Exercise 0.1: Compute the slope of a parametric curve: ten direct calculations

For each parametric curve, compute what the line asks for: the slope $\frac{dy}{dx}$ at the given parameter or point, the parameter of a horizontal or vertical tangent line, or the second derivative $\frac{d^{2}y}{dx^{2}}$ .

A single method: differentiate $x$ and $y$ separately with respect to $t$ , then form $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$ . The tangent line is horizontal when $y'(t)=0$ and $x'(t)\neq 0$ , vertical when $x'(t)=0$ and $y'(t)\neq 0$ . For the second derivative, differentiate the slope with respect to $t$ , then divide AGAIN by $x'(t)$ : $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)}$ .

a) $x=t^{2}+1$ , $y=t^{3}-2t$ : $\frac{dy}{dx}$ at $t=2$ .
b) $x=\sqrt{t}$ , $y=t-1$ : $\frac{dy}{dx}$ at $t=4$ .
c) $x=\ln t$ , $y=t^{2}$ : $\frac{dy}{dx}$ at $t=3$ .
d) $x=1+2\cos t$ , $y=3\sin t$ : $\frac{dy}{dx}$ at $t=\frac{\pi}{3}$ .
e) $x=e^{t}$ , $y=te^{-t}$ : $\frac{dy}{dx}$ at $t=-1$ .
f) $x=t^{2}-1$ , $y=t^{3}+t$ : $\frac{dy}{dx}$ at the point $(3, 10)$ .
g) $x=t^{3}-3t$ , $y=t^{2}-4t$ : the parameter $t$ where the tangent line is horizontal, then the $y$ -coordinate of the point.
h) Same curve as in g): the number of points with a vertical tangent line, then the $x$ -coordinate of the one whose parameter is positive.
i) $x=t^{2}+t$ , $y=t^{3}$ : $\frac{d^{2}y}{dx^{2}}$ at $t=1$ .
j) $x=2\sin t$ , $y=\cos(2t)$ : $\frac{d^{2}y}{dx^{2}}$ at $t=\frac{\pi}{6}$ .

Type your answers, the page tells you right or wrong 0/12

a)

\frac{dy}{dx}

at

t=2

b)

\frac{dy}{dx}

at

t=4

c)

\frac{dy}{dx}

at

t=3

d)

\frac{dy}{dx}

at

t=\frac{\pi}{3}

e)

\frac{dy}{dx}

at

t=-1

f)

\frac{dy}{dx}

at the point

(3, 10)

g)

Parameter

t

y

-coordinate of the point

h)

Number of points

x

-coordinate for

t>0

i)

\frac{d^{2}y}{dx^{2}}

at

t=1

j)

\frac{d^{2}y}{dx^{2}}

at

t=\frac{\pi}{6}

Show the solution

Answers

a) $\frac{dy}{dx}=\frac{3t^{2}-2}{2t}$ ; $\frac{5}{2}$
b) $\frac{dy}{dx}=2\sqrt{t}$ ; $4$
c) $\frac{dy}{dx}=2t^{2}$ ; $18$
d) $\frac{dy}{dx}=-\frac{3}{2}\cot t$ ; $-\frac{\sqrt{3}}{2}\approx-0.866$
e) $\frac{dy}{dx}=(1-t)e^{-2t}$ ; $2e^{2}\approx 14.778$
f) $t=2$ , $\frac{dy}{dx}=\frac{3t^{2}+1}{2t}$ ; $\frac{13}{4}$
g) $t=2$ ; point $(2, -4)$
h) $t=\pm 1$ , two points; $(-2, -3)$ for $t=1$
i) $\frac{d^{2}y}{dx^{2}}=\frac{6t(t+1)}{(2t+1)^{3}}$ ; $\frac{4}{9}$
j) $\frac{d^{2}y}{dx^{2}}=-1$ for every $t$

The move that decides everything: differentiate $x$ and $y$ separately WITH RESPECT TO $t$ , then take the quotient, $y'$ on top; you never look for $y$ as a function of $x$ . For the second derivative, differentiate the slope with respect to $t$ , then divide a second time by $x'(t)$ .

a)

$x'(t)=2t$ and $y'(t)=3t^{2}-2$ , so $\frac{dy}{dx}=\frac{3t^{2}-2}{2t}$ . At $t=2$ : $\frac{10}{4}=\frac{5}{2}$ . Pitfall: the quotient upside down, $\frac{x'}{y'}=\frac{2}{5}$ .

b)

$x'(t)=\frac{1}{2\sqrt{t}}$ and $y'(t)=1$ , so $\frac{dy}{dx}=\frac{1}{\frac{1}{2\sqrt{t}}}=2\sqrt{t}$ . At $t=4$ : $\frac{dy}{dx}=4$ .

c)

$x'(t)=\frac{1}{t}$ and $y'(t)=2t$ , so $\frac{dy}{dx}=2t\cdot t=2t^{2}$ . At $t=3$ : $\frac{dy}{dx}=18$ . Pitfall: writing $\frac{2t}{t}=2$ , forgetting that $x'(t)$ is $\frac{1}{t}$ and not $t$ .

d)

$x'(t)=-2\sin t$ and $y'(t)=3\cos t$ , so $\frac{dy}{dx}=-\frac{3\cos t}{2\sin t}=-\frac{3}{2}\cot t$ . At $t=\frac{\pi}{3}$ : $-\frac{3}{2}\cdot\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{2}\approx-0.866$ . The constant $1$ in $x(t)$ disappears when differentiating: it shifts the ellipse without changing its slopes.

e)

$x'(t)=e^{t}$ and, by the product rule, $y'(t)=e^{-t}-te^{-t}=(1-t)e^{-t}$ . So $\frac{dy}{dx}=\frac{(1-t)e^{-t}}{e^{t}}=(1-t)e^{-2t}$ . At $t=-1$ : $2e^{2}\approx 14.778$ . Pitfall: differentiating $te^{-t}$ as $-te^{-t}$ only.

f)

First find the parameter: $t^{2}-1=3$ gives $t=\pm 2$ , and $y(2)=10$ whereas $y(-2)=-10$ , so $t=2$ . $\frac{dy}{dx}=\frac{3t^{2}+1}{2t}$ , hence at $t=2$ : $\frac{13}{4}$ . Pitfall: keeping $t=-2$ , which gives the point $(3, -10)$ .

g)

$y'(t)=2t-4=0$ gives $t=2$ , and $x'(2)=3\cdot 4-3=9\neq 0$ : the tangent line is indeed horizontal. The point is $\left(x(2), y(2)\right)=(2, -4)$ , with $y$ -coordinate $-4$ .

h)

$x'(t)=3t^{2}-3=0$ gives $t=1$ and $t=-1$ . Check $y'$ : $y'(1)=-2\neq 0$ and $y'(-1)=-6\neq 0$ , so there are TWO vertical tangent lines, at the points $(-2, -3)$ for $t=1$ and $(2, 5)$ for $t=-1$ . For the positive parameter, the $x$ -coordinate is $1-3=-2$ . Pitfall: concluding without checking $y'\neq 0$ ; if $x'$ and $y'$ vanish together, the point is singular.

i)

$\frac{dy}{dx}=\frac{3t^{2}}{2t+1}$ . Differentiate it with respect to $t$ : $\frac{6t(2t+1)-2\cdot 3t^{2}}{(2t+1)^{2}}=\frac{6t^{2}+6t}{(2t+1)^{2}}$ , then divide by $x'(t)=2t+1$ : $\frac{d^{2}y}{dx^{2}}=\frac{6t(t+1)}{(2t+1)^{3}}$ . At $t=1$ : $\frac{12}{27}=\frac{4}{9}$ . Pitfall: $\frac{y''(t)}{x''(t)}=\frac{6t}{2}=3$ , which means nothing.

j)

$x'(t)=2\cos t$ and $y'(t)=-2\sin(2t)=-4\sin t\cos t$ , so $\frac{dy}{dx}=-2\sin t$ . Differentiate with respect to $t$ : $-2\cos t$ , then divide by $x'(t)=2\cos t$ : $\frac{d^{2}y}{dx^{2}}=-1$ , for every $t$ such that $\cos t\neq 0$ . Pitfall: $\frac{y''(t)}{x''(t)}=\frac{-4\cos(2t)}{-2\sin t}$ equals $2$ at $\frac{\pi}{6}$ .

