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AP CALCULUS AB CHAPTER 6: DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING SECTION 6.4: EXPONENTIAL GROWTH AND DECAY 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 dy ky Understanding the differential equation dx gives us new insight into exponential growth and decay. Separable Differential Equation dy f y g x is dx called separable. We separate the variables by writing it in the form 1 dy g x dx. f y A differential equation of the form The solution is found by antidifferentiating each side with respect to its thusly isolated variable. Example Solving by Separation of Variables Solve for y if dy x y and y 3 when x 0. dx 2 2 dy x y dx y dy x dx y dy x dx x y C 3 Apply the initial conditions to find C. 1 x 1 3 C So, y and y 3 3 3 1 x This solution is valid for the continuous section of the function that goes through the point (0, 3), 2 2 2 2 2 2 3 1 3 1 3 that is, on the domain ,1 . Section 6.4 – Exponential Growth and Decay Law of Exponential Change If y changes at a rate proportional to the amount present dy and y = y0 when t = 0, ky dt then y y e kt , 0 where k>0 represents growth and k<0 represents decay. The number k is the rate constant of the equation. Section 6.4 – Exponential Growth and Decay From Larson: Exponential Growth and Decay Model If y is a differentiable function of t such that y>0 and y’=kt, for some constant k, then where C = initial value of y, and k = constant of proportionality y Ce (see proof next slide) kt Section 6.4 – Exponential Growth and Decay Derivation of this formula: dy ky dt dy kdt y dy y kdt ln y kt C e kt C y e kt eC y C1e kt y Section 6.4 – Exponential Growth and Decay This corresponds with the formula for Continuously Compounded Interest A t A0e rt This also corresponds to the formula for radioactive decay kt y y0e , k 0 Continuously Compounded Interest If the interest is added continuously at a rate proportional to the amount in the account, you can model the growth of the account with the initial value problem: dA Differential equation: rA dt Initial condition: A(0) A O The amount of money in the account after t years at an annual interest rate r: A(t ) A e . rt O Example Compounding Interest Continuously Suppose you deposit $500 in an account that pays 5.3% annual interest. How much will you have 4 years later if the interest is (a) compounded continuously? (b) compounded monthly? Let A 500 and r 0.053. O a. A(4) 500e 0.053 4 618.07 0.053 b. A(4) 500 1 12 12 4 617.79 Example Finding Half-Life Find the half-life of a radioactive substance with decay equation y y e . Hint: When will the quantity be half as much? 1 The half-life is the solution to the equation y e y . 2 1 Solve algebraically e 2 1 - kt ln 2 1 1 ln 2 t - ln k 2 k Note: The value t is the half-life of the element. It depends - kt O kt O - kt only on the value of k . O Section 6.4 – Exponential Growth and Decay The formula for Derivation: half-life of a radioactive substance is 1 y0 e y 0 2 y0 e kt 1 y0 y0 2 y0 kt e ln 2 half-life k kt 1 2 1 kt ln 2 kt ln 2 ln 2 t k Newton’s Law of Cooling The rate at which an object's temerature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium. If T is the temperature of the object at time t , and T is the S surrounding temperature, then dT k T T . dt Since dT d (T - T ), rewrite (1) S S d T T k (T T ) dt Its solution, by the law of exponential change, is S S T - T T T e , kt S O S Where T is the temperature at time t 0. O (1) Section 6.4 – Exponential Growth and Decay Another version of Newton’s Law of Cooling (where H=temp of object & T=temp of outside medium) dH k H T dt to find H as a function of time k H T dt dH H T H T dH H T kdt ln H T kt C1 e kt C1 H T e kt eC1 H T Ce kt H T H Ce kt T 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. o S 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 Example Using Newton’s Law of Cooling Use time for L1 and T-Ts for L2 to fit an exponential regression equation to the data. This formula is T-Ts. 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 Use exponential regression to find that T - 4.5 61.66 0.9277 t is a model for the t , T - T t , T 4.5 data. Thus, S T 4.5 61.66 0.9277 is a model of the t , T data. t (a) At time t 0 the temperature was T 4.5 61.66 0.9277 66.16 C. 0 (b) The figure below shows the graphs of y 8 and y T 4.5 61.66 0.9277 t Section 6.4 – Exponential Growth and Decay Resistance Proportional to Velocity It is reasonable to assume that, other forces being absent, the resistance encountered by a moving object, such as a car coasting to a stop, is proportional to the object’s velocity. The resisting force opposing the motion is Force = mass acceleration = m dv . dt We can express that the resisting force is proportional to velocity by writing m dv dv k kv or v dt dt m k 0. This is a differential equation of exponential change, t m v v0e . k