
And the wheel goes 'round...
A sawmill is powered by a water wheel of diameter \( 10 \) m. A flowing river turns the wheel at a rate of one revolution per minute. At what speed is a paddle on the wheel rising when it is \( 2 \) m vertically from the top of the wheel?

And one flew off...
Two particles travel together counterclockwise at a speed of \( 1 \) m/s around a unit circle with equation \( x^2 + y^2 = 1 \).
Let one particle be named \( P \) and the other be named \( Q \). At the point \( (1, 0) \), after \( t = 2\pi \) seconds have passed, particle \( Q \) flies tangentially (i.e., vertically) off the circle, heading upwards at the same speed (assume the particles are not affected by gravity). Particle \( P \) continues along its circular path. How fast are the particles separating at \( t = 3\pi \) s?

A Bounded Graph
Consider the function \( f(x) = x  2\cos(x) \) on the interval \( \left[ 2\pi, 2\pi \right] \).
 Sketch the graphs of \( y = x \) and \( y = 2\cos(x) \) on \( \left[ 2\pi, 2\pi \right] \). How many points of intersection (i.e., zeros of \( f \)) occur on this interval?
 Determine any relative extremes and points of inflection of \( y = f(x) \) on the interval \( \left[ 2\pi, 2\pi \right] \).
Sketch the graph of \( f(x) \) on \( \left[ 2\pi, 2\pi \right] \).
Hint: First note where \( f(x) = 0 \) (using part a) and where \( f(x) = x \).
 Explain why the sketch from part c) lies between the lines \( y = x  2 \) and \( y = x + 2 \). Add these lines to the sketch, showing clearly where \( f(x) \) meets these lines.
 Challenge: Make a sketch, similar to the one in part d), for the general case \( y = x  c \cos(x) \) for each of the cases
 \( 0 \lt c \lt 1 \),
 \( c = 1 \), and
 \( c \gt 1 \).

Implicit Curves
Sometimes implicit equations define curves in the plane. For example, \( x^2 + y^2 = 4 \) defines a circle centred at the origin of radius \( 2 \).
There is also the possibility that more than one curve is defined. For example, the equation \( x^2 = y^2 \) implies \( x^2  y^2 = 0 \), or \( (x  y)(x + y) = 0 \), which means that both \( y = x \) and \( y = x \) satisfy the equation, defining two straight lines.
For each of the following equations, determine any curves defined. Illustrate your solutions with suitable sketches.
 \( \cot(x) = \tan(y) \), where \( x \) and \( y \) are acute angles \( \left( 0 \lt x, y \lt \frac{\pi}{2} \right) \).
 \( \sin{(2x)} = 2y\sin(x) \)
 \( e^{8x^2y} = e^{y^4} \)
 \( \ln( y^2 ) = \ln(4) + 2\sin(\pi x) \)

Challenge: A computergenerated plot of points satisfying the equation \( x \ln(y) = y \ln(x) \) (or \( y^x = x^y \)) on the region \( x \gt 0, y \gt 0 \) is shown.
 State the equation of the linear solution.
 Explain the obvious symmetry of the other solution.
 Given that you know \( \dfrac{\ln(x)}{x} = k \) has
 one solution if \( k \leq 0 \) or \( k = \frac{1}{e} \), or
 two solutions if \( 0 \lt k \lt \frac{1}{e} \)
explain why \( P \) has coordinates \( (e, e) \) and why the second, nonlinear solution cannot meet the lines \( x = 1 \) or \( y = 1 \).

A MaxedOut Triangle
A triangle with vertices \( O~(0, 0) \), \( P~(x, y) \) (on the curve \( y = x^2e^{ax} \), where \( a \gt 0 \)), and \( Q~(x, 0) \) has area \( A(x) \).
Suppose that \( x = x_m \) maximizes \( A(x) \) and that \( x = x^{*} \) maximizes the height of the curve.
 Determine the value of \( x^{*} \) in terms of \( a \).
 Determine the value of \( x_m \) in terms of \( a \).
 Which of the two values, \( x^{*} \) or \( x_m \), is greater? Think about why this is so, given the geometry of the curve \( y = x^2e^{ax} \) (a sample curve is shown for some value of \( a \gt 0 \)).

