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theEric
 3 years ago
Hi! This is a financebased 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.
theEric
 3 years ago
Hi! This is a financebased 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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theEric
 3 years ago
Best ResponseYou've already chosen the best response.1I 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?

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Oh, answers: (a) \(\ln(2)\div r\) year (b) \(9.90\) years (c) \(8.66\%\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it something like this?\[ 2S_0=S_0(1+r)^T \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then solve for \(T\) first the first problem.

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1That 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 \(S(0)=S_0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh it's compounded continuously, I see.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you used a differential equation to solve this?

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah... 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is your original differential equation?

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1I 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay then your answer for (a) is correct.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Have 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. :)

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Hi! 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!

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Alright, 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.

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Then 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!

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Oh wait, maybe I was on the right track! One second!

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Now 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! :)

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Yes! You are welcome! The other two should follow up pretty nicely from your new equation now. :)

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Is part (b) a followup to part (a)? Should I leave \(S=S_0\), do you think?

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, I think they want you to simply apply your new equation. I was able to get the correct answers doing so.

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1(b) \(T=\dfrac{\ln(2)}{.07}\approx 9.90\) Thank you very much!

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1Thank you for checking the answers before I got to them, I mean! :D

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Yep! I am glad to have been helpful! :)

theEric
 3 years ago
Best ResponseYou've already chosen the best response.1At 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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