The force of interest δ(t) is a function of time and, at any time t (measured in years),
is given by the formula:
δ(t) =
0.07 0.005t , 0 t < 5
0.06 0.003t, 5 t < 10
0.03, t > 10
(a) Calculate the accumulated amount at time t = 15 of $100 invested at time
t = 0.
A=$100 x A(0,5) x A(5,10) x A(10,15)
=100×e∫05(0.07−0.005t)dt×e∫510(0.06−0.003t)dt×e0.03×5=100×e0.07t−0.005t22∣05×e0.06t−0.003t22∣510×e0.15=100×e0.35−0.0625×e0.45−0.2925×e0.15=100×e0.2875×e0.1575×e0.15=100×1.3331×1.1706×1.1618=$181.30=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=100×e∫05(0.07−0.005t)dt×e∫510(0.06−0.003t)dt×e0.03×5=100×e0.07t−20.005t2∣05×e0.06t−20.003t2∣510×e0.15=100×e0.35−0.0625×e0.45−0.2925×e0.15=100×e0.2875×e0.1575×e0.15=100×1.3331×1.1706×1.1618=$181.30
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