Independent solution

How to solve this Time-Varying Force of Interest question

Setup

Setup

The fifth year after time zero is the interval from t = 4 through t = 5.

1+i5=exp(45dtt+8)1+i_5=\exp\left(\int_4^5\frac{dt}{t+8}\right)

Model

Model

The antiderivative is the logarithm of t plus 8.

45dtt+8=ln13ln12\int_4^5\frac{dt}{t+8}=\ln13-\ln12

Compute

Compute

Exponentiating gives the one-year accumulation factor.

1+i5=13121+i_5=\frac{13}{12}
i5=112=0.083333i_5=\frac1{12}=0.083333

Answer

Answer

The effective rate in year 5 is 8.3%, choice D.

i58.3%(D)\boxed{i_5\approx8.3\%\quad\text{(D)}}