Questions and Answers #2 Exponential Models

Questions and Answers Sheet 2 Exponential Models Question #1 The exponential growth models describe the population of the indicated country, A, in millions, t years after 2006.Uganda’s growth rate is approximately 3.8 times that of Canada’s.Determine whether the statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. Canada: 𝐴 = 33.1𝑒 0.009𝑡 Uganda: 𝐴 = 28.2𝑒 0.0034𝑡 Answer: Given Data The exponential growth rate of population of Canada is 𝐴 = 33.1𝑒 0.009𝑡 The exponential growth rate of population of Uganda is 𝐴 = 28.2𝑒 0.034𝑡 The exponential constant of the exponential growth rate of population of Canada is 𝑘𝑐 = 0.009 The exponential constant of the exponential growth rate of population of Uganda is 𝑘𝑢 = 0.034 The expression for the ratio of the exponential constant of the exponential growth rate of population of Canada to exponential constant of the exponential growth rate of population of Uganda is, 𝑘𝑢 𝑘𝑐 = 0.034 0.009 𝑘𝑢 = 3.777𝑘𝑐 ≈ 3.80𝑘𝑐 Hence the statement is true that Uganda’s growth rate is approximately 3.8 times that of Canada’s. Question #2 The exponential growth models, 𝐴 = 33.1𝑒 0.009𝑡 (Canada) and 𝐴 = 28.2𝑒 0.034𝑡 (Uganda) describe the population of the indicated country, A, in millions, t years after 2006. Use this information to determine whether the statement is true or false: ''The models indicate that in 2013, Uganda’s population will exceed Canada’s''. If the statement is false, make the necessary change(s) to produce a true statement. Answer: Determine whether the statement ''The models indicate that in 2013, Uganda’s population will exceed Canada’s'' is true or false The exponential growth model for Canada is 𝐴 = 33.1𝑒 0.009𝑡 and exponential growth model for Uganda is 𝐴 = 28.2𝑒 0.034𝑡 . Since t is after 2006 years, so for the year 2013, t=7 𝐴 = 33.1𝑒 0.009(7) = 33.1𝑒 0.063 =35.25 Therefore, the population of Canada in 2013 is 35.25 million. Calculate the population of Uganda. 𝐴 = 28.2𝑒 0.034(7) = 28.2𝑒 0.238 =35.777 Therefore, the population of Uganda in 2013 is 35.777 million. Thus, the population of Uganda is more than Canada's population in 2013. Hence, the given statement is true. Question #3 Find the exponential model 𝑦 = 𝑎𝑒 𝑏𝑥 that fits the points (0, 2) and (4, 3) Answer: Use the first point to find a. 𝑦 = 𝑎𝑒 𝑏𝑥 2 = 𝑎𝑒 𝑏⋅0 2 =\𝑎 ⋅ 1 a=2 Use the second point to find b. 𝑦 = 2𝑒 𝑏𝑥 3 = 2𝑒 4𝑏 3 = 𝑒 4𝑏 2 ln 3 2 = ln 𝑒 4𝑏 4𝑏 = ln 3 1 2 3 4 2 𝑏 = ln Write the equation for the exponential model. 𝑥 3 𝑦 = 2𝑒 4 ln 2 𝑥 3 Result: 𝑦 = 2𝑒 4 ln 2 Question #4 I need help with exponential functions. I know that the derivative of 𝑒 𝑥 is 𝑒 𝑥 , but wolfram alpha shows a different answer to my function below. If you, for example, take the derivative of 𝑒 −2𝑥 do you get −2𝑒 −2𝑥 or 𝑒 −2𝑥 ? Answer: You have to use the chain rule here. Writing 𝑓(𝑥) = 𝑒 𝑥 and g(x)=-2x we have ℎ(𝑥) ≔ 𝑓(𝑔(𝑥)) = 𝑒 −2𝑥 , hence by the chain rule ℎ′ (𝑥) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥) Now 𝑓 ′ (𝑥) = 𝑒 𝑥 , hence 𝑓 ′ (𝑔(𝑥)) = 𝑒 −2𝑥 , and g'(x)=-2 , this gives ℎ′ (𝑥) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥) = 𝑒 −2𝑥 ⋅ (−2) = −2𝑒 −2𝑥 Question #5 Differentiate, please: 𝑦 = 𝑥 𝑥 . Answer: Use logarithmic diferentiation: ln 𝑦 = ln 𝑥 𝑥 Diferentiaate both sides with respect to x: 1 𝑑𝑦 𝑦 𝑑𝑥 1 𝑑𝑦 𝑦 𝑑𝑥 = ln 𝑥 + 𝑥 1 𝑥 = ln 𝑥 + 1 Multiply by y: 𝑑𝑦 𝑑𝑥 = 𝑦(ln 𝑥 + 1) Now 𝑦 = 𝑥 𝑥 , thus, 𝑑𝑦 𝑑𝑥 = 𝑥 𝑥 (ln 𝑥 + 1) Question #6 Solve 𝑒 𝑥 = 0 Answer: That is 0 is the only value that 𝑒 𝑥 cannot take Thus, we can say, that there is no x such that 𝑒 𝑥 = 0 Question #7 How to graph 𝑦 = 3−𝑥 Answer: Firstly, plug in values for x and y. Plot the dots (x,y) . Connect these dots. The resulting graph should look like this (like a reverse 𝑦 = 𝑒 𝑥 ): graph 𝑦 = 3−𝑥 [−10,10, −5,5] Question #8 Find the derivative of 𝑦 = 𝑒 5𝑥 Answer: The derivative of 𝑒 𝑥 is simply 𝑒 𝑥 . However, x has a coefficient, so you have to use the chain rule. If 𝑦 = 𝑒 5𝑥 by the chain rule, the derivative will be equal to the derivative of 𝑒 5𝑥 with respect to 5𝑥 , multiplied by the derivative of 5𝑥 with respect to x. 𝑑𝑦 𝑑 [5𝑥] × 𝑒 5𝑥 = 𝑑𝑥 𝑑𝑥 𝑑𝑦 = 5𝑒 5𝑥 𝑑𝑥 Question #9 Furthermore why is it that 𝑒 𝑥 is used in exponential modelling? Why aren't other exponential functions used, i.e. 2𝑥 , etc? Answer: Apply the usual definition of the derivative: 𝑑 𝑥 𝑒 𝑥+ℎ − 𝑒 𝑥 𝑒 = lim ℎ→0 𝑑𝑥 ℎ Next, apply the usual laws of indices: lim 𝑒 𝑥+ℎ−𝑒 𝑥 ℎ ℎ→0 = lim 𝑒 𝑥 (𝑒 ℎ −1) ℎ ℎ→0 = 𝑒 𝑥 lim ℎ→0 𝑒 ℎ −1 ℎ If we're able to show that (𝑒 ℎ − 1)/ℎ → 1 as ℎ → 0 then we're done. This is easier said than done. We need to employ a little trickery. Let us first start by defining 1 𝑡= ℎ 𝑒 −1 so that lim ℎ→0 𝑒 ℎ −1 ℎ = lim 1 𝑡 = 𝑒 ℎ − 1, and hence 1 + 1/𝑡 = 𝑒 ℎ . Finally, we have: ℎ = ln(1 + 1/𝑡). Hence: = lim 1/𝑡 𝑡→∞ ln(1+1/𝑡) 1 𝑡→∞ 𝑡 ln(1+1/𝑡) 1 lim 𝑡→∞ ln|(1+1/𝑡)𝑡 | 1 = ln(𝑒) =1 This proof uses the definition that 𝑒 = lim (1 + 1/𝑡)𝑡 . To appreciate this definition, consider compound 𝑡→∞ interest added yearly, monthly, weekly, daily, hourly, etc. Adding less and less interest, more and more frequently corresponds to letting t tends towards infinity in (1 + 1/𝑡)2 . Allowing a system to grow and decay on an infinitesimally small time scale gives rise to 𝑒 𝜏 . This definition also benefits from a lack of circular reasoning: to be able to define 𝑒 𝑥 as a power series required us knowing how to differentiate it and get its Taylor series. Either that, or it was an amazingly, infinitely-lucky guess. Question #10 I have a real valued number 𝑦𝑡 . At each time step t, 𝑦𝑡 is multiplied by (1 + 𝜖) with probability p and multiplied by (1 − 𝜖) with probability 1-p . What is the expected value of 𝑦𝑡+𝑛 ? What is the variance? Answer: It is quite simple if you use independence in a more direct way and the method works for any distribution. Assuming that 𝑦𝑛 = ∏1≤𝑘≤𝑛 𝑋𝑘 where the 𝑋𝑘 are i.i.d. factors. By independence: 𝐸𝑦𝑛 = 𝐸 ∏ 𝑋𝑘 = ∏ o𝐸𝑋𝑘 = (𝐸𝑋𝑘 )𝑛 𝑘 𝑘 valid for any distribution. In the specific (Bernoulli) example: 𝐸𝑋𝑘 = (1 + 𝜖)𝑝 + (1 − 𝜖)(1 − 𝑝) . Similarly 𝐸𝑦𝑛2 = 𝐸 ∏𝑘 𝑋𝑘2 = ∏𝑘 o𝐸𝑋𝑘2 = (𝐸𝑋𝑘2 )𝑛 again valid for any distribution. In our case: 𝐸𝑋𝑘2 = (1 + 𝜖)2 𝑝 + (1 − 𝜖)2 (1 − 𝑝) . In particular, 𝑣𝑎𝑟 = (Ⅎo{𝐸}𝑋_𝑘^2)^𝑛 − ((𝐸𝑋𝑘 )2𝑛 and you may carry on from there, in order to calculate limits etc... (e.g. 𝑛 → ∞, 𝜖𝑛 → 𝜆 gives a nice limit). The actual distribution of 𝑦𝑛 is in general quite complicated. Question #11 How to simplify (𝑒 3 )2 𝑒 2 ? Answer: First, condense the term in parenthesis using the following exponentiation rule: (x ) a b ( ) 2 = x ab e3 e = e32e = e6e Now that the simplification is complete, we can use these two exponentiation rules: a = a1 and x a  xb = x a +b e6e = e6e1 = e6+1 = e7