Sliding along...
Note: This question assumes familiarity with Newton's laws of motion. Expositions can be found in most high school physics courses/textbooks.
A child on a toboggan is pulled along a horizontal surface. The force \( F \) exerted on the rope is given by \[ F( \theta ) = \dfrac{\mu W}{\mu \sin(\theta) + \cos(\theta)} ~ \text{for} ~ 0 \leq \theta \leq \tfrac{\pi}{2} \] where \( \mu \) is a positive constant satisfying \( 0 \lt \mu \lt 1 \), \( W \) is the combined weight of the child and toboggan, and \( \theta \) is the angle of the rope to the horizontal, as shown in the diagram.
 Evaluate \( F(0) \) and \( F\left( \frac{\pi}{2} \right) \). Explain why these values make sense, physically.
 Find the absolute extremes of \( F(\theta) \) on the interval \( 0 \leq \theta \leq \frac{\pi}{2} \).
 What value of \( F \) is easiest on the person pulling the toboggan?
 Assuming the toboggan moves horizontally, what value of \( F \) gives the child the fastest ride?

How square is this?
Two smaller squares, \( S_1 \) and \( S_2 \), lie completely within a larger square, \( S \).
\( S \) has side length \( 1 \).
The smaller of the two inner squares, \( S_1 \), lies flush with the bottom left corner of the largest square, and has side length \( x \).
The larger of the two inner squares, \( S_2 \), has side length \( y \). \( S_2 \) lies on an angle such that each of its vertices lies on exactly one side of \( S \). The vertex of \( S_2 \) which lies on the bottom side of \( S \), side \( AB \), is labelled as point \( P \); the vertex of \( S_2 \) which lies on the rightmost side of \( S \) is labelled \( Q \). \( S_2 \) also has one of its sides resting on the topright corner of \( S_1 \).
Let \( \theta = \angle{QPB} \). Then, as \( P \) moves along \( AB \), the values of \( \theta \), \( x \), and \( y \) will vary.
 Show that, in terms of angle \( \theta \), \[ y = \dfrac{1}{\sin(\theta) + \cos(\theta)}, \quad x = \dfrac{1}{2} \left( \dfrac{\sin{(2\theta)}}{\sin{(2\theta)} + 1} \right) \] for \( 0 \leq \theta \leq \frac{\pi}{2} \).
Determine the values of \( x \) and \( y \) for which \( x^2 + y^2 \) is a minimum, i.e., the values that minimize the area of the shaded region.
Hint: Show, after some simplification, that \( x^2 + y^2 \) has the form \( \dfrac{1}{4} \left( \dfrac{u + 2}{u + 1} \right)^2 \), where \( u \) is a simple expression in \( \theta \). Then write \( \dfrac{u + 2}{u + 1} \) in a simpler form.

To Row or To Jog — that is the question...
Standing at \( A \), on the edge of a circular pond as shown in the diagram, you observe a toddler straight across at \( B \), dangerously close to the edge of the water.
Thinking quickly, you realize there are three possibilities:
 Using your rowboat at point \( A \), you can row to some point \( P \) on the edge of the pond, located at an angle \( \theta \) to \( A \) as shown (with \( \theta \) to be determined by you), and then jog the rest of the way to \( B \), or
 Jog the entire way around the edge of the pond from \( A \) to \( B \), or
 Row directly from \( A \) to \( B \) across the pond
If you can jog \( 1.5 \) times faster than you can row, which of the three options will get you from \( A \) to \( B \) in the least time? If it is the first option, find the angle \( \theta \) which optimizes your route.

