Question 18
This differential equation can be used to determine the current $I$ (amperes) at time $t$ (seconds) with voltage $V$ (volts) in an electric circuit containing a resistance $R$ (ohms):
$$k\frac{dI}{dt}+RI=V$$
where $k$, $R$ and $V$ are positive constants and $t\ge0$.
Assuming that there is no current in the electric circuit initially, show that the size of the current can never be greater than $\dfrac{V}{R}$.
The current creeps up towards V/R
Watch the current being drawn from its differential equation. The white gap shows how far it still is from $\frac{V}{R}$.
The dashed orange line is a horizontal asymptote. You show this in the proof by writing the gap as an exponential.
Try to push the current past V/R
Change the constants and the time. The curve is $I=\frac{V}{R}\left(1-e^{-\frac{R}{k}t}\right)$, which is what you get at the end of the proof.
Work through the proof one mark at a time
Try each step yourself before you reveal it. There are no numbers, so keep $k$, $R$ and $V$ as letters all the way through.
Move $RI$ across so that the derivative is on its own, then separate the variables.
$$k\frac{dI}{dt}=V-RI\quad\Rightarrow\quad\int\frac{k}{V-RI}\,dI=\int1\,dt$$
Writing $\int\frac{1}{V-RI}\,dI=\int\frac{1}{k}\,dt$ is just as good.
Marker: correctly uses the separation of variables method to set up indefinite integrals
The derivative of $V-RI$ with respect to $I$ is $-R$, so you get a log with a factor of $-\frac{1}{R}$ in front.
$$-\frac{k}{R}\ln\left|V-RI\right|=t+c$$
You keep follow-through marks from here if you slipped earlier.
Marker: develops a general solution of the differential equation
There is no current initially, so $I=0$ when $t=0$.
$$-\frac{k}{R}\ln\left|V\right|=0+c\quad\Rightarrow\quad c=-\frac{k}{R}\ln(V)$$
You can drop the absolute value because $V>0$. Your constant may look different if you put $c$ somewhere else, and that is fine.
Marker: uses the given condition to determine an expression for the constant of integration
Put $c$ back in and make $\ln\left|V-RI\right|$ the subject. Multiply everything by $-\frac{R}{k}$.
$$\ln\left|V-RI\right|=-\frac{R}{k}t+\ln(V)$$
Marker: rearranges the relationship to express $\ln\left|V-RI\right|$ as the subject of the equation
Take the exponential of both sides, and use index laws to split off $e^{\ln(V)}=V$.
$$V-RI=e^{-\frac{R}{k}t+\ln(V)}=Ve^{-\frac{R}{k}t}$$
At $t=0$ the left side is $V$, which is positive, so you take the positive branch of the absolute value.
Marker: expresses the relationship as an exponential function
Now you use the fact that an exponential is always positive. $V>0$ and $e^{-\frac{R}{k}t}>0$ for every $t$, so the right side is always positive.
$$Ve^{-\frac{R}{k}t}>0\ \Rightarrow\ V-RI>0\ \Rightarrow\ RI<V\ \Rightarrow\ I<\frac{V}{R}$$
So the size of the current can never be greater than $\frac{V}{R}$.
You do not get this mark if you just write $I=\frac{V}{R}$ without a reason. You can also argue that as $t\to\infty$, $V-RI\to0^{+}$, so $I$ approaches $\frac{V}{R}$ from below.
Marker: considers the value of $I$ over time to determine the required limit
Each phrase of the question gave you something
Solving the differential equation is familiar, but it is only the first five marks. The last mark asks you to turn your solution into an inequality, which means explaining why one side is always positive.
I would check your last line first. The QCAA like “show that … can never” questions, and a bare $I=\frac{V}{R}$ at the end does not show that the current stays below it.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.