Webwork 11.5 Hints

AttachmentSize
F2010M2300_webwork-11-5-hints.pdf93.53 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Webwork 11.5 Hints – Wednesday, November 3, 2010

Half-Life Example

  1. I'll modify an example I've used before. Assume that at time t=0 I consume 300mg of caffeine (I believe this is the largest amount legally allowed in a single dose). The amount of caffeine in my system Q is then a function of time that satisfies some proportionality constant k. That is, the amount of caffeine in my system at a time t is a solution to the differential equation

    Q^{{\prime}}=\frac{dQ}{dt}=kQ.

    We don't know what k is, but maybe we know that the half-life of caffeine in my system is 1.5 hours. So, what we want to do is as follows: The differential equation above represents caffeine decay in my system in general. In the specific situation (corresponding to an initial condition) when I digest 300mg at time t=0, there is a specific solution. We need to find that specific solution. Let's do that, and then solve for k.

    \displaystyle\frac{dQ}{dt} \displaystyle=kQ
    \displaystyle\frac{1}{Q}\frac{dQ}{dt} \displaystyle=k
    \displaystyle\int\frac{1}{Q}\frac{dQ}{dt}\, dt \displaystyle=\int k\, dt
    \displaystyle\ln|Q| \displaystyle=kt+C
    \displaystyle Q \displaystyle=Be^{{kt}}

    Now, we know that at t=0 we have Q=300mg. So, we can solve for k in our specific solution to the given differential equation. We have 300=Be^{{k\cdot 0}} and so B=300. We know that at t=1.5 (t is in hours here) we will have half of our original amount. So, we solve for k as follows.

    \displaystyle\frac{1}{2}300 \displaystyle=300e^{{1.5k}}
    \displaystyle\frac{1}{2} \displaystyle=e^{{1.5k}}
    \displaystyle\ln\left(\frac{1}{2}\right) \displaystyle=1.5k
    \displaystyle\frac{2}{3}\ln\left(\frac{1}{2}\right) \displaystyle=k
    \displaystyle-\frac{2}{3}\ln(2) \displaystyle=k

© 2011 Jason B. Hill. All Rights Reserved.