# OSH/4/Solution

< OSH‎ | 4
1. [3 pts] $E'(t)$ is decreasing for $t<8$ and increasing for $t>8$ with a minimum at $t=8$ (1 pt). At the start of a shift, as suggested in the article referenced in problem statement, the error rate might decrease as a resident learns more about the condition of the patients currently under their care (1 pt). Later during a shift, the error rate might increase due to fatigue (1 pt).
2. [1 pt] I think the optimal time for shift change is when $E'(t)$ reaches a minimum at $t=8$.
3. [2 pts] The total number of errors at time $t$ is given by
$E(t) = 4t-\frac{1}{2}t^2+\frac{1}{48}t^3$ (1 pt for antiderivative, 1 pt for realizing $E(0)=0$ - showing this implicitly is fine).
4. [2 pts] The average rate of change is given by
$A(t) = \frac{E(t)-E(0)}{t-0} = 4 - \frac{1}{2}t + \frac{1}{48}t^2$ (1 pt for correct form, 1 pt for simplifying correctly).
5. [3 pts]
Option 1 Find the absolute minimum of $A(t)$.
Finding critical points of $A(t)$, we solve the equation
$A'(t)= -\frac{1}{2} + \frac{1}{24}t =0$ (1 pt).
Thus, $t=12$ is the only critical point (1 pt). Because $A(t)$ is a upward facing quadratic, we know $t=12$ is the location of the absolute minimum of $A(t)$ (1 pt - for any valid justification that it's the absolute min).
Option 2 Find the point along the graph of $E(t)$ at which the secant line connecting the origin and $(t,E(t)$ has the smallest slope.
Ploting the graph of $E(t)$, we can see that this will occur at a value of $t$ for which the tangent line to $E(t)$ goes through the origin (1 pt for making it clear this is what they're doing). The tangent to $E(t)$ at $t=a$ is $y=E'(a)(t-a)+E(a)$. This goes through the origin when
$0=E'(a)(0-a)+E(a)$ or equivalently when ...
$0=-4a+a^2-\frac{1}{16}a^3 + 4a-\frac{1}{2}a^2+\frac{1}{48}a^3$
$0=-\frac{2}{48}a^3 +\frac{1}{2}a^2$
$a=0$ or $a=12$ (1 pt). The tangent line at the origin has a slope of 4 (and corresponds to an obviously ridiculous scenario) and the tangent line at 12 has a slope of 1 so the minimum average error rate over one shift occurs when the shift length is chosen to be $t=12$ (1 pt).
6. [5 pts] See this desmos page for an interactive plot. Marking: Axes should be labeled $t$ and $E(t)$ with a few values marked for scale (1 pt). The function should be increasing everywhere (1 pt). The inflection point at $(8,32/3)$ should have a zero slope (1 pt) (red dot in Desmos). There should be a secant line (blue line) from the origin to $(12,12)$ (orange dot) whose slope gives the average rate of change over a 12 hour shift (1 pt). The secant line should actually be tangent to the graph at the $(12,12)$ (1 pt).
7. [1 pt] My expectation does not match the result, which indicates that a shift should extend 4 hours beyond the lowest error rate. During the period from $t=8$ until $t=12$, even though the resident on the first shift is making more errors than they did previously, they accummulate fewer errors than a new resident just starting their shift at $t=8$ would in their first 4 hours.

Subtotal: 17 pts.

1. [6 pts] $S(0)=0$. This means that no cancer patient will survive without the drug treatment. As $q\to\infty$, we can approximate
$\dfrac{q}{k_1+q}\approx\dfrac{q}{q}=1$ and $\dfrac{q}{k_2+q}\approx\dfrac{q}{q}=1$ so
$\lim_{q\to\infty}S(q) = 1-1=0.$ This means that any cancer patient getting a very large dose is unlikely to survive, presumably due to toxicity. (Note that we need $k_1<k_2$ or else $S(q)$ would be a negative number for all values of $q$ which makes no sense since it's supposed to be the fraction of surviving patients.)
Students should:
• (1 pt) Compute $S(0)$.
• (2 pts) The interpretation of $S(0)$. (1 pt for interpreting $S=0$ and 1 pt for also including an interpretation of $q=0$ in the statement).
• (1 pt) Compute the limit as $q\to\infty$ either by approximating the function for large $q$ or by a more formal limit calculation.
• (2 pts) The interpretation of $\lim_{q\to\infty}S(0)$. (1 pt for interpreting $S\to 0$ and 1 pt for also including an interpretation of $q\to\infty$ in the statement).
2. [4 points ] From (a), we know the general shape of $S(q)$, and that its maximum will occur at a critical point (rather than an endpoint). So, we calculate: $S'(q)=\frac{k_1}{(k_1+q)^2}-\frac{k_2}{(k_2+q)^2}$ Since $S'(q)$ exists everywhere in our model domain, we find its zeroes.
\begin{align*}0&=\frac{k_1}{(k_1+q)^2}-\frac{k_2}{(k_2+q)^2}\\ \frac{k_1}{(k_1+q)^2}&=\frac{k_2}{(k_2+q)^2}\\ k_1(k_2^2+2k_2q+q^2)&=k_2(k_1^2+2k_1q+q^2)\\ k_1k_2^2+2k_1k_2q+k_1q^2&=k_1^2k_2+2k_1k_2q+k_2q^2\\ k_1k_2(k_2-k_1)&=q^2(k_2-k_1)\\ k_1k_2&=q^2\\ q&=\sqrt{k_1k_2} \end{align*}
(In this calculation, we made use of the fact that $k_1 \neq k_2$ and $q \ge 0$.) So, the optimal dosage is $q=\sqrt{k_1k_2}$.
Students should:
• (1 pt) Compute $S'(q)$.
• (1 pt) Find the critical point of $S(q)$.
• (1 pt) Justify that this is a maximum.
• (1 pt) Interpret this as the optimal dosage.
3. [3 pts] As we found in (b), if $k_1=25$, the survival rate is maximized by a dosage of $q_*(k_2)=\sqrt{25k_2}=5\sqrt{k_2}$. However, the given information is that we must add the constraint $q_*(k_2) \le 45.$ So, the function giving the optimal dose in this case is: $q_*(k_2)=\begin{cases} 5\sqrt{k_2} & \text{ if }k_2 \le 81\\ 45 & \text{ if }k_2 > 81 \end{cases}$ So, the graph that gives the optimum dosage, based on the patient's $k_2$ value, is:
2 points: explanation
1 point: graph

Subtotal: 13 pts

Communication: 2 pts
Presentation: 1 pt
Total: 33 pts.