Chapter 10 · Parametric Equations and Polar Coordinates
10.1 Parametric Equations
Describe a curve by where a moving point is at each instant — $x=f(t)$, $y=g(t)$ — and capture loops, cusps and motion that no graph $y=f(x)$ can.
A graph $y=f(x)$ tells us where a curve is, but nothing about how it was drawn. Yet many curves in science and engineering are drawn by something moving: the tip of a robot arm, a satellite, a drone following a flight plan, a ball in the air. For these it is natural to give the position at each instant $t$ — the horizontal coordinate $x(t)$ and the vertical coordinate $y(t)$ — as two separate functions of the same variable. The pair of functions records the path and the timing, and it has no trouble with paths that loop, double back, or go straight up.
You first met such curves in Calculus I, Section 2.5, where we used them only to compute slopes. In this section we study the curves themselves: how to plot them, which way they run, how to turn them into ordinary Cartesian equations (and when that is a bad idea), and how to build parametric equations for real motions — a rolling wheel, a kicked football, and the beautiful Lissajous figures of an oscilloscope.
Curves Defined by Parametric Equations
📘 Definition 10.1 — Parametric equations and parametric curves
Let $f$ and $g$ be continuous functions on an interval $I$. The equations
$$x=f(t),\qquad y=g(t),\qquad t\in I$$
are called parametric equations, and $t$ is the parameter. As $t$ runs through $I$, the point $\big(f(t),g(t)\big)$ traces out a set of points in the plane called a parametric curve (or plane curve).
If $I=[a,b]$, the point $\big(f(a),g(a)\big)$ is the initial point and $\big(f(b),g(b)\big)$ is the terminal point. The direction in which the curve is traced as $t$ increases is its orientation; we show it with arrows drawn on the curve.
The parameter is often time, but it need not be: it may be an angle, a distance, or just a convenient label. Notice also that every ordinary graph is a parametric curve — the graph of $y=f(x)$ is traced by $x=t$, $y=f(t)$ — so parametric curves include everything we have drawn so far, and much more.
✏️ Example 1 — Plotting a parametric curve and its orientation
Sketch the curve $x=t^2-4$, $y=2t$, $-2\le t\le3$. Mark the initial and terminal points and the orientation, then find a Cartesian equation for the curve.
💡 Show solution▶
Table Choose some values of $t$ and compute the points:
$t$
$-2$
$-1$
$0$
$1$
$2$
$3$
$x=t^2-4$
$0$
$-3$
$-4$
$-3$
$0$
$5$
$y=2t$
$-4$
$-2$
$0$
$2$
$4$
$6$
Plot Join the points in order of increasing $t$. The curve starts at the initial point $(0,-4)$, moves left to $(-4,0)$, then turns and moves right and up to the terminal point $(5,6)$. Because $y=2t$ always increases, the orientation is upward along the whole curve.
Eliminate $t$ From $y=2t$ we get $t=\dfrac y2$. Substituting into the first equation, $$x=\left(\frac y2\right)^2-4=\frac{y^2}{4}-4.$$ This is a parabola opening to the right with vertex $(-4,0)$; our curve is the piece with $-4\le y\le6$.
Check At $t=3$: $\dfrac{6^2}{4}-4=5=x$. ✓ (Select this curve in Interactive 1 to watch it being traced.)
Eliminating the Parameter
Converting a pair of parametric equations into one equation in $x$ and $y$ is called eliminating the parameter. It helps us recognise the shape — a line, a circle, a parabola — but, as we shall see, some information is lost on the way.
🧭 Problem-solving strategy — Eliminating the parameter
Solve and substitute. If one equation is easy to solve for $t$ (for example $x=2t+1$ or $y=\sqrt t$), solve it and substitute into the other.
Use an identity. If $t$ appears inside $\sin$ and $\cos$, isolate $\cos t$ and $\sin t$ and use $\cos^2t+\sin^2t=1$ (or $\sec^2t-\tan^2t=1$, or a double-angle formula).
Restrict. Find the possible values of $x$ and $y$ from the parametric equations (for example $x=\sqrt t\ge0$, $x=\cos t\in[-1,1]$, $x=e^t\gt0$) and state them with the Cartesian equation.
Check one or two points from the original equations.
