Chapter 4 Additional Derivative Topics

Additional Derivative Topics In this chapter, we will explore these topics: Review exponential and log functions Find derivatives of exponential and log functions Find derivatives of products and quotients Use the Chain Rule to find derivatives Use Implicit Differentiation to find derivatives Solve related rates problems Find Elasticity of Demand Barnett/Ziegler/Byleen Business Calculus 12e

Chapter 4 Additional Derivative Topics Section 1 The Constant e and Continuous Compound Interest

Objectives for Section 4.1 e and Continuous Compound Interest Students will review exponential and log applications involving e and ln simple interest compound interest continuous compound interest Barnett/Ziegler/Byleen Business Calculus 12e

The Constant e Reminder: By definition, e  2.718 281 828 459 … e is defined as either one of the following limits: Let’s verify these limits using our calculators… 𝑒= lim 𝑥→∞ 1+ 1 𝑥 𝑥 𝑒= lim 𝑥→0 1+𝑥 1 𝑥 Barnett/Ziegler/Byleen Business Calculus 12e

e  2.718 281 828 459 … 𝑦= 1+ 1 𝑥 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→∞, 𝑦→𝑒 𝑦= 1+ 1 𝑥 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→∞, 𝑦→𝑒 Barnett/Ziegler/Byleen Business Calculus 12e

e  2.718 281 828 459 … 𝑦= 1+𝑥 1 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→0, 𝑦→𝑒 𝑦= 1+𝑥 1 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→0, 𝑦→𝑒 Barnett/Ziegler/Byleen Business Calculus 12e

Evaluating Expressions Examples showing: e 6  e  e  9 𝑒 .05 𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑒 𝑥 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 Evaluating Expressions

Interest Formulas P = principal (initial deposit) r = annual interest rate t = time in years n = number of time interest is compounded per year A = amount in account at the end of the time period Simple Interest: Compound interest: Continuously compounded interest: 𝐴=𝑃(1+𝑟𝑡) 𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡 𝐴=𝑃 𝑒 𝑟𝑡 Barnett/Ziegler/Byleen Business Calculus 12e

Simple vs Compound Interest 𝑆𝑖𝑚𝑝𝑙𝑒 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 − 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑 Year Account Balance Interest Earned 1 $1000 1000(.05) = $50 2 $1050 3 $1100 4 $1150 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 − 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝒅𝒐𝒆𝒔 𝑖𝑛𝑐𝑙𝑢𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑 Year Account Balance Interest Earned 1 $1000 1000(.05) = $50 2 $1050 1050(.05) = $52.50 3 $1102.50 1102.50(.05) = $55.13 4 $1157.63 1157.63(.05) = $57.88 Barnett/Ziegler/Byleen Business Calculus 12e

Example 1 When you were born, your Grandma put $1,000 in a bank for you, at 5% interest. Calculate the amount in the account after 20 years using: Simple interest Interest compounded quarterly Interest compounded continuously (My Grandma) Yuri Sasaki 1900−1997 Barnett/Ziegler/Byleen Business Calculus 12e

Example 1 (continued) Principal = $1,000 Rate = 5% Time = 20 years Simple interest: Compounded quarterly: Compounded continuously: 𝐴=1000 1+.05∙20 = $2000 𝐴=1000 1+ .05 4 4(20) = $2701.48 𝐴=1000 𝑒 .05(20) = $2718.28 Barnett/Ziegler/Byleen Business Calculus 12e

Example 2 IRA After graduating from USC, Tammy Trojan lands a great job with Google. In her first year, she buys a $3,000 Roth IRA and invests it in a mutual fund that grows at 12% per year, compounded continuously. She plans to become semi-retired in 35 years. What will be its value when she “retires”? In her second year she buys an identical Roth IRA for $3,000. What will be its value in 34 years? What advice can be learned from this, regarding when you should start saving for retirement? Barnett/Ziegler/Byleen Business Calculus 12e

Example 2 (continued) Principal = $3000 Rate = 12% Time = 35 years At the end of 35 years, the first account will have: At the end of 34 years, the second account will have: The lesson learned is once you have a full-time job, you should start saving for retirement as early as possible. Save a little every month and be amazed at how quickly it adds up! 𝐴=3000 𝑒 .12(35) = $200,058.99 𝐴=3000 𝑒 .12(34) = $177,436.41 Barnett/Ziegler/Byleen Business Calculus 12e

Example 3 Computing Growth Time How long will it take an investment of $5,000 to grow to $8,000 if it is invested at 5% compounded continuously? A=$8000 P=$5000 rate = 5% time=? 𝐴=𝑃 𝑒 𝑟𝑡 8000=5000 𝑒 .05𝑡 8 5 = 𝑒 .05𝑡 ln 8 5 .05 =𝑡 ln 8 5 = ln 𝑒 .05𝑡 𝑡≈9.4 ln 8 5 =.05𝑡∙ ln 𝑒 It will take about 9.4 years for the money to grow to $8000. ln 8 5 =.05𝑡 Barnett/Ziegler/Byleen Business Calculus 12e

Example 4 How long will it take an investment to triple if it is invested at 6.5% compounded every 4 months? A=3 P=1 rate = 6.5% n = 3 time=? 𝑙𝑜𝑔 3 = 3𝑡∙𝑙𝑜𝑔 1+ .065 3 𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡 3=1 1+ .065 3 3𝑡 log 3 3 log 1+ .065 3 =𝑡 𝑙𝑜𝑔 3 = 𝑙𝑜𝑔 1+ .065 3 3𝑡 𝑡≈17.1 It will take about 17.1 years for the money to triple. Barnett/Ziegler/Byleen Business Calculus 12e

Example 5 A CD will pay $50,000 at maturity 5 years from now. How much will you have to pay for the CD now if the money is 6.4% compounded continuously? A=50,000 P=? rate = 6.4% time=5 𝐴=𝑃 𝑒 𝑟𝑡 50000=𝑃 𝑒 .064(5) 50000 𝑒 .064(5) =𝑃 You will pay $36,307.45 for the CD in order for it to be worth $50,000 in 5 years. 𝑃≈$36,307.45 Barnett/Ziegler/Byleen Business Calculus 12e

Example 6 A family paid $99,000 for a house. Fifteen years later, the house sold for $195,000. If interest was compounded continuously, what annual rate of interest did the original investment earn? A=195,000 P=99,000 rate = ? time=15 𝐴=𝑃 𝑒 𝑟𝑡 195000=99000 𝑒 𝑟(15) 𝑙𝑛 195 99 15 =𝑟 195000 99000 = 𝑒 15𝑟 𝑟≈.0452 𝑙𝑛 195 99 = ln 𝑒 15𝑟 The interest rate was 4.52%. 𝑙𝑛 195 99 = 15r∙ln 𝑒 Barnett/Ziegler/Byleen Business Calculus 12e

Homework #4-1: Pg 215 (9, 13, 14, 17, 19, 21, 27, 29, 33, 35, 37) Barnett/Ziegler/Byleen Business Calculus 12e