Pg. 130: 1,2,3,7,10,14,15,17,19,26

Lesson 10 - 3.4#1 Continuous Growth (e) - Match graphs of exponential with different values for b: Exponential growth: y = a(b)t , where b > 1 Note: the larger the value of b the faster the function grows: So 3 > e > 2

Lesson 10 - 3.4#2 Exponential decreasing function (e) - Match graphs of exponential with different values for k: For exponential with continuous percent rate (e) for the form: y = a(e)t , where e = 2.7142.., and k any real number except 0, larger values of k makes the function grows faster. So fastest percent rate for k is (a) Graph III (b) Graph II (c) Graph IV (d) Graph I

Lesson 10 - 3.4#3 For population P = 25,000 when t = 0 and continuous rate, k of 7.5% per year: Exponential increasing function: Pt = P0(e)kt (a) Formula is Pt = 25000(e)0.075 t (b) Yearly rate is r, so find r when Or rate is 7.79%, Yearly rate must be more to catch up to a continuos rate that grows faster each year.

Lesson 10 - 3.4#7 Given nominal rate (yearly interest rate) find balance after t = 1. Rate = 1%, Compounded weekly (n = 52) yields $505.03 Compounded every minutes (n = 52600) yields $ 505.03 Compounded continuously (e) yields $ 505.03 Lesson 10 - 3.4#10 Given nominal rate (yearly interest rate) find balance after t = 1. Rate = 8%, effective annual rate => (1 + (0.01/52)^52 = 1.0100491 or r = 0.0100491 So effective rate is 1.0049%

Lesson 10 - 3.4#14 Match function without a calculator for y = ex , y = e-x , and y = -ex . Note: for y = e kx , when k is positive it is an increasing function and when k is negative it is a decreasing function. When the coefficient of e is negative is suggest that the graph has been flipped across the x-axis.

Lesson 10 - 3.4#15 Find t for exponential growth problem, given . When will V = 3000? or find t when Lesson 10 - 3.4#17 Find t for exponential doubling problem, given . When will V = 2(537) = 1074? or find t when Lesson 10 - 3.4#19 Comparing two exponential growth investments (assume V0 = $1): Investment 1: pays 5.3% interest and compounds continuously yields after 1 year: Investment 2: pays 5.5% and compounds annually yields after 1 year: (so Investment 2 is better - greatest effective annual rate)

Lesson 10 - 3.4#26 Effective annual rate , r' for growth rate of (1 + r') of different compounding periods: (b) corrected Given: nominal rate = 4.2%. (a) Effective annual rate is 4.2% when compounding is annual since, (1 + 0.042)1 = 1.042 (b) Effective annual rate is 4.28% when compounding is monthly since, (c) Effective annual rate is 4.29% when compounding is continuously since,