✏️ Example 2 — Eliminating the parameter
Find a Cartesian equation for each curve and describe it, including any restriction on $x$ or $y$.
a $x=2t+1$, $y=3-t$, $t\in\mathbb R$
b $x=3\cos t$, $y=2\sin t$, $0\le t\le2\pi$
c $x=\sqrt t$, $y=2t+3$, $t\ge0$
💡 Show solution▶
(a) Solve the first equation: $t=\dfrac{x-1}{2}$. Then $$y=3-\frac{x-1}{2}=\frac72-\frac x2.$$ A straight line with slope $-\tfrac12$. As $t$ increases, $x$ increases, so the line is traced from left to right.
(b) Isolate the trigonometric functions: $\cos t=\dfrac x3$, $\sin t=\dfrac y2$. Then $\cos^2t+\sin^2t=1$ gives $$\frac{x^2}{9}+\frac{y^2}{4}=1.$$ An ellipse centred at the origin with semi-axes $3$ (horizontal) and $2$ (vertical). It starts at $(3,0)$, passes $(0,2)$ at $t=\tfrac\pi2$, so it is traced once counter-clockwise.
(c) From $x=\sqrt t$ we get $t=x^2$, so $y=2x^2+3$. But $x=\sqrt t$ is never negative, so the curve is only the right half of this parabola: $$y=2x^2+3,\qquad x\ge0.$$ It starts at $(0,3)$ and moves up and to the right.
Check (b) at $t=\tfrac\pi4$: $x=\tfrac{3}{\sqrt2}$, $y=\sqrt2$, and $\tfrac{9/2}{9}+\tfrac{2}{4}=\tfrac12+\tfrac12=1$. ✓
⚠️ What eliminating the parameter throws away
1. Only part of the curve. The equations $x=\sin t$, $y=\sin^2t$ give $y=x^2$ — but since $-1\le\sin t\le1$, only the arc with $-1\le x\le1$ is ever reached. Always state the restriction.
2. Orientation and repetition. The same equations trace that arc back and forth forever: right from $(0,0)$ to $(1,1)$, back to $(-1,1)$, and so on. The equation $y=x^2$ says nothing about this.
3. Timing. $x=t$, $y=t^2$ and $x=t^3$, $y=t^6$ both give the whole parabola $y=x^2$, traced left to right — but at very different speeds. A Cartesian equation describes the path; parametric equations describe the motion.
Interactive 1 — The Parametric Tracer
Pick a curve and let $t$ run (or drag the slider). The dashed path is the whole curve; the solid part is what has been traced so far, and the red arrow shows the direction of motion. The last readout substitutes the current point into the Cartesian equation — it stays at $0$ all the way along, which is exactly what “eliminating the parameter” promises.
$t$—
$(x,y)$—
Cartesian equation—
left side − right side—
One Curve, Many Parametrisations
A curve does not have just one set of parametric equations. Any choice of functions that visits the same points gives a parametrisation of the curve, and different choices can trace it in different directions, at different speeds, or a different number of times. For example, each of the following traces the unit circle $x^2+y^2=1$:
$x=\cos t,\ y=\sin t,\ 0\le t\le2\pi$once, counter-clockwise, starting at $(1,0)$
$x=\cos t,\ y=-\sin t,\ 0\le t\le2\pi$once, clockwise, starting at $(1,0)$
$x=\sin t,\ y=\cos t,\ 0\le t\le2\pi$once, clockwise, starting at the top $(0,1)$
$x=\cos2t,\ y=\sin2t,\ 0\le t\le2\pi$twice, counter-clockwise, at double speed
The standard building blocks are worth memorising. The line segment from $P(x_0,y_0)$ to $Q(x_1,y_1)$ is $$x=x_0+(x_1-x_0)t,\quad y=y_0+(y_1-y_0)t,\qquad 0\le t\le1,$$ which is at $P$ when $t=0$ and at $Q$ when $t=1$. The circle with centre $(h,k)$ and radius $r$ is $x=h+r\cos t$, $y=k+r\sin t$, and the ellipse $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ is $x=h+a\cos t$, $y=k+b\sin t$ (both counter-clockwise for $0\le t\le2\pi$).
✏️ Example 3 — Finding parametric equations
Find parametric equations for each curve, with a parameter interval.
a The line segment from $P(1,-2)$ to $Q(4,3)$.
b The circle with centre $(2,-1)$ and radius $3$, traced once clockwise starting from its highest point.
c The same segment as in (a), but traced from $Q$ to $P$.
