the temperature probe was removed. • the time when the temperature • probe reading will be 8 C. o S o Example Using Newton’s Law of Cooling

  14. Example Using Newton’s Law of Cooling

  15. 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= 4.5 + 61.66 0.9277 is a model of the (t,T), data. ( ) t • At time t = 0 the temperature was • T= 4.5 + 61.66(0.9277)» 66.16 C o (b) The figure below shows the graphs of y = 8 and y =T = 4.5 + 61.66(0.9277) t Example Using Newton’s Law of Cooling - ( ) t Use exponential regression to find that T - 4.5 = 61.66 0.9277 is a model for the (t, T – T) = ( t,T - 4.5) data. Thus, S

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