Check: when $t$ can be eliminated, the Cartesian slope must agree. In b), $t=x^{2}$ gives $y=x^{2}-1$ and $y'=2x=2\sqrt{t}$ ; in c), $t=e^{x}$ gives $y=e^{2x}$ and $y'=2e^{2x}=2t^{2}$ ; in j), $\cos(2t)=1-2\sin^{2}t$ gives $y=1-\frac{x^{2}}{2}$ , whose second derivative is indeed $-1$ .

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Exercise 1: From the parametrization to the Cartesian equation, and what gets lost on the way

Eliminating the parameter turns a system of two equations into a single equation in $x$ and $y$ . The operation is legitimate, but it systematically forgets three things: the orientation of the motion, the domain actually reached, and the number of passes.

a) Eliminate $t$ from $x=2t-1$ , $y=t^{2}+3$ . Which curve do you get, and is it traced in full?
b) Eliminate $t$ from $x=3\cos t$ , $y=2\sin t$ with $t\in[0, 2\pi]$ . Give the direction of motion and the starting point.
c) Eliminate $t$ from $x=\sec t$ , $y=\tan t$ with $t\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ . Does the Cartesian equation describe exactly the curve?
d) Eliminate $t$ from $x=t^{2}$ , $y=t^{4}$ . Compare with the parabola $y=x^{2}$ : what is missing?
e) State the three pieces of information that eliminating the parameter destroys, and give one of the examples above for each.

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Answers

a) $y=\frac{(x+1)^{2}}{4}+3$ , whole parabola
b) $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ , start $(3, 0)$ , counterclockwise
c) $x^{2}-y^{2}=1$ , right branch only
d) Half-parabola $x\geq 0$ , points reached twice
e) Orientation, domain, number of passes

a)

From $x=2t-1$ we get $t=\frac{x+1}{2}$ , hence $y=\left(\frac{x+1}{2}\right)^{2}+3=\frac{(x+1)^{2}}{4}+3$ . It is a parabola with vertex $(-1, 3)$ . Since $t$ ranges over $\mathbb{R}$ , $x=2t-1$ also ranges over $\mathbb{R}$ : the whole parabola is traced, from left to right. Nothing is lost here, except the orientation, as for the ellipse in b).

b)

$\cos t=\frac{x}{3}$ and $\sin t=\frac{y}{2}$ , so $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ : an ellipse with semi-axes 3 and 2. The starting point is $t=0$ , that is $(3, 0)$ . At $t=\frac{\pi}{2}$ we are at $(0, 2)$ : the motion is COUNTERCLOCKWISE, and the ellipse is traced exactly once.

c)

$\sec^{2}t-\tan^{2}t=1$ gives $x^{2}-y^{2}=1$ , a hyperbola with two branches. But on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ we have $\cos t>0$ , so $x=\sec t\geq 1$ : only the RIGHT branch is traced. The Cartesian equation describes twice too much curve. This is the classic error: $x^{2}-y^{2}=1$ is not the curve, it is a curve that contains it.

d)

$y=t^{4}=\left(t^{2}\right)^{2}=x^{2}$ . But $x=t^{2}\geq 0$ : the curve is the HALF-parabola $y=x^{2}$ with $x\geq 0$ , and moreover each point with $x>0$ is reached TWICE, for $t$ and for $-t$ . Here both the domain and the one-to-one property are lost.

e)

Elimination destroys: (1) the ORIENTATION, the direction and the starting point of the motion, illustrated in b; (2) the DOMAIN, the set of points actually reached, illustrated in c and d; (3) the NUMBER OF PASSES, illustrated in d where each point is reached twice. In practice: after eliminating $t$ , always go back to the parametrization to find the range of $x$ and of $y$ .

Exercise 2: The slope, horizontal tangent lines and vertical tangent lines

Let $C$ be the curve defined by $x=t^{2}$ and $y=t^{3}-3t$ , $t\in\mathbb{R}$ . It is the archetype of the chapter: it has a vertical tangent line, two horizontal tangent lines, and it crosses itself.

a) Compute $x'(t)$ , $y'(t)$ and $\dfrac{dy}{dx}$ in terms of $t$ . For which value of $t$ does the quotient make no sense?
b) Find the points where the tangent line is horizontal.
c) Find the point where the tangent line is vertical, and justify that it is indeed a vertical tangent line and not a singular point.
d) Write the equation of the tangent line at the point with parameter $t=2$ .
e) Does the curve have a point where $x'(t)$ and $y'(t)$ vanish simultaneously? What can you conclude about its shape?

Show the solution

Answers

a) $\frac{dy}{dx}=\frac{3t^{2}-3}{2t}$ , $t\neq 0$
b) $(1, -2)$ and $(1, 2)$
c) $(0, 0)$ , vertical tangent line
d) $y=\frac{9}{4}x-7$
e) No: the curve is smooth

a)

$x'(t)=2t$ and $y'(t)=3t^{2}-3$ , so $\dfrac{dy}{dx}=\dfrac{3t^{2}-3}{2t}$ for $t\neq 0$ . The quotient makes no sense at $t=0$ , precisely where the tangent line will be vertical.

b)

Horizontal tangent: $y'(t)=0$ and $x'(t)\neq 0$ . $3t^{2}-3=0$ gives $t=\pm 1$ , and $x'(\pm 1)=\pm 2\neq 0$ . For $t=1$ : $x=1$ , $y=1-3=-2$ , point $(1, -2)$ . For $t=-1$ : $x=1$ , $y=-1+3=2$ , point $(1, 2)$ . Two distinct points above the same $x$ : impossible for the graph of a function, and perfectly normal here.

c)

Vertical tangent: $x'(t)=0$ and $y'(t)\neq 0$ . $2t=0$ gives $t=0$ , and $y'(0)=-3\neq 0$ : the point $(0, 0)$ has a vertical tangent line. It is not a singular point because only one of the two derivatives vanishes; the slope $\dfrac{3t^{2}-3}{2t}$ tends to $-\infty$ as $t\to 0^{+}$ and to $+\infty$ as $t\to 0^{-}$ , which is the behaviour of a true vertical tangent line.

d)

At $t=2$ : $x=4$ , $y=8-6=2$ , and $\dfrac{dy}{dx}=\dfrac{12-3}{4}=\dfrac{9}{4}$ . The tangent line is $y-2=\dfrac{9}{4}(x-4)$ , that is $y=\dfrac{9}{4}x-7$ .

e)

$x'(t)=2t$ vanishes only at $t=0$ , where $y'(0)=-3\neq 0$ . The two derivatives therefore never vanish together: the curve is SMOOTH everywhere, it has no cusp. Its shape is a loop: the branch $t<0$ rises, passes through $(1, 2)$ , comes back down to the origin, and the branch $t>0$ leaves again as its mirror image. The point $(3, 0)$ , reached for $t=\pm\sqrt{3}$ , is a double point studied in exercise 4.

