Using logarithms to solve Newton’s Law of Cooling Recall Newton’s Law of Cooling in which the ratio of differences in temperature decays exponentially: Elementary Functions T ( t ) − T a T 0 − T a = e − kt Part 3, Exponential Functions & Logarithms Lecture 3.6a, Applications of Logarithms Here k is a constant that depends on the liquid and the environment. We solve for the temperature T at time t . Dr. Ken W. Smith T ( t ) = T a + ( T 0 − T a ) e − kt Sam Houston State University (Here we use the notation of the textbook by Stitz and Zeager.) In most applications of this law, we need to solve for k and/or t . Since 2013 both k and t are exponents, we must use logarithms. Smith (SHSU) Elementary Functions 2013 1 / 21 Smith (SHSU) Elementary Functions 2013 2 / 21 Applications of logarithms Applications of logarithms A worked example. Suppose the water in a hot tub is heated to 150 ◦ . After the heater is So the equation for T at time t is turned off, the hot tub takes an hour to cool to 120 ◦ . The temperature of the surrounding air is 80 ◦ . Use Newton’s Law of Cooling: T ( t ) = 80 + (150 − 80) e t ln(4 / 7) . T ( t ) = T a + ( T 0 − T a ) e − kt We may simplify this by replacing (150 − 80) by 70 . We can also (if we to find the temperature of the hot tub after 3 hours. wish) replace e t ln(4 / 7) by ( e ln( 4 7 ) ) t = ( 4 7 ) t so that Solution. At time t = 1 we have T ( t ) = 80 + 70 e t ln(4 / 7) = 80 + 70( 4 7 ) t . 120 = 80 + (150 − 80) e − k Subtracting 80 from both sides and then dividing by 70 gives At time t = 3 , 40 = 70 e − k T (3) = 80 + 70(4 / 7) 3 ≈ 80 + 13 . 0612 = 93.0612 4 7 = e − k . Taking the natural log of both sides we have ln( 4 7 ) = − k Smith (SHSU) Elementary Functions 2013 3 / 21 Smith (SHSU) Elementary Functions 2013 4 / 21 k = − ln(4 / 7) = ln(7 / 4) .
Applications of logarithms Radioactive half-life Now let us suppose that we want to know the time t this hot tub will cool A radioactive isotope decays at random, but with a probability computable down to 85 ◦ . over an interval of time. The proportion of radioactive material that decays in a particular interval depends on the isotope. Since the amount Solution. of material that decays is proportional to the total amount of material, Here we need to solve for t in the equation radioactive decay is an example of exponential decay. 85 = 80 + 70( 4 7 ) t . Here is an example. A tree that died in the forest 83 years ago is examined for carbon 14. We assume that the ratio amount of carbon 14 in the tree 70 = 1 5 Subtract 80 from both sides and divide by 70 and simplify 14 so that when it died is compatible with that in the atmosphere at the time but 14 = (4 / 7) t 1 now, 83 years later, we discover that one percent of the carbon 14 has Taking the natural log of both sides we have decayed to nitrogen 14. We can model this by an exponential decay equation: ln( 1 14 ) = ln(( 4 7 ) t ) . P ( t ) = P 0 e − kt Rewriting the lefthand side as ln( 1 14 ) = ln 1 − ln 14 = − ln 14 and t ) = t ln 4 rewriting the righthand side as ln( 4 7 gives us where P ( t ) represents the mass of carbon 14 at time t and P 0 represents 7 the mass of carbon 14 at time t = 0 . (Here we assume that k is positive, − ln 14 = t ln 4 7 . so that the exponential function is decaying, not growing.) So t = − ln 14 7 = ln 14 4 ≈ 4 . 7158 hours. In this case, with t = 83 , we have that P ( t ) = 0 . 99 P 0 . So our equation is ln 4 ln 7 0 . 99 P 0 = P 0 e − 83 k . Smith (SHSU) Elementary Functions 2013 5 / 21 Smith (SHSU) Elementary Functions 2013 6 / 21 Radioactive half-life Applications of logarithms Sample Problem. Suppose the amount of carbon 14 in a specimen can We discovered that be measured by the equation above. From the equation, what is the 0 . 99 P 0 = P 0 e − 83 k . half-life of carbon 14? We may divide both sides of this equation by P 0 to obtain Solution. In this case, we solve for t in the equation 0 . 99 = e − 83 k 2 P 0 = P 0 e − 0 . 000121 t . 1 and then take the natural log of both sides Dividing both sides by P 0 we have 1 2 = e − 0 . 000121 t . ln(0 . 99) = − 83 k. Taking the natural log of both sides we have Therefore k = − ln 0 . 99 ≈ 0 . 000121 83 ln( 1 So the amount of carbon 14 t years after the death of the specimen can be 2 ) = − 0 . 000121 t, approximated by or (using the power rule of logs) P ( t ) = P 0 e − 0 . 000121 t . − ln(2) = − 0 . 000121 t, ln(2) t = 0 . 000121 ≈ 5728 . 5 years . (The correct half-life for carbon 14 (according to Wikipedia is 5730 ± 40 Smith (SHSU) Elementary Functions 2013 7 / 21 Smith (SHSU) Elementary Functions 2013 8 / 21
