The charge in coulombs that passes through a wire after $t$ seconds is given by the function:
$Q(t) = t^3 – 2t^2 + 5t + 2$.
Determine the average current during the first two seconds.
The answer is $5\textrm{A}$.
The question is in the calculus part of the problem set. I'm guessing you have to apply calculus, but I don't know why or what is the rationale for applying it. What I've been doing is plugging the $t$'s for $2$. I get $12\textrm{C}$ then I convert it to Amperes, to which I get $6$. I just don't see the why you should get the slope for this?
Best Answer
The calculus part uses the integral, not the derivative (slope). The average of a function $f(x)$ for $a\le x\le b$ is defined to be
$$\bar f=\frac 1{b-a}\int_a^b f(x)\,dx$$
Therefore you are to calculate
$$\frac 12\int_0^2 (t^3-2t^2+5t+2)\,dt$$
Here's the rationale. It is easy to calculate the average of finitely many values: just add the values and divide by how many there are. However, averaging a function over an interval involves infinitely many values. How do you add them or divide by infinity? The intuitive way to do that is to divide each value by an infinitely large value, which is the same as multiplying each value by an infinitely small value which we'll call $\frac {dx}{b-a}$. Then we add all those infinitely many values by integration. Simplifying that gives us the integral I showed you.
At least, that's the intuitive rationale that Leibniz used, and many engineers still use today. My mathematical soul shudders at this train of thought, but I admit it has been useful to many engineers over many years. Arbitrary use of this reasoning can lead you into trouble, so be careful with it.