Chapter 11 – 2b Derivatives of Exponential and Logarithmic ...
Chapter 11 – 2b Derivatives of Exponential and Logarithmic FunctionsReviewddxex=ex, ddxax=axlnaExponential growth and decay: Qt=Q0ert where r > 0 indicates growth and r < 0 indicates decay.ddxlnx=1x, ddxlogbx=1lnb1x3464560-3175Example 11. Find the derivative of fx=lncx.fx=lncx=lnc+lnx ? f'x=1xExample 12. The salvage value S (in dollars) of a company airplane after t years is estimated to be2854960140970St=300,0000.9tDepreciation is the change in the salvage value relative to time. A) What is the rate of depreciation at t = 1 year?S't=300,000ln0.90.9tS'1=300,000ln0.90.91 =-$28,447.34 per yearB) What is the rate of depreciation at t = 5 years?S'5=300,000ln0.90.95=-$18,664.30 per yearC) What is the rate of depreciation at t = 10 years?S'10=300,000ln0.90.910=-$11,021.08 per yearExample 13. Cholera bacteria divide every half hour. If we start with a colony of 10 bacteria, after t hours the number of bacteria will be326771069215Nt=10?4tA) What is the instantaneous rate of change in the number of bacteria after 1 hour?N't=10?ln4?4tN'1=10?ln4?41≈55 bacteria per hourB) What is the instantaneous rate of change in the number of bacteria after 5 hours?N'5=10?ln4?45≈14,196 bacteria per hourC) What is the instantaneous rate of change in the number of bacteria after 10 hours?N'1=10?ln4?410≈14,536,350 bacteria per hourExample 14. A study at a children’s hospital found the following relationship between systolic blood pressure (P, in millimeters of mercury) and weight (x, in pounds):359664093980Px=17.51+lnx 10≤x≤100A) What is the instantaneous rate of change in pressure relative to weight when the weight is 40 pounds?P'x=17.5xP'40=17.540=0.4375 millimeters of mercury per poundB) What is the instantaneous rate of change in pressure relative to weight when the weight is 90 pounds?P'90=17.590≈0.1944 millimeters of mercury per poundExample 15. A mathematical model for world population growth is given by3451225226695P(t)=P0ertwhere P0 is the population at time t = 0, P is the population at time t, r is the continuous compound rate of growth, and t is the time in years. The current world population is approximately 6.9 billion and the current growth rate is about 1.3% compounded continuously.A) Based on this model, how long will it take to reach a world population of 8 billion people?Pt=6.9e0.013t8=6.9e0.013t ? e0.013t=86.9 ? 0.013t=ln86.9 ? t=ln86.90.013=11.38 yearB) What is the current instantaneous rate of change in the population relative to time?P't=6.90.013e0.013t=0.0897e0.013tP'0=0.0897e0.013*0≈0.0897 billion; an increase of 89,700,000 people per year.C) What will be the instantaneous rate of change at t = 5 years?P'5=0.0897e0.013*5≈0.0957 billion; an increase of 95,700,000 people per year. ................
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