💡 Show solution▶
(a) Here $x_1-x_0=4-1=3$ and $y_1-y_0=3-(-2)=5$, so $$x=1+3t,\quad y=-2+5t,\qquad 0\le t\le1.$$ Check: $t=0$ gives $(1,-2)$ and $t=1$ gives $(4,3)$. ✓
(b) The highest point is $(2,-1+3)=(2,2)$. Swapping the roles of sine and cosine starts us at the top and runs clockwise: $$x=2+3\sin t,\quad y=-1+3\cos t,\qquad 0\le t\le2\pi.$$ Check: $t=0$ gives $(2,2)$; $t=\tfrac\pi2$ gives $(5,-1)$, the rightmost point — moving from the top to the right is clockwise. ✓ And $(x-2)^2+(y+1)^2=9\sin^2t+9\cos^2t=9$. ✓
(c) Start at $Q$ and move towards $P$: $$x=4-3t,\quad y=3-5t,\qquad 0\le t\le1.$$ (Equivalently, replace $t$ by $1-t$ in part (a).)
The Cycloid
Paint a dot on the edge of a bicycle tyre and roll the bicycle forward. The dot does not move in a circle — the wheel is travelling — and it does not move in a straight line. Its path, a series of arches touching the road, is called a cycloid. No single equation $y=f(x)$ describes it conveniently, but parametric equations come almost for free if we use the angle the wheel has turned as the parameter.
The wheel has turned through angle θ. Because it rolls without slipping, the green arc on the rim and the green stretch of road have the same length rθ.
📐 Theorem 10.1 — Parametric equations of a cycloid
A circle of radius $r\gt0$ rolls without slipping along the $x$-axis, and $P$ is a fixed point on the circle that starts at the origin. If $\theta$ is the angle (in radians) through which the circle has turned, then $P$ is at
$$x=r(\theta-\sin\theta),\qquad y=r(1-\cos\theta),\qquad \theta\in\mathbb R.$$
One arch corresponds to $0\le\theta\le2\pi$; it has width $2\pi r$ and height $2r$.
💡 Show the derivation▶
Centre Rolling without slipping means that the length of road covered equals the length of rim that has touched it: $|OT|=\text{arc }PT=r\theta$, where $T$ is the current contact point. So the centre is at $C=(r\theta,\,r)$.
Point relative to centre At $\theta=0$, $P$ is directly below $C$. The wheel rolls to the right, so it turns clockwise, and after turning through $\theta$ the radius $CP$ has swung back by $\theta$ from the downward vertical. Its horizontal component is $-r\sin\theta$ (to the left) and its vertical component is $-r\cos\theta$ (downward when $\theta$ is small).
Check $\theta=\pi$ (half a turn): $P=(\pi r,2r)$, the top of the wheel. $\theta=2\pi$: $P=(2\pi r,0)$, back on the road one circumference along. ✓
✏️ Example 4 — Following a point on a rolling wheel
A wheel of radius $2$ m rolls along level ground. A point $P$ on its rim starts on the ground at the origin.
a Write parametric equations for the path of $P$.
b Find the position of $P$ after the wheel has turned through $\tfrac\pi2$, $\pi$ and $2\pi$.
c How far has the wheel's centre travelled when $P$ first returns to the ground, and how high does $P$ rise in between?
💡 Show solution▶
(a) By Theorem 10.1 with $r=2$: $\;x=2(\theta-\sin\theta)$, $y=2(1-\cos\theta)$.
(b) $$\begin{aligned}\theta=\tfrac\pi2:&\quad \big(2(\tfrac\pi2-1),\,2(1-0)\big)=(\pi-2,\,2)\approx(1.142,\,2)\ \text{m},\\ \theta=\pi:&\quad \big(2\pi,\,4\big)\approx(6.283,\,4)\ \text{m},\\ \theta=2\pi:&\quad \big(4\pi,\,0\big)\approx(12.566,\,0)\ \text{m}.\end{aligned}$$ Notice that after a quarter turn the centre has moved $\pi\approx3.14$ m but $P$ has moved only $1.14$ m to the right: near the ground the point lags behind and mostly rises.
(c) $y=0$ again when $\cos\theta=1$, first at $\theta=2\pi$. The centre is at $x=r\theta$, so it has travelled $2\cdot2\pi=4\pi\approx12.57$ m (one circumference). The greatest height is $y=2(1-\cos\pi)=4$ m — the diameter.
