# Thread: differential equation- wordy question 2-

1. ## 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-

2. 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

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# a bettery is being charged, the charging rate is modeled by dq/dt=k(Q-q)

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