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theEric

  • 2 years ago

Hi! This is a finance-based problem, which might be my difficulty. I have the answers from the book. Here is the prompt: "Suppose that a sum \(S_0\) is invested at an annual rate of return \(r\) and compounded continuously. (a) Find the time \(T\) required for the original sum to double in value as a function of \(r\). (b) Determine \(T\) if \(r=7\%\). (c) Find the return rate that must be achieved if the initial investment is to double in \(8\) years.

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  1. theEric
    • 2 years ago
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    I think I should start out by finding \(S'(t)\), but that's where I think I'm wrong. It just comes from understanding the situation, I know.. I was thinking \(S'=S\ r\). Is that correct?

  2. theEric
    • 2 years ago
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    Oh, answers: (a) \(\ln(2)\div r\) year (b) \(9.90\) years (c) \(8.66\%\)

  3. wio
    • 2 years ago
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    Is it something like this?\[ 2S_0=S_0(1+r)^T \]

  4. wio
    • 2 years ago
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    Then solve for \(T\) first the first problem.

  5. theEric
    • 2 years ago
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    That doesn't look familiar... I used separation of variables to get \(S=e^{r\ t}\ C\), and then I was going to treat it like an initial value problem where \(S(0)=S_0\)

  6. wio
    • 2 years ago
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    Oh it's compounded continuously, I see.

  7. wio
    • 2 years ago
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    So you used a differential equation to solve this?

  8. theEric
    • 2 years ago
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    Yeah... I think that's what the class needs me to do! So, I need to think about what the rate is. Is it just the the \(S\) times the \(r\)? I was less sure when I asked... I think I'll see if I can get to a correct answer...

  9. wio
    • 2 years ago
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    What is your original differential equation?

  10. theEric
    • 2 years ago
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    I was thinking to use \(S'=S\ r\), and then from there I got to \(S=e^{rt}C\). I'm sorry, I'm doing a couple things at once, now. Don't worry about getting back to me - I'm only half here!

  11. wio
    • 2 years ago
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    Okay then your answer for (a) is correct.

  12. AccessDenied
    • 2 years ago
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    Have you had luck with this problem so far? I think what you've had so far seems like it's on the right track, just curious. :)

  13. theEric
    • 2 years ago
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    Hi! I'm actually confused! The answers are from the book. I typed up my work here, but I hit a button and it all went away.. So I'm typing it up in notepad and will paste it here in a moment. Thanks for checking!

  14. AccessDenied
    • 2 years ago
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    Alright, and yes that's a good idea. I do that from time to time as well in case of emergencies or when it feels laggy typing LaTeX stuff.

  15. theEric
    • 2 years ago
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    Then I did \(S(0)=S_0=e^{r\ (0)}C=C\implies S_0=C\) Then \(S=e^{r\ t}S_0\). I just realized I did some work but it doesn't pertain to the question!

  16. theEric
    • 2 years ago
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    Oh wait, maybe I was on the right track! One second!

  17. theEric
    • 2 years ago
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    Now I try to solve for \(T\). \(S=e^{r\ T}S_0\\\implies\dfrac{S}{S_0}=e^{r\ T}\\\implies\ln\left(\dfrac{S}{S_0}\right)=\ln(e^{r\ T})=r\ T\\\implies T=\ln\left(\dfrac{S}{S_0}\right)\ r^{-1}\) \(S=2S_0\\\implies T=\ln\left(\dfrac{2S_0}{S_0}\right)\ r^{-1}=\ln(2)\ r^{-1}\) Mission one, success! Thanks for sticking around to see my accomplishment! :)

  18. AccessDenied
    • 2 years ago
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    Yes! You are welcome! The other two should follow up pretty nicely from your new equation now. :)

  19. theEric
    • 2 years ago
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    Is part (b) a follow-up to part (a)? Should I leave \(S=S_0\), do you think?

  20. theEric
    • 2 years ago
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    \(S=2S_0\), I mean.

  21. AccessDenied
    • 2 years ago
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    yeah, I think they want you to simply apply your new equation. I was able to get the correct answers doing so.

  22. theEric
    • 2 years ago
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    (b) \(T=\dfrac{\ln(2)}{.07}\approx 9.90\) Thank you very much!

  23. theEric
    • 2 years ago
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    Thank you for checking the answers before I got to them, I mean! :D

  24. AccessDenied
    • 2 years ago
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    Yep! I am glad to have been helpful! :)

  25. theEric
    • 2 years ago
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    :)

  26. theEric
    • 2 years ago
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    At this point, I found the answers! Thank you everyone! Just for completion, I'll show my work to get (c). Maybe this can help someone else, if they search for it or something. Or maybe if someone wants to use it as a reference! \(r=\ln(2)\ t^{-1}\) and \(t=8\) so \(r=\dfrac{\ln(2)}{8}=.08664...\approx.0866=8.66\%\) \(\huge\color{#44BB33}{\large\ \ o\ o\ \\\smile}\)

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