Chapter 5 Applications of the Exponential and Natural Logarithm Functions

Chapter Outline Exponential Growth and Decay Compound Interest Applications of the Natural Logarithm Function to Economics Further Exponential Models

§ 5.1 Exponential Growth and Decay

Section Outline Exponential Growth The Exponential Growth and Decay Model Exponential Growth in Application Exponential Decay Exponential Decay in Application

Exponential Growth Definition Example Exponential Growth: A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant

Exponential Growth & Decay Model

Exponential Growth in Application EXAMPLE (World’s Population) The world’s population was 5.51 billion on January 1, 1993 and 5.88 billion on January 1, 1998. Assume that at any time the population grows at a rate proportional to the population at that time. In what year will the world’s population reach 7 billion? SOLUTION Since the “oldest” information we’re given corresponds to 1993, that will serve as our initial time. Therefore the year 1993 will be the year t = 0 and the population at time t = 0 is 5.51 (measured in billions). Therefore, the year 1998 will be year t = 5 and the population at time t = 5 is 5.88 (measured in billions). Since the population grows at a rate proportional to the size of the population, we can use the exponential growth model P(t) = P0ekt to describe the population of the world. Since P0 is the initial quantity, P0 = 5.51. Therefore, our formula becomes

Exponential Growth in Application CONTINUED Now we use the other given information (5.88 billion in 1998) to determine k. This is our function so far. When t = 5, the population is 5.88 billion people. Divide. Rewrite in logarithmic form. Solve for k. Therefore, our formula to model this situation is Now we can determine when the world’s population will be 7 billion.

Exponential Growth in Application CONTINUED This the derived function. Replace P(t) with 7. Divide. Rewrite in logarithmic form. Solve for t. Therefore, the world’s population will be 7 billion people about 18.36 years after our initial year, 1993. This would be the year 1993 + 18.36 = 2011.36. That is, around the year 2011. The graph is given below.

Exponential Growth in Application CONTINUED

Exponential Decay Definition Example Exponential Decay: A quantity, such that, at every instant the rate of decrease of the quantity is proportional to the amount of the quantity at that instant

Exponential Decay in Application EXAMPLE (Radioactive Decay) Radium-226 is used in cancer radiotherapy, as a neutron source for some research purposes, and as a constituent of luminescent paints. Let P(t) be the number of grams of radium-226 in a sample remaining after t years, and suppose that P(t) satisfies the differential equation (a) Find the formula for P(t). (b) What was the initial amount? (c) What is the decay constant? (d) Approximately how much of the radium will remain after 943 years? (e) How fast is the sample disintegrating when just one gram remains? Use the differential equation.

Exponential Decay in Application CONTINUED (f) What is the weight of the sample when it is disintegrating at the rate of 0.004 grams per year? (g) The radioactive material has a half-life of about 1612 years. How much will remain after 1612 years? 3224 years? SOLUTION (a) Since the function y = Cekt satisfies the differential equation y΄ = ky, the function P(t) = Cekt = Ce-0.00043t (where k = -0.00043). Since for the function y = Cekt, C is always the initial quantity (at time t = 0), C = 12 (since P(0) = 12). Therefore, our function is

Exponential Decay in Application CONTINUED (b) We were given P(0) = 12. Therefore the initial amount is 12 grams. (c) Since our exponential decay function is , the decay constant, being the coefficient of t, is -0.00043. (d) To determine approximately how much of the radium will remain after 943 years, we will evaluate the function at t = 943. This is the decay function. Evaluate the function at t = 943. Simplify. Therefore, after 943 years, there will be approximately 8 grams remaining.

Exponential Decay in Application CONTINUED (e) To determine how fast the sample is disintegrating when just one gram remains, we must first recognize that this is a situation concerning the rate of change of a quantity, namely the rate at which the radium is disintegrating. This of course involves the derivative function. This function was given to us and is . Now we will determine the value of the derivative function at P(t) = 1 (when one gram remains). This is the derivative function. Replace P(t) with 1. So, when there is just one gram remaining, the radium is disintegrating at a rate of 0.00043 grams/year. (f) To determine the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, we must determine P(t) when P΄(t) = -0.004.

Exponential Decay in Application CONTINUED This is the derivative function. Replace P΄(t) with -0.004. Solve for P(t). So, the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, is 9.3 grams. (g) To determine how much of the radium will remain after 1612 years, that is one half-life, we will simply recognize that after one half-life, half of the original amount of radium will be disintegrated. That is, 12/2 = 6 grams will be disintegrated and therefore 6 grams will remain. After 3224 years, two half-lives, half of what was remaining at the end of the first 1612 years (6 grams) will remain. That is, 6/2 = 3 grams. These results can be verified using the formula for P(t).

§ 5.2 Compound Interest

Section Outline Compound Interest: Non-Continuous Compound Interest: Continuous Applications of Interest Compounded Continuously

Compound Interest: Non-Continuous P = principal amount invested m = the number of times per year interest is compounded r = the interest rate t = the number of years interest is being compounded A = the compound amount, the balance after t years

Compound Interest Notice that as m increases, so does A. Therefore, the maximum amount of interest can be acquired when m is being compounded all the time - continuously.

Compound Interest: Continuous P = principal amount invested r = the interest rate t = the number of years interest is being compounded A = the compound amount, the balance after t years

Compound Interest: Continuous EXAMPLE (Continuous Compound) Ten thousand dollars is invested at 6.5% interest compounded continuously. When will the investment be worth $41,787? SOLUTION We must first determine the formula for A(t). Since interest is being compounded continuously, the basic formula to be used is Since the interest rate is 6.5%, r = 0.065. Since ten thousand dollars is being invested, P = 10,000. And since the investment is to grow to become $41,787, A = 41,787. We will make the appropriate substitutions and then solve for t. This is the formula to use. P = 10,000, r = 0.065, and A = 41,787.

