Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 19

The equation $Q(t) = t^3 - 2t^2 + 6t+2$ represents the quantity of change in coulombs(C) that has passed through a point in a wire up to time $t$ in seconds. Find the current when (a) $t = 0.55$ and (b) = $15$. At what time is the current lowest? The unit of current is ampere $\displaystyle 1A = 1 \frac{c}{s}$
The equation for current can be determined by taking the derivative of the equation $Q(t)$ with respect to $t$

$
\begin{equation}
\begin{aligned}
Q'(t) &= \frac{d}{dt}(t^3) - 2 \frac{d}{dt} (t^2) + 6 \frac{d}{dt} (t) + \frac{d}{dt}(2)\\
\\
Q'(t) &= 3t^2 - 2(2t) + 6(1) + 0\\
\\
Q'(t) &= 3t^2 - 4t + 6
\end{aligned}
\end{equation}
$


a.) when $t = 0.5 s$

$
\begin{equation}
\begin{aligned}
Q'(0.5) &= 3(0.5)^2 - 4(0.5)+6\\
&= 4.75 \text{ Amperes}
\end{aligned}
\end{equation}
$

b.) when $t = 1s$

$
\begin{equation}
\begin{aligned}
Q'(1) &= 3(1)^2 - 4(1) +6\\
&= 5 \text{ Amperes}
\end{aligned}
\end{equation}
$


The time where the current is lowest is where $Q'(t) =0$

$
\begin{equation}
\begin{aligned}
Q'(t) = 0 &= 3t^2 - 4t + 6\\
0 &= 3t^2 - 4t + 6
\end{aligned}
\end{equation}
$


By using quadratic formula,

$
\begin{equation}
\begin{aligned}
t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
t &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4 (3) (6)}}{2(3)}\\
\\
t &= \frac{4 \pm \sqrt{-56}}{6}\\
\\
t_1 &= \frac{2}{3} + \frac{\sqrt{-56}}{6} \text{ and } t_2 = \frac{2}{3} - \frac{\sqrt{-56}}{6}
\end{aligned}
\end{equation}
$


Therefore, the current is lowest at $\displaystyle t = \frac{2}{3}s$ (the real part of the solution.)