Time to Replace your Battery!

Calculus Level 2

You have a faulty laptop battery whose charge gauge acts very strangely. Particularly, after every minute, its estimate of how much time you have left before it dies drops by $d$ minutes, where $d$ is a positive number. The gauge estimates the rate at which you deplete the battery by finding the average rate of change of your battery charge thus far with respect to time. Your battery takes exactly $m$ minutes to die. The function $f(t)$ , where $t$ is a positive number of minutes, describes the amount of battery charge used (i.e. $f(0) = 0$ when your battery is charged, and $f(m) = 1$ when your battery is dead). Which is the correct form of $f(t)$ ?

f(t) = \frac{t}{(1-d)t+dm}

f(t) = \frac{dt}{(d-1)t+m}

f(t) = \frac{t}{m(t-d)}

f(t) = \frac{t}{(d+1)t-dm}

2 solutions

Paolo Lammens
Jun 11, 2018

Let $g(t)$ be the gauge's estimate, in minutes, of how much battery time remains. At $t=0$ , this estimate equals some fixed initial amount $a$ . Each minute, the estimate drops by $d$ minutes. So far, $g(t)=a-dt$ . We can solve for $a$ by establishing that $g(m)=0\,$ : $g(m)=a-dm=0 \Rightarrow a=dm$ . Putting this together, we get $g(t)=dm-dt$ .

We are also told that this estimate is based on the average rate of change of the battery thus far (not the actual instantaneous rate of change). Said rate of change is just $\frac{f(t)}{t}$ . So the estimate $g(t)$ is calculated as $\frac{1-f(t)}{\frac{f(t)}{t}}=\frac{t(1-f(t))}{f(t)}$ .

Therefore:

$\displaystyle g(t)=dm-dt=\frac{t(1-f(t))}{f(t)} \quad \Rightarrow \quad f(t)=\frac{t}{dm-dt+t}=\boxed{\displaystyle \frac{t}{dm+t(1-d)}}\,\Box$

No differential equations needed!

How do you get the "estimate g(t) is calculated as..." bit?

Ali Al-Ali - 2 years, 7 months ago

it will be more understandable if you put $md$ and $td$ instead of $dm$ and $dt$ :-)

Bostang Palaguna - 5 months, 2 weeks ago

Michael Lee
May 17, 2014

The average rate of change of battery life is $f(t)/t$ . Thus, the time remaining is estimated as $T(t) = (1-f(t))\cdot t/f(t)$ . Then, we have $T'(t) = -d \Rightarrow \left(f(t)-tf'(t)\right)/f(t)^2-1 = -d$ . Rearrange this to get $f'(t)/\left(f(t)((d-1)f(t)+1)\right) = 1/t$ and integrate both sides with respect to $t$ to get $\log(f(t))-\log((d-1)f(t)+1) = log(t)+c_1$ . Then, $f(t) = \frac{-e^{c_1}t}{(d-1)e^{c_1}t-1}$ . Plug in $f(m) = 1 \Rightarrow \frac{-e^{c_1}m}{(d-1)e^{c_1}m-1} = 1 \Rightarrow c_1 = -\log(md)$ . Then, $f(t) = \frac{-t}{md\left(t/m-t/md-1\right)} = \frac{t}{(1-d)t+md}$ . Q.E.D.

The problem says "The gauge estimates the rate at which you deplete the battery by finding the average rate of change of your remaining battery charge with respect to time." I think instead of the remaining battery charge, that should say the battery charge used already. (If you use the average rate of change of the remaining battery charge, then in fact the gauge will always accurately say how much time is left.)

The equation $T(t) = (1 - f(t)) \cdot f(t)/t$ should be $T(t) = (1 - f(t)) \cdot \frac{t}{f(t)}.$ Also, since you know this function decreases at a constant rate of $-d$ , you can simply set $T(t) = (1 - f(t)) \cdot \frac{t}{f(t)} = c - dt.$ You can then sovle for $f(t)$ , then use the condition $f(m) = 1$ to solve for $c$ .

Jon Haussmann - 7 years ago

You're right. My mistake(s)!

Michael Lee - 7 years ago

Time to Replace your Battery!

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