Question

Q2. The force of interest at any time t, (measured in years) is given by

 0.07—0.005t ,0 ≤ t < 5 0.06—0.003t, 5≤ t < 10

δ(t) =

(a) Calculate the accumulated amount at time t = 15 of $100 invested at time

0.03, t > 10 t = 0.



 0.07−0.005t, 0≤t<5 0.06 − 0.003t, 5 ≤ t < 10

δ(t) =

What is the total accumulated in value at any time t (> 0) of investments of $100 at

0.03, t ≥ 10 times 0, 4 and 6?



Answer 1

$a)100 \times A(0,5) \times A(5,10) \times A(10,15) \\A=100\times e^{\int_0^5(0.07-0.005t)dt} \times e^{\int_5^{10}(0.06-0.003t)dt} \times e^{0.03 \times 5}\\ =100 \times e^{0.07t-\frac{0.005t^2}{2}|_0^5} \times e^{0.06t-\frac{0.003t^2}{2}|_5^{10}} \times e^{0.15}\\ =100 \times e^{0.35-0.0625} \times e^{0.45-0.2925} \times e^{0.15}\\ =100 \times e^{0.2875} \times e^{0.1575} \times e^{0.15}\\ =100 \times 1.3331 \times 1.1706 \times 1.1618\\ =\$ 181.30$

at time t=0

$A=100 \times e^{0.07 \times 5} \times e^{0.06 \times 5} \times e^{0.073\times 5}\\ A=100 \times 1.4191 \times 1.3499 \times 1.1618\\ A=\$ 222.56$

at time t=4

$=100\times e^{\int_4^5(0.07-0.005t)dt} \times e^{\int_5^{10}(0.06-0.003t)dt} \times e^{0.03 \times 5}\\ =100 \times e^{0.07t-\frac{0.005t^2}{2}|_4^5} \times e^{0.06t-\frac{0.003t^2}{2}|_5^{10}} \times e^{0.15}\\ =100 \times e^{0.2875-0.24} \times e^{0.45-0.2925} \times e^{0.15}\\ =100 \times e^{0.0475} \times e^{0.1575} \times e^{0.15}\\ =100 \times 1.0486 \times 1.1706 \times 1.1618\\ =\$ 142.61$

at time 6

$=100 \times e^{\int_6^{10}(0.06-0.003t)dt} \times e^{0.03 \times 5}\\ =100 \times e^{0.06t-\frac{0.003t^2}{2}|_6^{10}} \times e^{0.15}\\ =100 \times e^{0.45-0.27} \times e^{0.15}\\ =100 \times e^{0.18} \times e^{0.15}\\ =100 \times 1.1972 \times 1.1618\\ =\$ 139.09\\$