Exercise 3: The second derivative, concavity and the formula that does not exist

We stay on $C$ : $x=t^{2}$ , $y=t^{3}-3t$ . Every year, one paper in three writes $\dfrac{d^{2}y}{dx^{2}}=\dfrac{y''(t)}{x''(t)}$ . This exercise shows how wrong that is, and why.

a) Starting from $\dfrac{dy}{dx}=\dfrac{3}{2}\left(t-\dfrac{1}{t}\right)$ , show that $\dfrac{d^{2}y}{dx^{2}}=\dfrac{3\left(t^{2}+1\right)}{4t^{3}}$ .
b) Study the sign of $\dfrac{d^{2}y}{dx^{2}}$ and conclude on the concavity of each branch.
c) Does the curve have an inflection point? Justify, and explain why the concavity can change without an inflection point.
d) Compute $\dfrac{y''(t)}{x''(t)}$ and compare its value at $t=1$ with that of $\dfrac{d^{2}y}{dx^{2}}$ . Explain in one sentence where the error comes from.
e) Study the curve $x=t^{2}$ , $y=t^{3}$ : show that $t=0$ is a cusp, and give its Cartesian equation.

Show the solution

Answers

a) $\frac{3(t^{2}+1)}{4t^{3}}$
b) Concave up for $t>0$ , concave down for $t<0$
c) No inflection: the second derivative never vanishes
d) $3$ versus $\frac{3}{2}$
e) Cusp; $y^{2}=x^{3}$

a)

Differentiate the slope with respect to $t$ : $\dfrac{d}{dt}\left[\dfrac{3}{2}\left(t-\dfrac{1}{t}\right)\right]=\dfrac{3}{2}\left(1+\dfrac{1}{t^{2}}\right)=\dfrac{3\left(t^{2}+1\right)}{2t^{2}}$ . Then divide by $x'(t)=2t$ : $\dfrac{d^{2}y}{dx^{2}}=\dfrac{3\left(t^{2}+1\right)}{2t^{2}}\cdot\dfrac{1}{2t}=\dfrac{3\left(t^{2}+1\right)}{4t^{3}}$ .

b)

The numerator $3\left(t^{2}+1\right)$ is strictly positive, so the sign is that of $t^{3}$ , that is, that of $t$ . For $t>0$ the branch is CONCAVE UP; for $t<0$ it is CONCAVE DOWN. The two branches meet at the origin.

c)

$\dfrac{d^{2}y}{dx^{2}}$ NEVER vanishes: $t^{2}+1>0$ . There is therefore no inflection point. Yet the concavity does change at $t=0$ , but through a CHANGE OF SIGN OF THE DENOMINATOR, not of the numerator: $t=0$ is the point with a vertical tangent line, where $\dfrac{d^{2}y}{dx^{2}}$ is not defined. An inflection point requires the second derivative to vanish WHILE CHANGING SIGN, at a point where it is defined; here it blows up. It is the same phenomenon as $y=\frac{1}{x}$ , concave down then concave up on either side of $x=0$ , with no inflection point.

d)

$y''(t)=6t$ and $x''(t)=2$ , so $\dfrac{y''(t)}{x''(t)}=3t$ , which equals 3 at $t=1$ . The true value is $\dfrac{3(1+1)}{4}=\dfrac{3}{2}$ . The two results differ by a factor of two. The error comes from the fact that $\dfrac{dy}{dx}$ is a function of $t$ that must be differentiated again WITH RESPECT TO $x$ , hence divided again by $x'(t)$ : the chain rule comes in twice, and the quotient of the second derivatives ignores it.

e)

For $x=t^{2}$ , $y=t^{3}$ : $x'(t)=2t$ and $y'(t)=3t^{2}$ BOTH vanish at $t=0$ . The slope $\dfrac{3t^{2}}{2t}=\dfrac{3t}{2}$ tends to 0 on both sides: the curve reaches the origin tangent to the $x$ -axis, turns back there and leaves symmetrically. It is a cusp of the first kind. Cartesian equation: $t^{2}=x$ so $y^{2}=t^{6}=x^{3}$ , the semicubical parabola $y^{2}=x^{3}$ .

Exercise 4: Double points, symmetries and reading the shape

A parametric curve can pass through the same point twice, with two different tangent lines. Finding these points means solving a system in two distinct parameters, which has no equivalent in the study of graphs of functions.

a) For the curve $x=t^{2}$ , $y=t^{3}-3t$ , look for two distinct parameters $t\neq s$ giving the same point. Deduce the double point.
b) Compute the two slopes at this double point, then the acute angle between the two tangent lines.
c) Consider the Lissajous curve $x=\cos t$ , $y=\sin 2t$ , $t\in[0, 2\pi]$ , drawn above. Show that it is symmetric about the $x$ -axis and about the $y$ -axis.
d) Find its double point and the two tangent lines at that point.
e) Find its vertical tangent lines, and check that this agrees with the figure.

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Answers

a) $t=\pm\sqrt{3}$ : double point $(3, 0)$
b) Slopes $\pm\sqrt{3}$ , angle $60°$
c) Symmetric about the $x$ -axis, about the origin, hence about the $y$ -axis
d) Tangent lines $y=2x$ and $y=-2x$
e) $(1, 0)$ and $(-1, 0)$

a)

We look for $t\neq s$ with $t^{2}=s^{2}$ and $t^{3}-3t=s^{3}-3s$ . The first equation forces $s=-t$ (the case $s=t$ is excluded). Substituting: $t^{3}-3t=-t^{3}+3t$ , that is $2t^{3}=6t$ , so $2t\left(t^{2}-3\right)=0$ . The solution $t=0$ would give $s=0$ , which is excluded; what remains is $t=\sqrt{3}$ and $s=-\sqrt{3}$ . The double point is $\left(3, 0\right)$ , since $x=3$ and $y=3\sqrt{3}-3\sqrt{3}=0$ .

b)

$\dfrac{dy}{dx}=\dfrac{3t^{2}-3}{2t}$ . At $t=\sqrt{3}$ : $\dfrac{9-3}{2\sqrt{3}}=\dfrac{6}{2\sqrt{3}}=\sqrt{3}$ . At $t=-\sqrt{3}$ : $\dfrac{6}{-2\sqrt{3}}=-\sqrt{3}$ . The angle between the two tangent lines satisfies $\tan\theta=\left|\dfrac{\sqrt{3}-\left(-\sqrt{3}\right)}{1+\sqrt{3}\left(-\sqrt{3}\right)}\right|=\left|\dfrac{2\sqrt{3}}{-2}\right|=\sqrt{3}$ , so $\theta=60^{\circ}$ . The two tangent lines make an angle of 60 degrees: the slopes $\pm\sqrt{3}$ correspond to $\pm 60^{\circ}$ with respect to the horizontal, which confirms the result by direct reading.

