# Problem of the MonthProblem 7: April 2024

Curves in the plane are often given as the set of points $$(x,y)$$ that satisfy some equation in $$x$$ andÂ $$y$$. 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 particleâ€™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 particleâ€™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 particleâ€™s position at time $$t$$ is $$(\cos(t),\sin(2t))$$.

1. Plot the position of the particle at $$t=\dfrac{k\pi}{12}$$ 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 $$0\leq t\leq 2\pi$$.

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 $$r=\dfrac{1}{4}$$, the points on the curve satisfy the equation $$\big(\sqrt[3]{x}\big)^2 + \big(\sqrt[3]{y}\big)^2=1$$.

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