Interactive 2 — Roll the Wheel, Draw the Cycloid
A wheel of radius $r=1$ rolls to the right. The green stretch of road always has the same length as the green arc of rim that has touched it — that is what makes the centre sit at $(\theta,1)$. Choose a point on the rim (a cycloid), inside the wheel like a reflector on a spoke (a curtate cycloid), or outside it like the flange of a train wheel (a prolate cycloid, which loops backwards).
angle turned $\theta$—
centre $(\theta,1)$—
point $P(x,y)$—
speed of $P$ (per radian)—
$P$ is moving…—
Modelling Motion: Projectiles and Lissajous Figures
Parametric equations are the natural language of motion. If an object is launched from height $h$ with speed $v_0$ at angle $\alpha$ above the horizontal, its initial velocity has horizontal part $v_0\cos\alpha$ and vertical part $v_0\sin\alpha$. Ignoring air resistance, nothing changes the horizontal velocity, while gravity reduces the vertical velocity at the constant rate $g$. Integrating each velocity once (as in Calculus I) gives the position.
📐 Theorem 10.2 — The path of a projectile
Suppose a projectile is launched at time $t=0$ from the point $(0,h)$ with speed $v_0$ at angle $\alpha$ above the horizontal, that the only force on it is gravity with constant acceleration $g$ downward, and that air resistance is negligible. Then, while it is in flight,
$$x=(v_0\cos\alpha)\,t,\qquad y=h+(v_0\sin\alpha)\,t-\tfrac12gt^2.$$
If $\cos\alpha\neq0$, eliminating $t$ shows that the path is a parabola: $\;y=h+x\tan\alpha-\dfrac{g\,x^2}{2v_0^2\cos^2\alpha}$.
✏️ Example 5 — A free kick
A football is kicked from the ground with speed $20$ m/s at $30^\circ$ above the horizontal. Take $g=9.8$ m/s² and ignore air resistance.
a Write parametric equations for the ball's position.
b When and where does it land?
c What is its greatest height?
d The crossbar of a goal is $2.44$ m high. If the goal line is $30$ m away, does the ball pass over the bar?
💡 Show solution▶
(a) $v_0\cos30^\circ=20\cdot\tfrac{\sqrt3}{2}=10\sqrt3$ and $v_0\sin30^\circ=20\cdot\tfrac12=10$, so $$x=10\sqrt3\,t,\qquad y=10t-4.9t^2\qquad(x,y\text{ in m},\ t\text{ in s}).$$
(b) $y=0$ when $t(10-4.9t)=0$, so $t=0$ (the kick) or $t=\dfrac{10}{4.9}=\dfrac{100}{49}\approx2.04$ s. Then $x=10\sqrt3\cdot\dfrac{100}{49}=\dfrac{1000\sqrt3}{49}\approx35.35$ m.
(c) $y$ is greatest when $\dfrac{dy}{dt}=10-9.8t=0$, i.e. $t=\dfrac{50}{49}\approx1.02$ s (half the flight time). Then $y=10\cdot\dfrac{50}{49}-4.9\left(\dfrac{50}{49}\right)^2=\dfrac{500}{49}-\dfrac{250}{49}=\dfrac{250}{49}\approx5.10$ m.
(d) The ball reaches $x=30$ when $t=\dfrac{30}{10\sqrt3}=\sqrt3\approx1.732$ s. Then $y=10\sqrt3-4.9\cdot3=17.32-14.70\approx2.62$ m. Since $2.62\gt2.44$, the ball clears the bar by about $18$ cm. (Using the Cartesian path $y=\dfrac{x}{\sqrt3}-\dfrac{4.9x^2}{300}$ gives the same height — but the parametric form also tells us when.)
When an object vibrates in two directions at once — a pendulum swinging in an elliptical path, a speaker cone, or the electron beam of an oscilloscope driven by two signals — its position is often $x=A\sin(at)$, $y=B\sin(bt+\delta)$. The resulting curves are called Lissajous figures. If the ratio $a:b$ is rational the curve closes up; engineers once read frequency ratios directly from their shapes.
✏️ Example 6 — A Lissajous figure-eight
Consider $x=\sin t$, $y=\sin2t$, $0\le t\le2\pi$. Find a Cartesian equation, find where the curve crosses itself, and find its highest points.
