Answer to Question #143732 in Calculus for Kate

Question

In deciding how long a resident’s shift in the emergency room should be, the Chief of Staff at Vancouver General Hospital would like to minimize the average rate at which errors are made. Let E(t) be the number of errors made by a resident from the start of a shift until t hours into the shift. The instantaneous rate of change of errors made is E′(t) = t3 − 3 t2 + 2. What is the average rate of change of E(t) from the start of a shift (t = 0) up until time t? Call it A(t).

Answer 1

Let us find the number $E(t)$ of errors made by a resident from the start of a shift until $t$ hours into the shift. Since $E'(t) = t^3 − 3 t^2 + 2$, we have that

$E(t)=\int(t^3-3t^2+2)dt=\frac{t^4}{4}-t^3+2t+ C$.

Taking into account that according to the defenition of $E(t),$ $E(0)=0$ and $E(0)=C$, we conclude that $C=0$ and thus $E(t)=\frac{t^4}{4}-t^3+2t$. Then the average rate of change of $E(t)$ from the start of a shift ($t = 0$ ) up until time $t$ is

$A(t)=\frac{E(t)}{t}=\frac{\frac{t^4}{4}-t^3+2t}{t}= \frac{t^3}{4}-t^2+2.$