OSH/2/Solution
From UBCMATH WIKI
Solution key

 The distance between point $(x,0)$ and $(2,3)$ is given by the distance formula (or, if you prefer, the Pythagorean theorem) as $d(x) = \sqrt{(x2)^2+(03)^2} = \sqrt{x^24x+13}$. The graph $y=d(x)$ is shown below. Note that as $x$ tends towards positive or negative infinity, $d(x)$ behaves similarly to $x2$. Furthermore, the closest we get to our destination is when $x=2$, at which point our distance is 3.
 From the map, we see Gas Station $A$ is at point $(40,10)$, and Gas Station $B$ is at point $(70,10)$. The point along the $x$axis that is equidistant from the two stations is $(55,0)$. At that point, our distance to the nearest station is $\sqrt{(5540)^2+(010)^2}=\sqrt{325}\approx 18$ km. The closest we ever get to a gas station is 10 km away; this occurs when $x=40$ and when $x=70$. We know the shape of the graph when $x\le55$, and when $x\ge55$, from part (a).
 $d_{\text{crow}}(x)$ is indeed a continuous function: at the point where the nearest gas station changes (that is, $x=55$), the nearest two gas stations are equidistant. Then $\lim\limits_{x \to 55^}d_{\text{crow}}(x)$ is the distance to station $A$, and $\lim\limits_{x \to 55^+}d_{\text{crow}}(x)$ is the distance to station $B$. Since these are the same, $\lim\limits_{x \to 55^}d_{\text{crow}}(x)=\lim\limits_{x \to 55^+}d_{\text{crow}}(x)=d_{\text{crow}}(55)$. That is, $\lim\limits_{x \to 55}d_{\text{crow}}(x)=d_{\text{crow}}(55)$, which is the definition of continuity. At points where the nearest gas station does not change, the function is also continuous.
 The function $d_{\text{crow}}$ is not differentiable at the point $x=55$. We see this from the sharp corner on the graph. Immediately before x=55 they are getting farther from the nearest station ($A$) at a nonzero rate (in fact, close to 1) and immediately after $x=55$ they are getting closer to the nearest station ($B$) at a nonzero rate (1). Because these are not equal, the function is not differentiable.

Using the map, we see gas station $A$ is at point $(40,10)$, and gas station $B$ is at point $(70,10)$.
 Along the section of road from $x=0$ to $x=40$, gas station $A$ is closer, because you'd have to pass its ramp to get to gas station $B$. The distance to station $A$ is $50x$.
 Along the section of road after $x=50$ until $x=70$, gas station $B$ is closer, because you'd have to pass its ramp to get to gas station $A$. The distance to station $B$ is $80x$.
 Suppose $40<x\le 50$.
 To get to gas station $A$, you have to take the following route:
 Go to the off ramp: $(50x)$ km
 Take the onramp back onto the highway in the other direction, and drive to the offramp leading to the station: $10$ km
 Go to the station: 10 km
So, all together, $70x$ km.
 To get to gas station $B$, you have to take the following route:
 Go to the off ramp: $(70x)$ km
 Go to the station: 10 km
So, all together, $80x$ km.
 To get to gas station $A$, you have to take the following route:
 Since $(70x)<(80x)$, when $40<x\le 50$, the closest gas station is Gas Station $A$, and its distance is $(70x)$ km.
 $d_{\text{drive}}$ has a jump discontinuity at $x=40$ and $x=50$. At $x=40$, traveling to the nearest gas station (station $A$) suddenly gets harder, because you've passed the first exit. After this exit, you need to travel all the way to the second exit and back to reach Station $A$. At $x=50$, you pass the second exit, and now Station $B$ is closer than Station $A$. Just before $x=50$, you are 20 km from Station $A$. Just after it, you are 30 km from station $B$ (also you're 60 km from Station $A$, since you'd have to drive all the way to the third exit and turn around). So at $x=50$, there is a jump discontinuity where the $y$value of your function jumps from 20 to 30.
 $d_{\text{drive}}$ is differentiable everywhere except $x=40$ and $x=50$ where it is discontinuous. Elsewhere inside its domain, the function has a constant derivative of $1$.


By definition,
 $\displaystyle B'(t)=\lim_{h \to 0} \frac{B(t+h)B(t)}{h} = \lim_{h \to 0} \frac{A\frac{t+h}{k+t+h}A\frac{t}{k+t}}{h}.$
 $\displaystyle B'(t)=\lim_{h \to 0} \frac{A\frac{(t+h)(k+t)t(k+t+h)}{(k+t)(k+t+h)}}{h} = \lim_{h \to 0} \frac{A\frac{(kt+t^2+kh+ht)(kt+t^2+ht)}{(k+t)(k+t+h)}}{h}$
 $\displaystyle = \lim_{h \to 0} \frac{A\frac{kh}{(k+t)(k+t+h)}}{h} = \lim_{h \to 0} \frac{k}{(k+t)(k+t+h)} = A\frac{k}{(k+t)^2}$.
 $B'(t)$ is positive for $t>0$. A positive slope means that the quantity of berries that the bear has collected is always increasing.
 A steep slope in $B(t)$ means it is easy to collect berries. A shallow slope means it is hard to collect berries.

By definition,
Allocation of points


Explaining how they got their graphs is not required but if they got it wrong because they misinterpreted the question and explained what they did, they can get some points.

 1 pt minimum at $(2,3)$
 1 pt straight lines growing without bound as $x \to \pm \infty$

 1 pt dual minima
 1 pt straight lines growing without bound as $x \to \pm \infty$
 1 pt cusp at $x=55$

 1 pt continuous, including where the nearest gas station changes. Students need to give an explanation, but don't have to use the formal definition of continuity.

 1 pt not differentiable at the cusp

 1 pt jump discontinuities at $x=40$ and $x=50$.
 1 pt correct maxima and minima
 1 pt piecewise linear

 1 pt correct answer plus explanation

 1 pt correct answer plus explanation



 1 pt  Plug the function in correctly to the limit expression.
 1 pt  Recognize the indeterminacy problem (possibly implicitly by fixing the problem). Explaining in words is not required as long as the procedure is clearly shown.
 1 pt  Finding a common denominator.
 1 pt  Arrive at the correct final answer.
 2 pt  Stating that $B'(t)$ is always positive  1 pt. Correct translation  1 pt.
 1 pt  Associating steep slope with ease of collecting and shallow slope with hard to collect  1 pt.

 presentation (1 pt) and communication (2 pts) across all questions.
Subtotal 12 pts.
Subtotal 7 points.
Subtotal 3 points.
Total 22 points.