Question

Show your work. Clearly identify your answer. Justify every step. 1. (5 points) The function A(t) graphed below gives the balance in a savings account after t years with interest compounded continuously. The second graph shows the derivative of A(t). (Pay attention to the units on the graphs.) u y 350 14 250 y = A(t) 10 y = A'(t) 150 6 50 2 t t 10 20. 30 10 20 30 (a) What is the balance after 20 years? (b) How fast is the balance increasing after 20 years? (c) Use the differential equation corresponding to exponential growth and the previous answers to determine the interest rate. (d) When is the balance $300? (e) When in the balance increasing at the rate of $12 per year?

Answer 1

Continuous compounding interest/Formula

P(t) = Poet

where P(t) = value at time t

P0= Original priciple value

r = annual rate of interest

t = time in years

6 350 B (3030) 14 & (30) 2 250 All A(2920] 200 8 P(20,8 150 6 2 st 20 30 10 20 30 az For balance after zoyears. From ast sroph at t=20 (Iroon point e) so balance ablo 20 deads = zuro $. A (20)=zo b) From second sraph ar pointP, Al (20) = 8. so balance chcocasios atarake or 8$ perseas. As eor diblerentiale A' (t)= Do. & (eo) = Ao erexr. Here 208 A£20)=200 to e and Al(20) 208 2 8 Ao e ora 8 dividir 2 = 20 by 0 Ao ezor Aberor 20

ch To luor" = rate ob interest 4% » From 1st graph at coher balance = 30 corros ponding point (at y=300) as B B (30, 3w) t=30 Balence will be 300 abler 30 years, Foor second sragh. As balance us increasiz $12 per year y=12 corresponding point 2 (30, 12) so y=12 at t=30 so Balence will be ancrecesin at a rako ob 12$ per year abler 30 years.