c)

Replace $t$ with $-t$ : $x(-t)=\cos(-t)=\cos t=x(t)$ and $y(-t)=\sin(-2t)=-y(t)$ . The point $(x,-y)$ therefore lies on the curve as soon as $(x,y)$ does: symmetry about the $x$ -axis. Now replace $t$ with $\pi-t$ : $x(\pi-t)=-\cos t=-x(t)$ and $y(\pi-t)=\sin(2\pi-2t)=-\sin 2t=-y(t)$ . We get $(-x,-y)$ : symmetry about the ORIGIN. Combined with the symmetry about the $x$ -axis, it implies symmetry about the $y$ -axis. The curve therefore has all three symmetries, as the figure confirms.

d)

The double point is the origin: $x=\cos t=0$ gives $t=\frac{\pi}{2}$ or $t=\frac{3\pi}{2}$ , and in both cases $y=\sin\pi=0$ or $y=\sin 3\pi=0$ . The slopes: $\dfrac{dy}{dx}=\dfrac{2\cos 2t}{-\sin t}$ . At $t=\frac{\pi}{2}$ : $\dfrac{2\cos\pi}{-\sin\frac{\pi}{2}}=\dfrac{-2}{-1}=2$ . At $t=\frac{3\pi}{2}$ : $\dfrac{2\cos 3\pi}{-\sin\frac{3\pi}{2}}=\dfrac{-2}{1}=-2$ . The two tangent lines at the origin are $y=2x$ and $y=-2x$ .

e)

Vertical tangent: $x'(t)=-\sin t=0$ with $y'(t)=2\cos 2t\neq 0$ . We get $t=0$ and $t=\pi$ , with $y'(0)=2$ and $y'(\pi)=2\cos 2\pi=2$ , both nonzero. The points are $(1, 0)$ and $(-1, 0)$ : the right and left ends of the figure eight, where the curve turns back smoothly with a vertical tangent line, exactly as in the figure. The horizontal tangent lines, for their part, come from $\cos 2t=0$ , that is $t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$ , giving the four peaks $\left(\pm\frac{\sqrt{2}}{2},\pm 1\right)$ .

Exercise 5: The cycloid: the parametrization you cannot do without

A point marked on the rim of a wheel of radius $a$ rolling without slipping along a line traces a cycloid: $x=a\left(t-\sin t\right)$ , $y=a\left(1-\cos t\right)$ , where $t$ is the angle through which the wheel has turned. Take $a=1$ .

a) Show that $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$ , then simplify this quotient using the half-angle formulas.
b) What happens at $t=2k\pi$ ? Give the nature of the point, and the limit of the slope on either side.
c) Compute the length of one arch, for $t\in[0, 2\pi]$ .
d) Compute the area between one arch and the $x$ -axis.
e) Why can we not eliminate $t$ to get an elementary Cartesian equation?

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Answers

a) $\frac{dy}{dx}=\cot\frac{t}{2}$
b) Cusp
c) $8$
d) $3\pi$
e) $t$ inside and outside the cosine: no elementary equation

a)

$x'(t)=1-\cos t$ and $y'(t)=\sin t$ , so $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$ . With $\sin t=2\sin\frac{t}{2}\cos\frac{t}{2}$ and $1-\cos t=2\sin^{2}\frac{t}{2}$ , the quotient becomes $\dfrac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^{2}\frac{t}{2}}=\cot\dfrac{t}{2}$ . The slope is therefore simply the cotangent of half the angle of rotation.

b)

At $t=2k\pi$ : $x'=1-\cos 2k\pi=0$ AND $y'=\sin 2k\pi=0$ . Both derivatives vanish: it is a SINGULAR point. The slope $\cot\frac{t}{2}$ tends to $+\infty$ as $t\to 0^{+}$ and to $-\infty$ as $t\to 2\pi^{-}$ : the curve comes in vertically, touches the axis and leaves vertically. It is a CUSP, physically the instant when the marked point touches the ground and its velocity is zero.

c)

$x'^{2}+y'^{2}=\left(1-\cos t\right)^{2}+\sin^{2}t=1-2\cos t+\cos^{2}t+\sin^{2}t=2-2\cos t=4\sin^{2}\dfrac{t}{2}$ . So $ds=2\left|\sin\frac{t}{2}\right|dt=2\sin\frac{t}{2}\,dt$ on $[0, 2\pi]$ , where $\sin\frac{t}{2}\geq 0$ . Then $L=\int_{0}^{2\pi}2\sin\dfrac{t}{2}\,dt=\left[-4\cos\dfrac{t}{2}\right]_{0}^{2\pi}=-4(-1)+4(1)=8$ . One arch measures $8a$ , that is eight times the radius: a result Christopher Wren obtained in 1658, one of the very first exact rectifications of a curve.

d)

$A=\int y\,dx=\int_{0}^{2\pi}\left(1-\cos t\right)\cdot\left(1-\cos t\right)dt=\int_{0}^{2\pi}\left(1-2\cos t+\cos^{2}t\right)dt$ . Term by term: $\int_{0}^{2\pi}1\,dt=2\pi$ , $\int_{0}^{2\pi}\cos t\,dt=0$ , and $\int_{0}^{2\pi}\cos^{2}t\,dt=\pi$ . So $A=2\pi-0+\pi=3\pi$ , that is $3\pi a^{2}$ : exactly three times the area of the rolling disk. Note that $x$ is indeed increasing on $[0, 2\pi]$ since $x'=1-\cos t\geq 0$ , which allows $\int y\,dx$ to be read as an area.

e)

From $y=1-\cos t$ we get $\cos t=1-y$ , so $t=\arccos\left(1-y\right)$ and $\sin t=\sqrt{1-\left(1-y\right)^{2}}=\sqrt{2y-y^{2}}$ . Substituting: $x=\arccos\left(1-y\right)-\sqrt{2y-y^{2}}$ . The parameter appears both INSIDE a trigonometric argument and OUTSIDE it, in the lone term $t$ : elimination therefore cannot produce an algebraic relation, only an expression with an inverse cosine, and even then valid one arch at a time. This is the typical situation where the parametrization is not an intermediate step but the only reasonable description of the curve.

Exercise 6: The area enclosed by a parametric curve

Three formulas give the same area for a simple closed curve traced once counterclockwise: $A=\oint x\,dy$ , $A=-\oint y\,dx$ , and the symmetric formula $A=\frac{1}{2}\oint\left(x\,dy-y\,dx\right)$ . This exercise puts them to the test.

a) On the ellipse $x=3\cos t$ , $y=2\sin t$ , $t\in[0, 2\pi]$ , compute $\frac{1}{2}\oint\left(x\,dy-y\,dx\right)$ and recover the area of the ellipse.
b) Compute the same area again with $-\oint y\,dx$ . What happens if the ellipse is traced in the other direction, with $x=3\cos t$ , $y=-2\sin t$ ?
c) The curve $x=t^{2}$ , $y=t^{3}-3t$ forms a closed loop for $t\in\left[-\sqrt{3},\sqrt{3}\right]$ . Show that $x\,y'-y\,x'=t^{4}+3t^{2}$ .
d) Compute the area of this loop.
e) Why is the symmetric formula the safest on an exam?

