Problem of the Month
Problem 7: April 2024

Curves in the plane are often given as the set of points $(x,y)$ that satisfy some equation in $x$ and $y$. For example, the set of points $(x,y)$ that satisfy $y=x^2$ is a parabola, the set of points $(x,y)$ that satisfy $y=3x+4$ is a line, and the set of points $(x,y)$ that satisfy $x^2+y^2=1$ is the circle of radius $1$ centred at the origin.

Another way to express a curve in the plane is using parametric equations. With this type of description, we introduce a third variable, $t$, called the parameter, and each of $x$ and $y$ is given as a function of $t$. This is useful for describing the position of a point that is moving around the plane. For example, imagine that an ant is crawling around the plane. If its $x$-coordinate at time $t$ is $x=x(t)$ and its $y$-coordinate at time $t$ is $y(t)$, then its position at time $t$ is $(x(t),y(t))$.

1. A particle’s position at time $t$ is $(x,y)=(1+t,-2+2t)$. That is, its $x$-coordinate at time $t$ is $1+t$ and its $y$-coordinate at time $t$ is $-2+2t$.

   1. Plot the position of the particle at $t=0$, $1$, $2$, $3$, and $4$.

   2. Show that every position the particle occupies is on the line with equation $y=2x-4$.

   3. Sketch the path of the particle from $t=0$ through $t=4$.

2. A particle’s position at time $t$ is $(\cos (t),\sin (t))$. Sketch the path of the particle from $t=0$ through $t=2\pi$.

3. A particle’s position at time $t$ is $(\cos (t),\sin (2t))$.

   1. Plot the position of the particle at $t={\displaystyle \frac{k\pi }{12}}$ for the integers $k=0$ through $k=24$. Sketch the path of the particle from $t=0$ through $t=2\pi$.

   2. Show that every position the particle occupies is on the curve with equation $y^2=4x^2-4x^4$.

4. Circle 1 is centred at the origin, Circle 2 is centred at $(2,0)$, and both circles have radius $1$. The circles are tangent at $(1,0)$. Circle 2 is "rolled" in the counterclockwise direction along the outside of the circumference of Circle 1 without slipping. The point on Circle 2 that was originally at $(1,0)$ (the point of tangency) follows a curve in the plane. Find functions $x(t)$ and $y(t)$ so that the points on this curve are $(x(t),y(t))$ for $0\le t\le 2\pi$.

   

5. The setup in this problem is similar to (d). Circle 1 is centred at the origin and has radius $2$ and Circle 2 is centred at $(1,0)$ and has radius $1$ so that the two circles are tangent at $(2,0)$. Circle 2 is rolled around the inside of the circumference of Circle 1 in the counterclockwise direction. Describe the curve in the plane followed by the point on Circle 2 that is initially at $(2,0)$.

   

6. Circle 1 is centred at the origin and has radius $1$. Circle 2 has radius $r<1$, is inside Circle 1, and the two circles are initially tangent at $(1,0)$. When Circle 2 is rolled around the inside of Circle 1 in the counterclockwise direction, the point on Circle 2 that was initially at $(1,0)$ will follow some curve in the plane.

   1. Show that when $r={\displaystyle \frac{1}{4}}$, the points on the curve satisfy the equation $(\sqrt[3]{x}{)}^2+(\sqrt[3]{y}{)}^2=1$.

   2. Show that the curves for $r={\displaystyle \frac{1}{3}}$ and $r={\displaystyle \frac{2}{3}}$ are exactly the same and that the point initially at $(1,0)$ travels this curve in opposite directions for the two radii.