505 4.6 Applications and Models of Exponential Growth and Decay SOLUTION (a) If y0 is the amount of radiocarbon present in a living thing, then 1 2 y0 is half this initial amount. We substitute and solve the given equation for t. y = y0 e-0.0001216t Given equation 1 2 y0 = y0 e-0.0001216t Let y = 1 2 y0 . 1 2 = e-0.0001216t Divide by y 0 . ln 1 2 = ln e-0.0001216t Take the natural logarithm on each side. ln 1 2 = -0.0001216t ln e x = x, for all x. ln 1 2 -0.0001216 = t Divide by -0.0001216. 5700 ≈t Use a calculator. The half-life is 5700 yr. (b) Solve again for t, this time letting the amount y = 1 4 y0 . y = y0 e-0.0001216t Given equation 1 4 y0 = y0 e-0.0001216t Let y = 1 4 y0 . 1 4 = e-0.0001216t Divide by y 0 . ln 1 4 = ln e-0.0001216t Take the natural logarithm on each side. ln 1 4 -0.0001216 = t ln e x = x; Divide by -0.0001216. t ≈11,400 Use a calculator. The charcoal is 11,400 yr old. S Now Try Exercise 23. EXAMPLE 6 Modeling Newton’s Law of Cooling Newton’s law of cooling says that the rate at which a body cools is proportional to the difference in temperature between the body and the environment around it. The temperature ƒ1t2 of the body at time t in appropriate units after being introduced into an environment having constant temperature T0 is ƒ1t2 = T0 + Ce-kt, where C and k are constants. A pot of coffee with a temperature of 100°C is set down in a room with a temperature of 20°C. The coffee cools to 60°C after 1 hr. (a) Write an equation to model the data. (b) Find the temperature after half an hour. (c) How long will it take for the coffee to cool to 50°C?
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