Compound Interest: Continuous CONTINUED Divide by 10,000. Rewrite the equation in logarithmic form. Divide by 0.065 and solve for t. Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years.

Compound Interest: Present Value Variables are defined the same as in Slide #21.

Compound Interest: Present Value EXAMPLE (Investment Analysis) An investment earns 5.1% interest compounded continuously and is currently growing at the rate of $765 per year. What is the current value of the investment? SOLUTION Since the problem involves a rate of change, we will use the formula for the derivative of That is, A΄ = rA. Since the investment is growing at a rate of $765 per year, A΄ = 765. Since the interest rate is 5.1%, r = 0.051. This is the given function. A΄ = 765 and r = 0.051. Solve for A. Therefore, the value of A for this situation is 15,000. We can now use this, and the present value formula, to determine P.

Compound Interest: Present Value CONTINUED This is the present value formula. A = 15,000, r = 0.051 and t = 1 (since we were given the rate of growth per year). Simplify. Therefore, the current value is $14,254.18.

§ 5.3 Applications of the Natural Logarithm Function to Economics

Section Outline Relative Rates of Change Elasticity of Demand

Relative Rate of Change Definition Example Relative Rate of Change: The quantity on either side of the equation is often called the relative rate of change of f (t) per unit change of t (a way of comparing rates of change for two different situations). An example will be given immediately hereafter.

Relative Rate of Change EXAMPLE (Percentage Rate of Change) Suppose that the price of wheat per bushel at time t (in months) is approximated by What is the percentage rate of change of f (t) at t = 0? t = 1? t = 2? SOLUTION Since we see that

Relative Rate of Change CONTINUED So at t = 0 months, the price of wheat per bushel contracts at a relative rate of 0.22% per month; 1 month later, the price of wheat per bushel is still contracting, but more so, at a relative rate of 0.65%. One month after that (t = 2), the price of wheat per bushel is contracting, but much less so, at a relative rate of 0.0087%.

Elasticity of Demand

Elasticity of Demand EXAMPLE (Elasticity of Demand) A subway charges 65 cents per person and has 10,000 riders each day. The demand function for the subway is (a) Is demand elastic or inelastic at p = 65? (b) Should the price of a ride be raised or lowered in order to increase the amount of money taken in by the subway? SOLUTION (a) We must first determine E(p).

Elasticity of Demand CONTINUED Now we will determine for what value of p E(p) = 1. Set E(p) = 1. Multiply by 180 – 2p. Add 2p to both sides. Divide both sides by 3. So, p = 60 is the point at which E(p) changes from elastic to inelastic, or visa versa.

Elasticity of Demand CONTINUED Through simple inspection, which we could have done in the first place, we can determine whether the value of the function E(p) is greater than 1 (elastic) or less than 1 (inelastic) at p = 65. So, demand is elastic at p = 65. (b) Since demand is elastic when p = 65, this means that for revenue to increase, price should decrease.

§ 5.4 Further Exponential Models

Section Outline Population Growth Equations Exponential Models in Application

Population Growth Equations

Exponential Models in Application EXAMPLE (Spread of News) A news item is spread by word of mouth to a potential audience of 10,000 people. After t days, people will have heard the news. The graph of this function is found below. (a) Approximately how many people will have heard the news after 7 days? (b) At approximately what rate will the news be spreading after 14 days? (c) Approximately when will 7000 people have heard the news? (d) Approximately when will the news be spreading at the rate of 600 people per day? (e) When will the news be spreading at the greatest rate?

Exponential Models in Application CONTINUED (f) Use and to determine the differential equation satisfied by f (t).

Exponential Models in Application CONTINUED SOLUTION (a) To determine approximately how many people will have heard the news after 7 days, we will evaluate f (7). So, after 7 days, we would expect about 2475 people to have heard the news. (b) To determine at approximately what rate the news will be spreading after 14 days, we will evaluate f ΄(14) (we use the derivative of the function since we seek a rate of change). We first use the quotient rule. This is the given function.

Exponential Models in Application CONTINUED Use the quotient rule. Simplify. Evaluate f ΄(14). So, after 14 days, the news will be spreading at approximately 526.76 people per day. (c) To determine approximately when 7000 people will have heard the news, we replace f (t) with 7000 and then solve for t.

Exponential Models in Application CONTINUED This is the given function. Replace f (t) with 7000. Multiply by the denominator. Distribute. Subtract. Divide. Rewrite in logarithm form. Divide.

Exponential Models in Application CONTINUED So, 7000 people will have heard the news after approximately 11.89 days. (d) To determine approximately when the news will be spreading at the rate of 600 people per day, we need to replace f ΄(t) with 600. However, since this will be a long, messy process (a good algebraic exercise for you), we will just look at the given graph where f ΄(t) = 600. As can be seen on the graph, the derivative has a value of 600 when t ≈ 6 or t ≈ 13.5.

Exponential Models in Application CONTINUED (e) The news will be spreading at the greatest rate when the rate (derivative) is greatest. That is, at t = 10 (or when f ΄΄(t) = 0).

Exponential Models in Application CONTINUED (f) Since we know satisfies the differential equation then we can rewrite f (t) in the form . Upon doing this, we will have defined M, k, and B for our function. We can then use to create a differential equation satisfied by f (t). This is the given function. Since the number 10,000 is by itself in the numerator, it must be that M = 10,000. And since the number 50 is the only coefficient of e-0.4t, it must be that B = 50. So we must now rewrite -0.4 in the form –Mk to determine k. That is

Exponential Models in Application CONTINUED Therefore, So, f (t) satisfies the differential equation