Show the solution

Answers

a) $6\pi$
b) $6\pi$ ; $-6\pi$ clockwise
c) $t^{4}+3t^{2}$
d) $\frac{24\sqrt{3}}{5}\approx 8.31$
e) No need for $x$ to be monotonic

a)

$x'=-3\sin t$ , $y'=2\cos t$ . Then $xy'-yx'=3\cos t\cdot 2\cos t-2\sin t\cdot\left(-3\sin t\right)=6\cos^{2}t+6\sin^{2}t=6$ . The integrand is constant: $A=\frac{1}{2}\int_{0}^{2\pi}6\,dt=\frac{1}{2}\cdot 6\cdot 2\pi=6\pi$ . We recover $\pi ab=\pi\cdot 3\cdot 2=6\pi$ .

b)

$-\oint y\,dx=-\int_{0}^{2\pi}2\sin t\cdot\left(-3\sin t\right)dt=6\int_{0}^{2\pi}\sin^{2}t\,dt=6\pi$ . Same result. With $y=-2\sin t$ , the motion becomes CLOCKWISE and each formula changes sign: we get $-6\pi$ . The area is the absolute value; the sign encodes the orientation. That is why you always write the interval of $t$ and check the direction by evaluating two points.

c)

$x=t^{2}$ , $y=t^{3}-3t$ , $x'=2t$ , $y'=3t^{2}-3$ . Then $xy'-yx'=t^{2}\left(3t^{2}-3\right)-\left(t^{3}-3t\right)\left(2t\right)=3t^{4}-3t^{2}-2t^{4}+6t^{2}=t^{4}+3t^{2}$ .

d)

$A=\dfrac{1}{2}\int_{-\sqrt{3}}^{\sqrt{3}}\left(t^{4}+3t^{2}\right)dt$ . The integrand is even, so $A=\int_{0}^{\sqrt{3}}\left(t^{4}+3t^{2}\right)dt=\left[\dfrac{t^{5}}{5}+t^{3}\right]_{0}^{\sqrt{3}}=\dfrac{9\sqrt{3}}{5}+3\sqrt{3}=\dfrac{24\sqrt{3}}{5}$ . Numerically $A\approx 8.31$ . The fact that the integrand $t^{4}+3t^{2}$ is positive everywhere confirms that the loop is traced counterclockwise.

e)

Because it does not require $x$ to be monotonic. $\int y\,dx$ can be read as an area under a curve only if $x$ increases; as soon as the curve goes back, as here where $x=t^{2}$ decreases then increases, the naive interpretation collapses, whereas the symmetric formula remains exact. It is also the only one of the three that treats $x$ and $y$ in the same way, hence the only one where a sign error shows immediately.

Exercise 7: Arc length in parametric form

The element of arc length $ds=\sqrt{x'^{2}+y'^{2}}\,dt$ is the only formula of the chapter that requires neither monotonicity nor orientation: it adds up lengths, all positive. The square root still has to simplify, which only happens in cases designed for it.

a) Recover the formula $L=\int_{a}^{b}\sqrt{1+\left(f'(x)\right)^{2}}\,dx$ as a special case of the parametrization $x=t$ , $y=f(t)$ .
b) Compute the length of the arc $x=t^{2}$ , $y=t^{3}$ for $t\in[0, 1]$ .
c) Compute the length of the spiral $x=e^{t}\cos t$ , $y=e^{t}\sin t$ for $t\in\left[0,\frac{\pi}{2}\right]$ .
d) Check that the unit circle parametrized by $x=\cos\left(t^{2}\right)$ , $y=\sin\left(t^{2}\right)$ , $t\in\left[0,\sqrt{2\pi}\right]$ , does have length $2\pi$ .
e) What does the length become if the unit circle is traced twice, with $t\in[0, 4\pi]$ ? Compare with the behaviour of the area, and conclude on the nature of each formula.

Show the solution

Answers

a) $x=t$ , $y=f(t)$
b) $\frac{13\sqrt{13}-8}{27}\approx 1.4397$
c) $\sqrt{2}(e^{\pi/2}-1)\approx 5.389$
d) $2\pi$
e) Length $4\pi$ ; the area formula would give $2\pi$ , which is wrong

a)

With $x=t$ and $y=f(t)$ : $x'=1$ and $y'=f'(t)$ , so $ds=\sqrt{1+\left(f'(t)\right)^{2}}\,dt$ . Since $t=x$ , we get exactly $L=\int_{a}^{b}\sqrt{1+\left(f'(x)\right)^{2}}\,dx$ . The Cartesian formula is therefore not a different formula: it is the case where the parameter is $x$ itself.

b)

$x'=2t$ , $y'=3t^{2}$ , so $x'^{2}+y'^{2}=4t^{2}+9t^{4}=t^{2}\left(4+9t^{2}\right)$ and, for $t\geq 0$ , $ds=t\sqrt{4+9t^{2}}\,dt$ . Let $u=4+9t^{2}$ , $du=18t\,dt$ : $L=\dfrac{1}{18}\int_{4}^{13}\sqrt{u}\,du=\dfrac{1}{18}\cdot\dfrac{2}{3}\left[u^{3/2}\right]_{4}^{13}=\dfrac{13\sqrt{13}-8}{27}\approx 1.4397$ .

c)

$x'=e^{t}\left(\cos t-\sin t\right)$ and $y'=e^{t}\left(\sin t+\cos t\right)$ . Expanding, $x'^{2}+y'^{2}=e^{2t}\left[\left(\cos t-\sin t\right)^{2}+\left(\sin t+\cos t\right)^{2}\right]=e^{2t}\left[2\cos^{2}t+2\sin^{2}t\right]=2e^{2t}$ . So $ds=\sqrt{2}\,e^{t}dt$ and $L=\sqrt{2}\int_{0}^{\pi/2}e^{t}dt=\sqrt{2}\left(e^{\pi/2}-1\right)\approx 5.389$ . The logarithmic spiral is the textbook case where the square root simplifies perfectly, because the speed is proportional to the radius.

d)

$x'=-2t\sin\left(t^{2}\right)$ and $y'=2t\cos\left(t^{2}\right)$ , so $x'^{2}+y'^{2}=4t^{2}$ and $ds=2t\,dt$ for $t\geq 0$ . Then $L=\int_{0}^{\sqrt{2\pi}}2t\,dt=\left[t^{2}\right]_{0}^{\sqrt{2\pi}}=2\pi$ . The point moves faster and faster, but the distance travelled is still the circumference: the change of parametrization has changed nothing.

e)

With $t\in[0, 4\pi]$ and the usual parametrization, $L=\int_{0}^{4\pi}1\,dt=4\pi$ : the length DOUBLES, which is correct, since we really do travel $2\pi$ of path twice. The area formula, on the other hand, would give $\frac{1}{2}\int_{0}^{4\pi}1\,dt=2\pi$ , twice the actual area $\pi$ : it is WRONG. The difference lies in the nature of the two formulas: length measures the PATH travelled, so it legitimately counts repetitions; area claims to measure a REGION of the plane, a notion that only makes sense if the curve is traced once. Remember: every area formula requires a single pass, no length formula does.

Exercise 8: The area of a surface of revolution

When a parametrized arc is rotated about an axis, each element $ds$ generates a frustum of a cone whose radius is the distance to the axis. Hence $S=2\pi\int\left(\text{distance to the axis}\right)ds$ . The whole art lies in not confusing the distance to the axis with the coordinate.

a) Recover the area of the sphere of radius $R$ by rotating $x=R\cos t$ , $y=R\sin t$ , $t\in[0,\pi]$ , about the $x$ -axis.
b) Let the arc be $x=3t-t^{3}$ , $y=3t^{2}$ , $t\in[0, 1]$ . Show that $ds=3\left(1+t^{2}\right)dt$ , then compute its length.
c) Compute the area of the surface generated by this arc rotating about the $x$ -axis.
d) Compute the area of the surface generated by the same arc rotating about the $y$ -axis.
e) Why is the formula written with $\left|y\right|$ and not with $y$ ? Give an arc for which forgetting the absolute value gives a zero area.

Show the solution

Answers

a) $4\pi R^{2}$
b) $ds=3(1+t^{2})dt$ , length $4$
c) $\frac{48\pi}{5}$
d) $11\pi$
e) Radius $=|y|$ ; without the absolute value, $0$

a)

$x'=-R\sin t$ , $y'=R\cos t$ , so $ds=R\,dt$ . The distance to the $x$ -axis is $\left|y\right|=R\sin t\geq 0$ on $[0,\pi]$ . Then $S=2\pi\int_{0}^{\pi}R\sin t\cdot R\,dt=2\pi R^{2}\left[-\cos t\right]_{0}^{\pi}=2\pi R^{2}\cdot 2=4\pi R^{2}$ . We recover Archimedes' formula.

b)

$x'=3-3t^{2}=3\left(1-t^{2}\right)$ and $y'=6t$ . Then $x'^{2}+y'^{2}=9\left(1-t^{2}\right)^{2}+36t^{2}=9\left(1-2t^{2}+t^{4}+4t^{2}\right)=9\left(1+t^{2}\right)^{2}$ . Since $1+t^{2}>0$ , $ds=3\left(1+t^{2}\right)dt$ . The length is $L=\int_{0}^{1}3\left(1+t^{2}\right)dt=3\left(1+\dfrac{1}{3}\right)=4$ . The perfect square under the root is no accident: the curve is built for it, like the cycloid and the astroid.

c)

About the $x$ -axis, the distance to the axis is $y=3t^{2}\geq 0$ . $S=2\pi\int_{0}^{1}3t^{2}\cdot 3\left(1+t^{2}\right)dt=18\pi\int_{0}^{1}\left(t^{2}+t^{4}\right)dt=18\pi\left(\dfrac{1}{3}+\dfrac{1}{5}\right)=18\pi\cdot\dfrac{8}{15}=\dfrac{48\pi}{5}\approx 30.16$ .

d)

About the $y$ -axis, the distance to the axis is $\left|x\right|=3t-t^{3}$ , positive on $[0, 1]$ . $S=2\pi\int_{0}^{1}\left(3t-t^{3}\right)\cdot 3\left(1+t^{2}\right)dt=6\pi\int_{0}^{1}\left(3t+3t^{3}-t^{3}-t^{5}\right)dt=6\pi\int_{0}^{1}\left(3t+2t^{3}-t^{5}\right)dt=6\pi\left(\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{6}\right)=6\pi\cdot\dfrac{11}{6}=11\pi\approx 34.56$ .

e)

Because the radius of a frustum is a DISTANCE, hence positive, whereas $y$ can change sign. Take the arc $x=t$ , $y=t$ for $t\in[-1, 1]$ and rotate it about the $x$ -axis: the surface generated is a double cone, of area $2\pi\int_{-1}^{1}\left|t\right|\sqrt{2}\,dt=2\pi\sqrt{2}\cdot 1=2\sqrt{2}\pi$ . Without the absolute value we would get $2\pi\sqrt{2}\int_{-1}^{1}t\,dt=0$ : a surface of zero area, although it is perfectly visible. Practical rule: always split the interval at the zeros of $y$ before integrating.

Exercise 9: Reparametrizing: the curve does not change, the integral sometimes does

The same curve has infinitely many parametrizations. Some quantities do not depend on the choice, others do. Knowing which ones is what separates a good paper from a paper that just applies formulas.

a) Let $t=\varphi(u)$ be a differentiable, strictly increasing change of parameter. Show that the arc length computed in the two parametrizations is the same.
b) Is the assumption of strict increase necessary? What happens if $\varphi$ is decreasing?
c) The unit circle traced once gives $\frac{1}{2}\oint\left(x\,dy-y\,dx\right)=\pi$ . Compute the same expression for $t\in[0, 6\pi]$ and interpret.
d) Arc length parametrization means using the length itself as the parameter. Check that $x=\cos s$ , $y=\sin s$ is the arc length parametrization of the unit circle, and say what characterizes it in general.
e) On the arch of the cycloid, the quantity $\int y\,dx$ can be read as the area under the curve. On the loop of exercise 6, it cannot. Which property of the parametrization makes the difference?

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Answers

a) Change of variable: the length is invariant
b) Same length if $\varphi$ decreases
c) $3\pi$ : three turns
d) $x'^{2}+y'^{2}=1$
e) The monotonicity of $x(t)$

a)

In the parametrization in $u$ , the curve is $\left(x\left(\varphi(u)\right),y\left(\varphi(u)\right)\right)$ and the chain rule gives $\dfrac{dx}{du}=x'\left(\varphi(u)\right)\varphi'(u)$ , and likewise for $y$ . Then $\sqrt{\left(\frac{dx}{du}\right)^{2}+\left(\frac{dy}{du}\right)^{2}}=\left|\varphi'(u)\right|\sqrt{x'^{2}+y'^{2}}=\varphi'(u)\sqrt{x'^{2}+y'^{2}}$ since $\varphi'>0$ . Integrating and setting $t=\varphi(u)$ , $dt=\varphi'(u)du$ , we get back exactly $\int\sqrt{x'^{2}+y'^{2}}\,dt$ : it is the change of variable theorem, nothing more.

b)

If $\varphi$ is decreasing, $\left|\varphi'\right|=-\varphi'$ and the bounds are swapped: the two signs cancel out and the length is still the same. What matters is therefore MONOTONICITY, not direction: the length is invariant under any monotonic change of parameter. On the other hand, if $\varphi$ is not monotonic, the curve is traced several times and the length increases.

c)

With $x=\cos t$ , $y=\sin t$ : $xy'-yx'=\cos^{2}t+\sin^{2}t=1$ , so $\frac{1}{2}\int_{0}^{6\pi}1\,dt=3\pi$ . We get THREE times the area of the disk, because the curve was traced three times. The area formula counts the turns: it actually computes the area weighted by the number of times the curve winds around each point, and it agrees with the geometric area only for a SIMPLE closed curve.

d)

$\left(\frac{dx}{ds}\right)^{2}+\left(\frac{dy}{ds}\right)^{2}=\sin^{2}s+\cos^{2}s=1$ : the speed is constantly equal to 1, so $\int_{0}^{s}1\,d\sigma=s$ , and the parameter is indeed the length travelled since $s=0$ . In general, a parametrization is by arc length if and only if $x'^{2}+y'^{2}=1$ everywhere, that is, if the point moves at unit speed. It is the canonical parametrization: it exists for every smooth curve, and it is the one in which the curvature formulas take their simplest form.

e)

The difference is the MONOTONICITY OF $x(t)$ . On the arch of the cycloid, $x'=1-\cos t\geq 0$ : $x$ increases, each $x$ -value is reached once, and $\int y\,dx$ is indeed the area under the curve in the usual sense. On the loop, $x=t^{2}$ decreases then increases: the curve passes back over the same $x$ -values, the contributions of the outward and return trips subtract from each other, and only the closed formula $\frac{1}{2}\oint\left(x\,dy-y\,dx\right)$ keeps a meaning. The question to ask yourself every time before writing $\int y\,dx$ : is $x$ monotonic on the interval?

Exercise 10: Synthesis problem: the astroid and its secret

The astroid $x=a\cos^{3}t$ , $y=a\sin^{3}t$ with $a=2$ is the curve traced by a point of a small circle of radius $\frac{a}{4}$ rolling inside a circle of radius $a$ . It brings the whole chapter together: cusps, length, area, surface of revolution, and a remarkable geometric property.

a) Give the Cartesian equation of the astroid, and locate its four cusps, justifying their nature.
b) Show that $ds=3a\left|\cos t\sin t\right|dt$ , then compute the total length of the curve.
c) Compute the area enclosed by the astroid.
d) Compute the area of the surface generated by rotating the whole astroid about the $x$ -axis.
e) Show that the tangent line at any point with parameter $t$ , $t$ not a multiple of $\frac{\pi}{2}$ , cuts off on the axes a segment of CONSTANT length equal to $a$ .

Show the solution

Answers

a) $x^{2/3}+y^{2/3}=a^{2/3}$ ; cusps $(\pm a, 0)$ , $(0, \pm a)$
b) $6a=12$
c) $\frac{3\pi a^{2}}{8}=\frac{3\pi}{2}$
d) $\frac{12\pi a^{2}}{5}=\frac{48\pi}{5}$
e) Segment of length $a=2$

a)

$\cos t=\left(\frac{x}{a}\right)^{1/3}$ and $\sin t=\left(\frac{y}{a}\right)^{1/3}$ , so $\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{a}\right)^{2/3}=1$ , that is $x^{2/3}+y^{2/3}=a^{2/3}$ . The derivatives are $x'=-3a\cos^{2}t\sin t$ and $y'=3a\sin^{2}t\cos t$ : they vanish SIMULTANEOUSLY when $\cos t=0$ or $\sin t=0$ , that is at $t=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$ , giving the four points $\left(\pm a,0\right)$ and $\left(0,\pm a\right)$ . These are singular points. The slope $\dfrac{dy}{dx}=\dfrac{3a\sin^{2}t\cos t}{-3a\cos^{2}t\sin t}=-\tan t$ tends to 0 on either side of $t=0$ : the curve comes in horizontally at $(a,0)$ , turns back there and leaves again. These are indeed four cusps, visible as sharp points on the figure.

b)

$x'^{2}+y'^{2}=9a^{2}\cos^{4}t\sin^{2}t+9a^{2}\sin^{4}t\cos^{2}t=9a^{2}\cos^{2}t\sin^{2}t\left(\cos^{2}t+\sin^{2}t\right)=9a^{2}\cos^{2}t\sin^{2}t$ . Hence $ds=3a\left|\cos t\sin t\right|dt$ . By symmetry, the total length is four times that of the quarter corresponding to $t\in\left[0,\frac{\pi}{2}\right]$ , where $\cos t\sin t\geq 0$ : $L=4\int_{0}^{\pi/2}3a\cos t\sin t\,dt=12a\left[\dfrac{\sin^{2}t}{2}\right]_{0}^{\pi/2}=6a$ . With $a=2$ , $L=12$ . The absolute value is essential: without it, $\int_{0}^{2\pi}3a\cos t\sin t\,dt=0$ .

c)

$xy'-yx'=a\cos^{3}t\cdot 3a\sin^{2}t\cos t-a\sin^{3}t\cdot\left(-3a\cos^{2}t\sin t\right)=3a^{2}\cos^{2}t\sin^{2}t\left(\cos^{2}t+\sin^{2}t\right)=3a^{2}\cos^{2}t\sin^{2}t=\dfrac{3a^{2}}{4}\sin^{2}2t$ . Then $A=\dfrac{1}{2}\int_{0}^{2\pi}\dfrac{3a^{2}}{4}\sin^{2}2t\,dt=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}\sin^{2}2t\,dt=\dfrac{3a^{2}}{8}\cdot\pi=\dfrac{3\pi a^{2}}{8}$ . With $a=2$ , $A=\dfrac{3\pi}{2}\approx 4.71$ , that is $\frac{3}{8}$ of the area of the disk of radius $a$ .

d)

Rotate the upper half, $t\in[0,\pi]$ , where $y=a\sin^{3}t\geq 0$ . $S=2\pi\int_{0}^{\pi}a\sin^{3}t\cdot 3a\left|\cos t\sin t\right|dt=6\pi a^{2}\int_{0}^{\pi}\sin^{4}t\left|\cos t\right|dt$ . By symmetry about $\frac{\pi}{2}$ , this equals $12\pi a^{2}\int_{0}^{\pi/2}\sin^{4}t\cos t\,dt=12\pi a^{2}\left[\dfrac{\sin^{5}t}{5}\right]_{0}^{\pi/2}=\dfrac{12\pi a^{2}}{5}$ . With $a=2$ : $S=\dfrac{48\pi}{5}\approx 30.16$ .

e)

The slope at $t$ is $-\tan t$ , and the point of tangency is $\left(a\cos^{3}t,a\sin^{3}t\right)$ . The tangent line is $y-a\sin^{3}t=-\tan t\left(x-a\cos^{3}t\right)$ . For $y=0$ : $x=a\cos^{3}t+\dfrac{a\sin^{3}t}{\tan t}=a\cos^{3}t+a\sin^{2}t\cos t=a\cos t\left(\cos^{2}t+\sin^{2}t\right)=a\cos t$ . For $x=0$ : $y=a\sin^{3}t+a\cos^{3}t\tan t=a\sin^{3}t+a\sin t\cos^{2}t=a\sin t$ . The two points of intersection are therefore $\left(a\cos t,0\right)$ and $\left(0,a\sin t\right)$ , and the length of the segment is $\sqrt{a^{2}\cos^{2}t+a^{2}\sin^{2}t}=a$ , independent of $t$ . In other words, the astroid is the envelope of the segments of length $a$ whose endpoints slide along the two axes: it is exactly the trace left by a ladder sliding down a wall, and it is the reason this curve appears in optimization problems such as the ladder carried around the corner of a hallway.

Exercise 11: A parametrized hyperbola: branch, tangent lines and asymptotes

We study the curve $x=t+\frac{1}{t}$ , $y=t-\frac{1}{t}$ for $t>0$ .

a) Compute $x+y$ and $x-y$ , then eliminate $t$ .
b) Show that $x\geq 2$ for every $t>0$ . Which part of the Cartesian curve is traced, and in which direction?
c) Compute $\frac{dy}{dx}$ . Find the point with a vertical tangent line. Does the curve have a horizontal tangent line?
d) Write the equation of the tangent line at the point with parameter $t=2$ .
e) Which lines does the curve approach as $t\to+\infty$ and as $t\to 0^{+}$ ?

