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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.

Tosh
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2 Answers2

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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 $$\beta=\frac{\log 120.74-\log{114}}{\log4-\log3}=\frac{\log\frac{120.74}{114}}{\log\frac{4}{3}}$$

miracle173
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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.

miracle173
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  • so $$\beta = \frac{1.2074-1.14}{\ln(2)-\ln(\frac{3}{2})}$$ – miracle173 Aug 21 '16 at 14:17
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    @miracle173 : I don't understand what you mean. The purpose is not to detail how to solve the linear system, the OP knows the integrals ... My last sentence is just friendly politeness :) –  Aug 21 '16 at 14:19
  • I wanted to point at the fact that your $\beta$ and th OP's $\beta$ have a different numerator. But as far as I can see your $\beta$ is wrong, too. I posted the correct $\beta$ in an answer. – miracle173 Aug 21 '16 at 14:50
  • @miracle173 : I computed something similar to your expression using a determinant resolver and the decimal values are the same. What did you find different ? wolfram to compute the expression while the mine with a decimal less is 0.23429 –  Aug 21 '16 at 15:08
  • check the numerators of $\beta$. yours is $1.2074−1.14$ mine is $\log 1.2074− \log 1.14$ – miracle173 Aug 21 '16 at 15:13
  • @miracle173 yes ! you are true ! it is fixed , TY ! –  Aug 21 '16 at 15:22
  • I think you should also fix the wrong format of the equations. I assume this format led to your wrong solution. – miracle173 Aug 21 '16 at 15:25