💡 Show solution▶
Identity $y=\sin2t=2\sin t\cos t=2x\cos t$. We cannot replace $\cos t$ by $\sqrt{1-x^2}$ because $\cos t$ can be negative, so square instead: $$y^2=4x^2\cos^2t=4x^2\left(1-x^2\right),\qquad -1\le x\le1.$$
Crossing The curve passes through $(0,0)$ when $\sin t=0$, i.e. at $t=0$, $t=\pi$ (and $t=2\pi$, which is the same as $t=0$). It goes through the origin twice in different directions — a figure-eight crossing itself at the origin.
Highest points $y=\sin2t$ is largest ($y=1$) when $2t=\tfrac\pi2$ or $\tfrac{5\pi}{2}$, i.e. $t=\tfrac\pi4$ or $\tfrac{5\pi}4$. These give $x=\sin\tfrac\pi4=\tfrac{\sqrt2}{2}$ and $x=-\tfrac{\sqrt2}{2}$: the points $\left(\pm\tfrac{\sqrt2}2,1\right)$.
Check At $\left(\tfrac{\sqrt2}{2},1\right)$: $4\cdot\tfrac12\cdot\tfrac12=1=y^2$. ✓ (Choose this curve in Interactive 1.)
Exercises
I
Sketch each curve (use a table of values), mark its initial point and orientation, then eliminate the parameter to find a Cartesian equation. State any restriction on $x$ or $y$.
Check: the traced curve (blue) must lie on your Cartesian curve (dashed red) — and may cover only part of it
$t$—
$(x,y)$—
left − right of answer—
💡 Show solutions▶
1 $t=\dfrac{x+1}{2}$, so $y=\dfrac{x+1}{2}+3=\dfrac x2+\dfrac72$. The segment from $(-5,1)$ to $(3,5)$, traced left to right; $-5\le x\le3$.
2 $t=x-1$, so $y=(x-1)^2$ with $-1\le x\le3$. An arc of the parabola with vertex $(1,0)$, from $(-1,4)$ down to the vertex and up to $(3,4)$; traced left to right.
3 $t=y$, so $x=3-y^2$: a parabola opening to the left with vertex $(3,0)$, the whole parabola, traced upward (bottom to top) since $y=t$ increases.
4 $x^2+y^2=25\cos^2t+25\sin^2t=25$. Since $y=5\sin t\ge0$ for $0\le t\le\pi$, this is the upper semicircle $y\ge0$, from $(5,0)$ counter-clockwise to $(-5,0)$.
5 $\cos t=\tfrac x2$, $\sin t=\tfrac y3$, so $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$. An ellipse (taller than wide) traced once counter-clockwise from $(2,0)$.
6 $\cos t=\dfrac{x-1}{2}$, $\sin t=\dfrac{y+2}{2}$, so $(x-1)^2+(y+2)^2=4$: the circle with centre $(1,-2)$ and radius $2$, once counter-clockwise from $(3,-2)$.
7 $t=x^2$ (because $x=\sqrt t\ge0$), so $y=1-x^2$ with $x\ge0$: the right half of a downward parabola, starting at $(0,1)$ and moving right and down.
8 $xy=e^te^{-t}=1$, so $y=\dfrac1x$ with $x\gt0$ (because $e^t\gt0$): only the branch in the first quadrant, traced left to right (as $t\to-\infty$ the point is high on the $y$-axis side; as $t\to\infty$ it runs out along the $x$-axis).
9 $y=\cos2t=2\cos^2t-1=2x^2-1$ with $-1\le x\le1$. The arc of the parabola from $(1,1)$ to $(-1,1)$, traced from right to left for $0\le t\le\pi$ and then back again for $\pi\le t\le2\pi$.
10 $\sec^2t-\tan^2t=1$ gives $x^2-y^2=1$, a hyperbola. On $-\tfrac\pi2\lt t\lt\tfrac\pi2$, $\sec t\ge1$, so only the right branch $x\ge1$ is traced, from bottom to top ($y=\tan t$ increases).
II
Finding and comparing parametrisations.
1 The line segment from $(-1,2)$ to $(5,-1)$.
2 The whole line through $(2,0)$ with slope $3$.
3 The circle $x^2+y^2=9$, traced once clockwise starting at $(3,0)$.
4 The circle with centre $(-2,1)$ and radius $4$, traced once counter-clockwise starting at $(2,1)$.
5 The ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, traced once counter-clockwise.
6 Give two different parametrisations of the curve $y=\ln x$, $x\gt0$, both traced with $x$ increasing.
7 Show that $x=t$, $y=t^2$ ($t\in\mathbb R$) and $x=t^3$, $y=t^6$ ($t\in\mathbb R$) trace the same curve, but $x=t^2$, $y=t^4$ ($t\in\mathbb R$) does not.
