Results 1 to 4 of 4

Math Help - Continuous compounding

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    7

    Continuous compounding

    I'm having a lot of trouble with my calculus homework. Does anyone know how to work this problem?
    Suppose the consumption of electricity grows at 7.6% per year, compounded continuously. Find the number of years before the use of electricity has tripled. Round the answer to the nearest hundredth.

    Possible Answers:
    14.46 , 39.47 , 0.14 , 1.45
    Last edited by hotblonde; March 11th 2009 at 12:37 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2009
    Posts
    7
    ANYBODY??? I figured the answer but am having problems with the formula. I came up with fv=p(1+r/n)^t, and cannot figure out t....my formula looks like this:

    3=1(1.076)^t

    help...........
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100
    I got 14.55 years - I will post how I did it in a minute - I have to get the equations into this box

    Your formula should be ...

    a(t) = lim as n approaches infinity ( a0(1+ r/n)^(nt) )
    where a0 is the initial amount of power used, r is the rate of growth (.076) and n is the number of times it is compounded

    Then you use L'Hopital's rule to find that the above function is the same as this

    a(t) = a0e^(rt)


    (Hopefully this helps a little)

    <br />
\mathop {\lim }\limits_{n \to \infty } \ln ((1 + \frac{r}<br />
{n})^{nt} ) = \mathop {\lim }\limits_{n \to \infty } (nt)\ln (1 + \frac{r}<br />
{n}) = \mathop {\lim }\limits_{n \to \infty } t \bullet \frac{{\ln (1 + \frac{r}<br />
{n})}}<br />
{{1/n}}<br />


    <br />
\begin{gathered}<br />
   = \mathop {\lim }\limits_{n \to \infty } t \bullet \frac{{\frac{1}<br />
{{1 + \frac{r}<br />
{n}}}}}<br />
{{ - 1/n^2 }}( - r/n^2 ) = \mathop {\lim }\limits_{n \to \infty } t\frac{n}<br />
{{n + r}}(r) = \frac{1}<br />
{1}rt \hfill \\<br />
  \mathop {\lim }\limits_{n \to \infty } e^{\ln f(x)}  = e^L  = e^{rt}  \hfill \\ <br />
\end{gathered} <br />


    a(t) = 3 = 1*e^(.076*t)
    ln(3) = .076*t
    t=ln(3)/.076 = 14.55 years
    Last edited by billa; March 11th 2009 at 03:31 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    23
    Anytime something is compounded continuously, use the  A=Ce^{kt} formula. Or, if it is money that is being compounded continuously, use  A=Pe^{rt} formula. Either way, the formulas are equivalent.

    Or like billa said, you can use the regular compound formula and just find the  {\lim x \rightarrow  \infty}  .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Formula for FVoAD with continuous compounding
    Posted in the Business Math Forum
    Replies: 1
    Last Post: October 7th 2011, 05:50 AM
  2. Replies: 0
    Last Post: February 23rd 2010, 04:05 AM
  3. Continuous compounding formula
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 19th 2009, 04:26 PM
  4. Continuous compounding problems
    Posted in the Business Math Forum
    Replies: 1
    Last Post: March 30th 2009, 04:10 PM
  5. Compound Interest for Continuous compounding
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 22nd 2006, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum