c) The force of interest at time \( t \) is given by: \[ \delta(t)=\left\{\begin{array}{cl}0.01+0.01 t, & 0 \leq t<4 \\ 0.15-0.003 t, & 4 \leq t<6 \\ 0.06 & t \geq 6\end{array}\right. \] 1. Find an expression for the value at time \( t=0 \) of a payment of 100 at time \( t \)

To find the present value of a payment of 100 at time \( t \) given the force of interest \( \delta(t) \), we need to use the relationship between the force of interest and the accumulated value. The present value \( V(0) \) of a payment \( P \) at time \( t \) can be expressed as: \[ V(0) = P e^{-\int_0^t \delta(s) \, ds} \] In this case, \( P = 100 \). We will need to evaluate the integral \( \int_0^t \delta(s) \, ds \) for different intervals of \( t \) based on the piecewise definition of \( \delta(t) \). ### Step 1: Determine the intervals for \( t \) 1. **For \( 0 \leq t < 4 \)**: \[ \delta(t) = 0.01 + 0.01 t \] 2. **For \( 4 \leq t < 6 \)**: \[ \delta(t) = 0.15 - 0.003 t \] 3. **For \( t \geq 6 \)**: \[ \delta(t) = 0.06 \] ### Step 2: Calculate the integral for each interval #### Case 1: \( 0 \leq t < 4 \) We need to calculate: \[ \int_0^t (0.01 + 0.01 s) \, ds \] Calculating the integral: \[ \int_0^t (0.01 + 0.01 s) \, ds = \int_0^t 0.01 \, ds + \int_0^t 0.01 s \, ds \] Calculating each part: 1. \(\int_0^t 0.01 \, ds = 0.01 t\) 2. \(\int_0^t 0.01 s \, ds = 0.01 \cdot \frac{s^2}{2} \bigg|_0^t = 0.01 \cdot \frac{t^2}{2} = 0.005 t^2\) Combining these results: \[ \int_0^t (0.01 + 0.01 s) \, ds = 0.01 t + 0.005 t^2 \] Thus, the present value is: \[ V(0) = 100 e^{-(0.01 t + 0.005 t^2)} \] #### Case 2: \( 4 \leq t < 6 \) We need to calculate: \[ \int_0^4 (0.01 + 0.01 s) \, ds + \int_4^t (0.15 - 0.003 s) \, ds \] The first integral (from 0 to 4) is: \[ \int_0^4 (0.01 + 0.01 s) \, ds = 0.01 \cdot 4 + 0.005 \cdot 4^2 = 0.04 + 0.08 = 0.12 \] Now for the second integral (from 4 to \( t \)): \[ \int_4^t (0.15 - 0.003 s) \, ds = \left(0.15 s - 0.0015 s^2\right) \bigg|_4^t \] Calculating this: \[ = \left(0.15 t - 0.0015 t^2\right) - \left(0.15 \cdot 4 - 0.0015 \cdot 16\right) \] \[ = 0.15 t - 0.0015 t^2 - (0.6 - 0.024) = 0.15 t - 0.0015 t^2 - 0.576 \] Combining both parts: \[ \int_0^t \delta(s) \, ds = 0.12 + 0.15 t - 0.0015 t^2 - 0.576 = 0.15 t - 0.0015 t^2 - 0.456 \] Thus, the present value is: \[ V(0) = 100 e^{-(0.15 t - 0.0015 t^2 - 0.456)} \] #### Case 3: \( t \geq 6 \) We need to calculate: \[ \int_0^4 (0.01 + 0.01 s) \, ds + \int_4^6 (0.15 - 0.003 s) \, ds + \int_6^t 0.06 \, ds \] The first integral (from 0 to 4) is \( 0.12 \) as calculated before. The second integral (from 4 to 6): \[ \int_4^6 (0.15 - 0.003 s) \, ds = \left(0.15 s - 0.0015 s^2\right) \bigg|_4^6 \] Calculating this: \[ = \left(0.15 \cdot 6 - 0.0015 \cdot 36\right) - \left(0.15 \cdot 4 - 0.0015 \cdot 16\right) \] \[ = (0.9 - 0.054) - (0.6 - 0.024) = 0.846 - 0.576 = 0.27 \] Now for the third integral (from 6 to \( t \)): \[ \int_6^t 0.06 \, ds = 0.06(t - 6) \] Combining all parts: \[ \int_0^t \delta(s) \, ds = 0.12 + 0.27 + 0.06(t - 6) = 0.39 + 0.06t - 0.36 = 0.06t + 0.03 \] Thus, the present value is: \[ V(0) = 100 e^{-(0.06t + 0.03)} \] ### Summary of Results 1. For \( 0 \leq t < 4 \): \[ V(0) = 100 e^{-(0.01 t + 0.005 t^2)} \] 2. For \( 4 \leq t < 6 \): \[ V(0) = 100 e^{-(0.15 t - 0.0015 t^2 -