8 Parametrise the parabola $y=x^2$ so that it is traced from right to left.
9 Suppose $x=f(t)$, $y=g(t)$, $a\le t\le b$ traces a curve $C$. Show that $x=f(a+b-s)$, $y=g(a+b-s)$, $a\le s\le b$ traces the same curve in the opposite direction.
5 $x=4\cos t$, $y=3\sin t$, $0\le t\le2\pi$; then $\tfrac{x^2}{16}+\tfrac{y^2}{9}=\cos^2t+\sin^2t=1$. ✓
6 For example $x=t$, $y=\ln t$, $t\gt0$; and $x=e^t$, $y=t$, $t\in\mathbb R$ (since $\ln e^t=t$). Both have $x$ increasing with $t$. (Many others work, e.g. $x=t^2$, $y=2\ln t$, $t\gt0$.)
7 In the first two, $y=x^2$ and $x$ takes every real value (since $t\mapsto t$ and $t\mapsto t^3$ are onto $\mathbb R$), so both give the whole parabola, left to right. For $x=t^2$, $y=t^4$ we still have $y=x^2$, but $x=t^2\ge0$: only the right half $x\ge0$, traced inwards to the origin ($t\lt0$) and then out again ($t\gt0$).
8 $x=-t$, $y=t^2$, $t\in\mathbb R$: as $t$ increases, $x$ decreases, and $y=(-t)^2=x^2$. ✓
9 As $s$ runs from $a$ to $b$, the number $u=a+b-s$ runs from $b$ down to $a$, taking every value in $[a,b]$. So the points $\big(f(u),g(u)\big)$ are exactly the points of $C$, visited in reverse order: the initial point is $\big(f(b),g(b)\big)$ and the terminal point is $\big(f(a),g(a)\big)$.
III
Cycloids and other rolling curves.
1 A wheel of radius $3$ rolls along the $x$-axis. Write the equations of the cycloid traced by a rim point starting at the origin, and find its position when $\theta=\tfrac\pi2,\ \pi,\ \tfrac{3\pi}2,\ 2\pi$.
2 For the cycloid $x=r(\theta-\sin\theta)$, $y=r(1-\cos\theta)$, find all values of $\theta$ at which $P$ is highest, and the coordinates of those points.
3Applied. A bicycle wheel of radius $0.35$ m rolls $10$ m along a flat road, and the tyre valve starts on the road. Through what angle has the wheel turned, and how high above the road is the valve now?
4 A point $P$ is fixed on a spoke at distance $b$ from the centre of a wheel of radius $r$, and starts directly below the centre. Show that as the wheel rolls, $x=r\theta-b\sin\theta$, $y=r-b\cos\theta$.
5Train wheels. For $r=1$ and $b=1.5$ in part 4 (a point on the flange, below the rail), find the values of $\theta$ in $[0,2\pi]$ for which $\dfrac{dx}{d\theta}\lt0$. What does this mean physically?
6 The astroid $x=4\cos^3t$, $y=4\sin^3t$ is traced by a point on a small circle rolling inside a larger one. Eliminate the parameter.
7 The involute of a circle, $x=\cos t+t\sin t$, $y=\sin t-t\cos t$, $t\ge0$, is the path of the end of a string unwound from a reel of radius $1$ (it is the shape of gear teeth). Show that the distance from the origin to the point at parameter $t$ is $\sqrt{1+t^2}$.
2 $y=r(1-\cos\theta)$ is largest, $y=2r$, when $\cos\theta=-1$: $\theta=(2k+1)\pi$, $k\in\mathbb Z$. Then $\sin\theta=0$, so the points are $\big((2k+1)\pi r,\ 2r\big)$ — the midpoints of the arches.
3 Rolling $10$ m means $r\theta=10$, so $\theta=\dfrac{10}{0.35}\approx28.571$ rad (about $4.55$ turns). Height $y=0.35\big(1-\cos28.571\big)\approx0.35(1+0.9562)\approx0.685$ m.
