Inactive Tutor answered 07/19/13
A) F(5) = 21,000 e^(0.088*5) = $32,606.85
B) F'(t) = 0.088*21,000 e^(0.088*t) => F'(13) = $5801.43/year
C) F'(t) = 0.088*21,000 e^(0.088*t) = 4,358.40 => t = 9.75 years
D) [F(8) - F(5)]/(8-5) = $3283.82/year
Mary B.
asked 07/19/13When an initial amount of money, in dollars, is invested into an account that earns interest continuously, the Future Value of the account after years is given by the formula: F(t)=Ae^rt where is the annual interest rate earned by the account. Let A=21,000 and r=8.8%.
A) What is the value of the account, in dollars, after 5 years? Give your answer rounded to two decimal places.
B) What is the exact instantaneous rate of change of the value of the account at exactly 13 years? Give your answer rounded to two decimal places.
C) At what time, in years, is the instantaneous rate of change of the value of the account increasing by $4,358.40 per year? If necessary, round your answer to two decimal places.
D) What is the average rate of change of the future value of the account between year 5 and year 8? (Round to the nearest penny/cent.)
Inactive Tutor answered 07/19/13
A) F(5) = 21,000 e^(0.088*5) = $32,606.85
B) F'(t) = 0.088*21,000 e^(0.088*t) => F'(13) = $5801.43/year
C) F'(t) = 0.088*21,000 e^(0.088*t) = 4,358.40 => t = 9.75 years
D) [F(8) - F(5)]/(8-5) = $3283.82/year
Inactive Tutor answered 07/19/13
Given: F(t) = Arrt ; A = 21,000 ; r = 8.8% = 8.8%/100% = 0.088
F(t) = 21000e0.088t
(A) value of account after 5 years ==> find F(t) when t=5
F(5) = 21000e0.088·5 = 32606.85 ==> $32606.85
(B) instantaneous rate of change of value of account at exactly 13 years
==> find F'(t) when t=13
F(t) = 21000e0.088t
F'(t) = (21000)e0.088t(0.088)
F'(t) = 1848e0.088t
F'(13) = 1848e0.088·13 = 5801.43
instantaneous rate of change at exactly 13 years is $5801.43
(C) time at which instantaneous rate of change of value of account is increasing by $4,358.40/yr
==> find t when F'(t)=4358.4
F'(t) = 1848e0.088t
1848e0.088t = 4358.4
e0.088t = 2.3584
ln(e0.088t) = ln(2.3584)
0.088t = ln(2.3584)
t = (ln(2.3584))/0.088
t = 9.75
instantaneous rate of change of value of account increasing by $4358.40/yr at 9.75 years
(D) average rate of change of future value of account between year 5 and year 8
==> find Δy/Δt when t1=5 and t2=8
Δy/Δt = (F(t2) - F(t1))/(t2 - t1)
= (F(8) - F(5))/(8 - 5)
= (F(8) - F(5))/3
==> F(8) = 21000e0.088·8 = 42458.30
F(5) = 21000e0.088·5 = 3283.85
Δy/Δt = (42458.30 - 32606.85)/3
= 9851.45/3
= 3283.82
Avg rate of change of future value of account between year 5 and year 8 is $3283.82
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