In the Limit...
 Evaluate each limit, or show that it does not exist.
 \( \displaystyle \lim_{x \to 0} e^{x \ln{\left\lvert x \right\rvert}} \)
 \( \displaystyle \lim_{x \to 0} \dfrac{e^{x \ln{\left\lvert x \right\rvert}}  1}{x} \)
Hence, comment on the differentiability at \( x = 0 \) of the function \[ f(x) = \begin{cases} \left\lvert x \right\rvert^x & x \neq 0 \\ 1 & x = 0 \end{cases} \]
 Given that \( \displaystyle \lim_{x \to 0} \dfrac{x^2 + \sin(bx) + \sin(x)}{ax^2 + 3x^4 + 2x^3} = 5 \), determine the values of \( a \) and \( b \).
 For what values of the constant \( a \) is \( \displaystyle \lim_{x \to \infty} \left( \dfrac{x + a}{x  a} \right)^x = e^{a^2} \)?

Falling Down, Down, Down...
A plausible model for the forces acting on a falling body near the Earth's surface is to describe them as a net force \[ F_{\text{net}} = F_{gr}  F_{r} \] where \( F_{gr} = mg \) is the mass of the body times the gravitational acceleration constant \( g \), and \( F_r \) is the resistance force created by atmospheric drag.
If we assume that \( F_r \) is proportional to the velocity of the object, then Newton's law of motion \[ (\text{mass}) \times (\text{acceleration}) = (\text{net force}) \] implies that the velocity, \( v(t) \), obeys the differential equation \[ m \dfrac{dv}{dt} = F_{gr}  F_r = mg  kv \] for some positive constant \( k \).
 Show that, if the body starts from rest, with \( v(0) = 0 \), then \( v(t) = \dfrac{mg}{k} \left( 1  e^{\frac{kt}{m}} \right) \) satisfies the given differential equation.
 Find \( \displaystyle \lim_{t \to \infty} v(t) \), and hence give a physical interpretation for the constant \( \dfrac{mg}{k} \).
 A value of \( k = 0 \) implies that the motion occurs in a vacuum (no atmospheric drag). Thinking of \( t \) as a constant (i.e., fixing a moment in time), use l'Hospital's rule to determine \( \displaystyle \lim_{k \to 0^{+}} v(t) \). Hence, determine the (wellknown) velocity for a body falling from rest in a vacuum.

Keeping Order
Suppose that \( \displaystyle \lim_{x \to \infty} f(x) = \infty \) and \( \displaystyle \lim_{x \to \infty} g(x) = \infty \). We say that "\( g \) grows faster than \( f \)" if \[ \lim_{x \to \infty} \dfrac{f(x)}{g(x)} = 0 \] and "\( g \) and \( f \) grow at the same rate" if \[ \lim_{x \to \infty} \dfrac{f(x)}{g(x)} = c \neq 0 \] That is, \( f(x) \approx cg(x) \) as \( x \to \infty \).
 Use l'Hospital's rule to show that
 \( \displaystyle \lim_{x \to \infty} \dfrac{\ln(x)}{x^p} = 0 \) for all \( p \gt 0 \), and
 \( \displaystyle \lim_{x \to \infty} \dfrac{x^p}{e^{ax}} = 0 \) for all \( a \gt 0 \), \( p \gt 0 \)
 Explain why, if \( g \) grows faster than \( f \), then \( e^g \) grows faster than \( e^f \) as \( x \to \infty \).
Order the following functions from slowest to fastest growing, as \( x \to \infty \):
\[ x^{\ln(x)}, x^{e}, x^{x}, 2^{x}, e^{x^2} \]
Hint: the results from parts a) and b) may be useful.

A Little Geometry...
In the diagram, \( P \) and \( Q \) are points on the circumference of a circle with center \( C \) and radius \( r \), and \( O \) is a point which lies outside the circle. The lines \( OP \) and \( OQ \) are tangent to the circle, and the line \( OR \) is perpendicular to the line \( PQ \), where \( R \) lies on \( PQ \).
Let \( x \) denote the portion of \( OR \) that lies within the circle, and let \( y \) denote the portion of \( OR \) which lies outside the circle. If \( \theta = \angle{PCQ} \), evaluate \( \displaystyle \lim_{\theta \to 0} ~ \dfrac{x}{y} \).