4 As in Theorem 10.1 the centre is at $(r\theta,r)$, because the rim rolls without slipping. The radius vector to $P$ has length $b$ and has turned clockwise by $\theta$ from straight down, so $\overrightarrow{CP}=(-b\sin\theta,-b\cos\theta)$. Adding: $x=r\theta-b\sin\theta$, $y=r-b\cos\theta$.
5 $\dfrac{dx}{d\theta}=1-1.5\cos\theta\lt0\iff\cos\theta\gt\tfrac23$, i.e. $0\le\theta\lt\cos^{-1}\tfrac23\approx0.841$ or $2\pi-0.841\lt\theta\le2\pi$ (about $5.442$). Near the bottom of its circle the flange point is moving backwards even though the train moves forwards — the path has little loops (see Interactive 2).
6 $\cos t=\left(\tfrac x4\right)^{1/3}$, $\sin t=\left(\tfrac y4\right)^{1/3}$, so $\left(\tfrac x4\right)^{2/3}+\left(\tfrac y4\right)^{2/3}=1$, i.e. $x^{2/3}+y^{2/3}=4^{2/3}$ $\left(\approx2.52\right)$.
7 $$\begin{aligned}x^2+y^2&=\left(\cos^2t+2t\sin t\cos t+t^2\sin^2t\right)+\left(\sin^2t-2t\sin t\cos t+t^2\cos^2t\right)\\&=\left(\cos^2t+\sin^2t\right)+t^2\left(\sin^2t+\cos^2t\right)=1+t^2,\end{aligned}$$ so the distance is $\sqrt{1+t^2}$. (The unwound string has length $t$ and is tangent to the reel of radius $1$: Pythagoras again.)
IV
Motion and modelling. Use $g=9.8$ m/s² and ignore air resistance.
1 A stone is thrown horizontally at $15$ m/s from the top of a $20$ m cliff. Write parametric equations, find when and how far from the foot of the cliff it lands, and find the Cartesian equation of its path.
2 A ball is kicked from the ground at $25$ m/s at $40^\circ$ above the horizontal. Find its time of flight, its range and its maximum height.
3 Show that the path in Theorem 10.2 with $h=0$ is $y=x\tan\alpha-\dfrac{gx^2}{2v_0^2\cos^2\alpha}$.
4 Show that, launched from ground level, the projectile lands at distance $R=\dfrac{v_0^2\sin2\alpha}{g}$, and that $R$ is greatest when $\alpha=45^\circ$.
5 Eliminate the parameter for the Lissajous curve $x=\sin t$, $y=\sin3t$. (Hint: $\sin3t=3\sin t-4\sin^3t$.)
6 For the Lissajous curve $x=\sin2t$, $y=\sin3t$, $0\le t\lt2\pi$, find the values of $t$ at which the curve passes through the origin, and explain why the curve closes up at $t=2\pi$.
7Applied. Two ships are tracked on a radar grid (units km, $t$ in hours). Ship A is at $x=3t$, $y=4t$; ship B is at $x=20-2t$, $y=5+t$. Do their paths cross? Do the ships collide?
💡 Show solutions▶
1 With the foot of the cliff at the origin: $x=15t$, $y=20-4.9t^2$. It lands when $4.9t^2=20$, $t=\sqrt{20/4.9}\approx2.02$ s, at $x=15\cdot2.0203\approx30.3$ m. Cartesian: $t=\tfrac{x}{15}$, so $y=20-\dfrac{4.9x^2}{225}\approx20-0.0218x^2$, $0\le x\le30.3$.
2 $x=25\cos40^\circ\,t\approx19.15t$, $y=25\sin40^\circ\,t-4.9t^2\approx16.07t-4.9t^2$. Flight time $T=\dfrac{2\cdot25\sin40^\circ}{9.8}\approx3.28$ s; range $\approx19.151\cdot3.2795\approx62.8$ m; maximum height $\dfrac{(25\sin40^\circ)^2}{2\cdot9.8}\approx13.2$ m (at $t\approx1.64$ s).
