Force of interest

Question

Suppose the force of interest over the time interval $[1,3]$ is given by $\delta (t) =\alpha +\beta t^{-1}$. If $100$ invested at $t=1$ grows to $120.74$ at $t=2$ and $100$ invested at $t=2$accumulates to $114.00$ at $t=3$. Find $\alpha$ and $\beta$.

$A(2)=e^{\int^2_1 (\alpha+\frac{\beta}{t})dt=\frac{120.74}{100}}$

$A(3)=e^{\int^3_2 (\alpha+\frac{\beta}{t})dt=\frac{114}{100}}$

$\rightarrow \alpha+\ln {2^\beta}=\ln{\frac{120.74}{100}}$ and

$\alpha + \ln{\frac{3}{2}^\beta}=\ln{\frac{114}{100}}$

Eliminating $\alpha$

$\beta=\frac{\ln{\frac{114}{100}}}{\ln{\frac{4}{3}}}$

But this does not yield the answer.

Answer 1

The expressions $$A(2)=e^{\int^2_1 (\alpha+\frac{\beta}{t})dt=\frac{120.74}{100}}$$ $$A(3)=e^{\int^3_2 (\alpha+\frac{\beta}{t})dt=\frac{114}{100}}$$ are meaningless. I think you wanted to write $$A(2)=e^{\int^2_1 (\alpha+\frac{\beta}{t})dt}=\frac{120.74}{100}$$ $$A(3)=e^{\int^3_2 (\alpha+\frac{\beta}{t})dt}=\frac{114}{100}$$ From this we get (by taking logarithms)

$$\int^2_1 (\alpha+\frac{\beta}{t})dt =\log{\frac{120.74}{100}}$$ $$\int^3_2 (\alpha+\frac{\beta}{t})dt=\log{\frac{114}{100}}$$ and further $$\alpha+ \beta \ln2 =\log{\frac{120.74}{100}} $$ $$\alpha+ \beta \ln\frac{3}{2} =\log{\frac{114}{100}} $$ This results in $$\beta=\frac{\log 120.74-\log{114}}{\log4-\log3}=\frac{\log\frac{120.74}{114}}{\log\frac{4}{3}}$$

Answer 2

You did the main , stating A(2) and A(3) and computing the integrals ...

$A(2)={e^{\int^2_1 (\alpha+\frac{\beta}{t})dt}= e^{{\alpha} + {\beta} \cdot \ln(2)} = \frac{120.74}{100}}$

$A(3)={e^{\int^3_2 (\alpha+\frac{\beta}{t})dt}= e^{{\alpha} + {\beta} \cdot \ln(3/2)}=\frac{114}{100}}$

Then we have merely a system of 2 linear equations with 2 unknowns ${\alpha}$ and ${\beta}$ :

• ${\alpha}+ {\beta} \cdot \ln(2) = \ln(1.2074)$
• ${\alpha} + {\beta} \cdot \ln(\frac{3}{2}) = \ln(1.14)$

and the solutions are ${\alpha} \approx 0.0500697$ and ${\beta} \approx 0.199668$

I'm looking for the error, not understanding well what you did after Eliminating $\alpha$ , but miracle173 had found them both in your calculus and the mine. Now it is fixed.