Results 1 to 2 of 2

Math Help - differential equation- wordy question 2-

  1. #1
    Junior Member
    Joined
    May 2006
    Posts
    37

    differential equation- wordy question 2-

    A battery is being charged. The charging rate is modelled by dq/dt=k(Q-q), where q is the charge in the battery( measured in ampere hours) at time t( measured in hours), Q is the maximum charge the battery can store and k is at constant of proportionality. The model is valid for q> and equal to o.4Q.
    a. It is given that q=xQ where x is a constant such that x is between 1 and 0.4- inclusive of 1 and 0.4. Solve the differential equation to find q in terms of t.

    b. It is noticed that the charging rate halves every 40 minutes. Show that k=3/2 in2- notice 3/2 in2 is different from 3/2in2.
    c. Charging is always stopped when q=0.95Q. If T is the time until charging is stopped, show that T=2in(20(1-x) over 3in2 for the values between 0.4 and 0.95 inclusive of these two values.

    Please explain if you can as it is a word question. Explanation has equal value to working, probably more-
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1
    Quote Originally Posted by kingkaisai2
    A battery is being charged. The charging rate is modelled by dq/dt=k(Q-q), where q is the charge in the battery( measured in ampere hours) at time t( measured in hours), Q is the maximum charge the battery can store and k is at constant of proportionality. The model is valid for q> and equal to o.4Q.
    a. It is given that q=xQ where x is a constant such that x is between 1 and 0.4- inclusive of 1 and 0.4. Solve the differential equation to find q in terms of t.
    I don't have much time today, but I can whip out a solution for a).

    Since q = xQ thus Q = \frac{q}{x} where x is a constant.

    So {dq}{dt} = k(Q - q) = k \left ( \frac{q}{x} - q} \right ) = k \left (\frac{1}{x} - 1 \right )q where k and x are constant.

    The general solution of such a linear differential equation is of the form
    q(t) = Ae^{bt}

    So \frac{dq}{dt} = Abe^{bt}

    Plugging these into your differential equation gives:
    Abe^{bt} = k \left (\frac{1}{x} - 1 \right ) Ae^{bt}

    Comparison shows that b = k \left (\frac{1}{x} - 1 \right ).

    So far we have: q(t) = A \cdot exp \left [ k \left (\frac{1}{x} - 1 \right ) t \right ]

    Now, we know that the maximum charge on the battery is Q. Thus the limit of q(t) as t goes to infinity is Q. Note that b is a negative number (given the range of acceptable x values.) Thus the limit of the exponential function as t gets very large is 1. ie.
    \lim_{t \to \infty} exp \left [ k \left (\frac{1}{x} - 1 \right ) t \right ] = 1

    Thus \lim_{t \to \infty} q(t) = A and we know q( \infty ) = Q according to the problem statement. Thus A = Q and your final solution is:
    q(t) = Q \cdot exp \left [ k \left (\frac{1}{x} - 1 \right ) t \right ]

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Wordy question, about rate of temperature increase.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 25th 2010, 01:03 PM
  2. Wordy question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 2nd 2009, 07:03 AM
  3. Another Differential Equation Question
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 13th 2009, 08:00 PM
  4. Differential Equation Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 13th 2009, 10:36 AM
  5. Differential equation- wordy question1- urgent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 1st 2006, 11:38 PM

Search Tags


/mathhelpforum @mathhelpforum