3 From $x=(v_0\cos\alpha)t$, $t=\dfrac{x}{v_0\cos\alpha}$. Substitute into $y=(v_0\sin\alpha)t-\tfrac12gt^2$: $$y=v_0\sin\alpha\cdot\frac{x}{v_0\cos\alpha}-\frac g2\cdot\frac{x^2}{v_0^2\cos^2\alpha}=x\tan\alpha-\frac{gx^2}{2v_0^2\cos^2\alpha}.$$
4 $y=0$ for $t\gt0$ when $t=\dfrac{2v_0\sin\alpha}{g}$. Then $R=v_0\cos\alpha\cdot\dfrac{2v_0\sin\alpha}{g}=\dfrac{v_0^2\cdot2\sin\alpha\cos\alpha}{g}=\dfrac{v_0^2\sin2\alpha}{g}$. For $0\lt\alpha\lt90^\circ$, $\sin2\alpha\le1$ with equality only at $2\alpha=90^\circ$, so $R$ is greatest at $\alpha=45^\circ$, where $R=\dfrac{v_0^2}{g}$.
5 $y=3\sin t-4\sin^3t=3x-4x^3$, with $-1\le x\le1$ — a cubic arc, traced back and forth.
6 We need $\sin2t=0$ and $\sin3t=0$: $t$ must be a multiple of $\tfrac\pi2$ and of $\tfrac\pi3$, hence a multiple of $\pi$. In $[0,2\pi)$: $t=0$ and $t=\pi$. The curve closes up because $\sin2t$ has period $\pi$ and $\sin3t$ has period $\tfrac{2\pi}{3}$; both divide $2\pi$, so $x(t+2\pi)=x(t)$ and $y(t+2\pi)=y(t)$.
7 Paths: A has $y=\tfrac43x$; for B, $t=\dfrac{20-x}{2}$ so $y=5+\dfrac{20-x}{2}=15-\dfrac x2$. They cross where $\tfrac43x=15-\tfrac x2$, i.e. $\tfrac{11}{6}x=15$, $x=\tfrac{90}{11}\approx8.18$ km, $y=\tfrac{120}{11}\approx10.91$ km. A is there at $t=\tfrac{x}{3}=\tfrac{30}{11}\approx2.73$ h, but B at $t=\tfrac{20-x}{2}=\tfrac{65}{11}\approx5.91$ h. The paths cross but the ships are there at different times — no collision. (A Cartesian equation could never tell you this.)
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Is $45^\circ$ really the best launch angle for a shot-putter, who releases the shot about $2$ m above the ground?
Why it matters. Sports scientists in athletics, javelin, basketball and football use projectile models to coach technique; a few degrees can be worth a metre at elite level.
What to do.
Sample case: release speed $v_0=13$ m/s from height $h=2.1$ m, $g=9.81$ m/s². Use Theorem 10.2 to write $x(t)$ and $y(t)$.
Solve $y(t)=0$ for the landing time and show the throw distance is $R(\alpha)=\dfrac{v_0\cos\alpha}{g}\left(v_0\sin\alpha+\sqrt{v_0^2\sin^2\alpha+2gh}\right)$.
Tabulate $R(\alpha)$ for $\alpha=30^\circ,32^\circ,\dots,50^\circ$ in a spreadsheet (or Interactive 1-style plot) and find the best angle to $0.1^\circ$.
Compare with $45^\circ$: how many centimetres does the better angle gain?
Tools: Spreadsheet or calculator; optionally a graphing tool to draw the paths.
Hand in: A half-page note with your table, the optimal angle, the gain over $45^\circ$, and one reason why real athletes release at even lower angles (around $35^\circ$–$38^\circ$).
Going further. Repeat for a basketball free throw (release $2.1$ m, hoop $3.05$ m high, $4.2$ m away): which angles and speeds make the ball pass through the hoop?
Instructor note: the optimum is about $41.9^\circ$ (from $\sin\alpha=v_0/\sqrt{2(v_0^2+gh)}$), giving $R\approx19.21$ m versus about $19.12$ m at $45^\circ$ — a gain of roughly $9$ cm. Real athletes can generate more release speed at lower angles, which shifts their personal optimum down.
🧪 Self-check quiz
0/5 correct
Q1. The curve $x=3\cos t$, $y=3\sin t$, $0\le t\le\pi$ is
Q2. Eliminating the parameter from $x=t+1$, $y=t^2$ gives
Q3. The parametric curve $x=\sin t$, $y=\sin^2t$, $t\in\mathbb R$, is
Q4. A point on the rim of a rolling wheel of radius $r$ starts on the ground. After the wheel has turned through $\pi$ radians, the point is at
Q5. A ball moves on $x=20t$, $y=15t-4.9t^2$ (metres, seconds). It returns to the ground at about