Applications of logarithms Elementary Functions In the next presentation we will continue with applications of logarithms to Part 3, Exponential Functions & Logarithms doubling times of investments and similar problems where we are, once Lecture 3.6b, Applications of Logarithms: Doubling Time again, solving for an exponent. (END) Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 9 / 21 Smith (SHSU) Elementary Functions 2013 10 / 21 Doubling time for an investment Doubling time for an investment Worked problems on doubling time. Find the doubling time for the following investments. 1 An annual interest rate of 9%, compounded monthly. Just as radioactive isotopes decay exponentially, investment of money Solution. Here the monthly interest rate is r = 0 . 09 / 12 = 0 . 0075 typically grows in proportion to the money invested and so financial and time n is measured in months, the length of the interest period. investments are an example of exponential growth. We solve for the exponent n in the equation 2 P 0 = P 0 (1 . 0075) n , In financial investments, we are often interested in the “doubling time” of an investment. or, after dividing by P 0 , It this case we solve for time where the future value of the investment is 2 = (1 . 0075) n . 2 P 0 where P 0 is the amount of our initial investment. Taking the natural log of both sides we have that Here are some sample problems. ln 2 = ln((1 . 0075) n ) . Using properties of logs, we may rewrite this as ln 2 = n ln(1 . 0075) ln 2 n = ln 1 . 0075 ≈ 92 . 766 . Here time is measured in months so the doubling time is almost 93 months or 7 years and 9 months . Smith (SHSU) Elementary Functions 2013 11 / 21 Smith (SHSU) Elementary Functions 2013 12 / 21
Doubling time for an investment Doubling time for an investment 2 An annual interest rate of 24%, compounded monthly. Solution. We solve for n in the equation 2 P 0 = P 0 (1 . 02) n . (Here r = 0 . 24 / 12 = 0 . 02 and time n is measured in months, the 3 An annual interest rate of 9%, compounded continuously. length of the interest period.) Dividing by P 0 and taking the natural Solution. We solve for t in the equation 2 = e 0 . 09 t where t is time log of both sides we have that measured in years. Taking the natural log of both sides we have that ln 2 = ln((1 . 02) n ) . ln 2 = 0 . 09 t and so t = ln 2 0 . 09 = 100 ln 2 which is about 7 . 702 years . 9 Using properties of logs, we may rewrite this as Notice that this answer is close to the answer in problem 1. ln 2 = n ln(1 . 02) . and so ln 2 n = ln 1 . 02 ≈ 35 . 0028 . Since time here is measured in months then the doubling time is 35 months or 2 years and 11 months . Smith (SHSU) Elementary Functions 2013 13 / 21 Smith (SHSU) Elementary Functions 2013 14 / 21 Doubling time for an investment The Rule of 72 The Rule of 72 In general, the doubling time for an investment solves for t in the exponential equation 2 = (1 + r ) t . 4 An annual interest rate of 24%, compounded continuously. If we apply the natural log to this equation, we find that Solution. We solve for t in the equation 2 = e 0 . 24 t . Taking the ln 2 t = ln(1+ r ) . natural log of both sides we have that ln 2 = 0 . 06 t and so t = ln 2 0 . 24 = 100 ln 2 For interest rates close to 8% , this formula can be accurately which is about 2 . 88 years . Notice that this answer 24 approximated by the formula is close to the answer in problem 2. 72 f ( t ) = 100 r. In other words, the doubling time is about what one would get by taking the interest rate, in percents, an dividing it into 72. Since it is fairly easy to divide numbers into 72, this rough rule of thumbs shows up in business discussions as “the rule of 72.” For example, when does a 4% interest rate double? Since 72 4 = 18 , a 4% Smith (SHSU) Elementary Functions 2013 15 / 21 Smith (SHSU) Elementary Functions 2013 16 / 21 rate should double in about 18 years.
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