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Answers

a) $x^{2}-y^{2}=4$
b) Right branch, traced from bottom to top
c) $\frac{t^{2}+1}{t^{2}-1}$ ; vertical at $(2, 0)$ ; no horizontal tangent
d) $y=\frac{5}{3}x-\frac{8}{3}$
e) $y=x$ and $y=-x$

a)

$x+y=2t$ and $x-y=\frac{2}{t}$ . Their product no longer depends on $t$ : $(x+y)(x-y)=4$ , that is $x^{2}-y^{2}=4$ , a hyperbola.

b)

$x-2=t+\frac{1}{t}-2=\frac{(t-1)^{2}}{t}\geq 0$ , so $x\geq 2$ , with equality for $t=1$ only. Only the RIGHT branch of the hyperbola is traced. For $0<t<1$ , $y<0$ ; for $t>1$ , $y>0$ : the point moves up along the branch, from bottom to top, passing through the vertex $(2, 0)$ for $t=1$ .

c)

$x'(t)=1-\frac{1}{t^{2}}$ and $y'(t)=1+\frac{1}{t^{2}}$ , so $\frac{dy}{dx}=\frac{t^{2}+1}{t^{2}-1}$ for $t\neq 1$ . Vertical tangent: $x'(t)=0$ and $y'(t)\neq 0$ , that is $t=1$ , at the point $(2, 0)$ , the vertex of the branch. No horizontal tangent line: $y'(t)=1+\frac{1}{t^{2}}$ never vanishes.

d)

At $t=2$ : $x=\frac{5}{2}$ , $y=\frac{3}{2}$ , and $\frac{dy}{dx}=\frac{5}{3}$ . Tangent line: $y-\frac{3}{2}=\frac{5}{3}\left(x-\frac{5}{2}\right)$ , that is $y=\frac{5}{3}x-\frac{8}{3}$ . Check: the point satisfies $\frac{25}{4}-\frac{9}{4}=4$ ✓.

e)

As $t\to+\infty$ , $\frac{1}{t}\to 0$ : $x\approx t$ and $y\approx t$ , and $y-x=-\frac{2}{t}\to 0$ . The curve approaches $y=x$ . As $t\to 0^{+}$ , $x\approx\frac{1}{t}$ and $y\approx-\frac{1}{t}$ , with $x+y=2t\to 0$ : it approaches $y=-x$ . These are the two asymptotes of the hyperbola, and the slope $\frac{t^{2}+1}{t^{2}-1}$ does tend to $1$ and then to $-1$ .

Exercise 12: The involute of a circle: unwinding a string

A string is wound around a circle of radius $1$ , then unwound while being kept taut. The end of the string traces the involute of the circle: $x=\cos t+t\sin t$ , $y=\sin t-t\cos t$ , where $t\geq 0$ is the angle unwound.

The figure below shows the situation.

a) Compute $x'(t)$ and $y'(t)$ and simplify them.
b) Show that $ds=t\,dt$ , then compute the length of the curve for $t\in[0, 2\pi]$ .
c) Show that $\frac{dy}{dx}=\tan t$ , then that the vector from the point of tangency $(\cos t, \sin t)$ to the point of the curve is perpendicular to the curve. Interpret.
d) Compute the distance from the point of the curve to the centre of the circle, in terms of $t$ , then for $t=2\pi$ .
e) Compare the length traced during the second turn, $t\in[2\pi, 4\pi]$ , with that of the first turn.

Show the solution

Answers

a) $x'=t\cos t$ , $y'=t\sin t$
b) $ds=t\,dt$ ; $2\pi^{2}\approx 19.74$
c) Slope $\tan t$ ; the string is perpendicular to the curve
d) $\sqrt{1+t^{2}}$ ; $\approx 6.362$
e) $6\pi^{2}$ : three times as long

a)

$x'(t)=-\sin t+\sin t+t\cos t=t\cos t$ and $y'(t)=\cos t-\cos t+t\sin t=t\sin t$ . The terms have cancelled in pairs.

b)

$x'^{2}+y'^{2}=t^{2}\cos^{2}t+t^{2}\sin^{2}t=t^{2}$ , so $ds=|t|\,dt=t\,dt$ for $t\geq 0$ . $L=\int_{0}^{2\pi}t\,dt=\frac{(2\pi)^{2}}{2}=2\pi^{2}\approx 19.74$ .

c)

$\frac{dy}{dx}=\frac{t\sin t}{t\cos t}=\tan t$ : the tangent line makes the angle $t$ with the horizontal. The vector from the point of tangency to the point of the curve is $(t\sin t, -t\cos t)$ ; its dot product with the tangent vector $(t\cos t, t\sin t)$ is $t^{2}\sin t\cos t-t^{2}\cos t\sin t=0$ . The taut string is therefore perpendicular to the curve it traces and, being perpendicular to the radius $(\cos t, \sin t)$ , it is tangent to the circle. Its length is $\sqrt{t^{2}\sin^{2}t+t^{2}\cos^{2}t}=t$ : exactly the arc unwound.

d)

$x^{2}+y^{2}=\cos^{2}t+2t\sin t\cos t+t^{2}\sin^{2}t+\sin^{2}t-2t\sin t\cos t+t^{2}\cos^{2}t=1+t^{2}$ . The distance is $\sqrt{1+t^{2}}$ , that is $\sqrt{1+4\pi^{2}}\approx 6.362$ for $t=2\pi$ : Pythagoras in the triangle formed by the centre, the point of tangency and the end of the string.

e)

$\int_{2\pi}^{4\pi}t\,dt=\frac{16\pi^{2}-4\pi^{2}}{2}=6\pi^{2}$ , three times the length of the first turn. The speed $\frac{ds}{dt}=t$ grows with the length of string already unwound: the spiral opens out and each turn is longer than the one before. It is the profile of most gear teeth: two involute teeth in mesh transmit the rotation at a constant speed ratio. The figure in the solution shows the circle and the first turn.

Exercise 13: The quadratic Bézier curve: three points are enough

In computer-aided design, a quadratic Bézier curve is defined by three control points $P_{0}$ , $P_{1}$ , $P_{2}$ : $B(t)=(1-t)^{2}P_{0}+2t(1-t)P_{1}+t^{2}P_{2}$ , for $t\in[0, 1]$ . Take $P_{0}=(0, 0)$ , $P_{1}=(2, 4)$ and $P_{2}=(4, 0)$ , as in the figure.

a) Write $x(t)$ and $y(t)$ .
b) Eliminate $t$ and identify the curve.
c) Compute the slopes at $t=0$ and at $t=1$ . Show that both tangent lines pass through $P_{1}$ .
d) What is the highest point of the curve? Compare its height with that of $P_{1}$ .
e) Compute the area between the curve and the $x$ -axis, then compare it with the area of the triangle $P_{0}P_{1}P_{2}$ .

Show the solution

Answers

a) $x=4t$ , $y=8t(1-t)$
b) $y=2x-\frac{x^{2}}{2}$
c) Slopes $2$ and $-2$ , tangent lines through $P_{1}$
d) $(2, 2)$ , half the height of $P_{1}$
e) $\frac{16}{3}$ , that is $\frac{2}{3}$ of the triangle

a)

$x(t)=2t(1-t)\times 2+t^{2}\times 4=4t-4t^{2}+4t^{2}=4t$ and $y(t)=2t(1-t)\times 4=8t(1-t)$ .

b)

$t=\frac{x}{4}$ , so $y=8\cdot\frac{x}{4}\left(1-\frac{x}{4}\right)=2x-\frac{x^{2}}{2}$ : an arc of a parabola, from $x=0$ to $x=4$ .

c)

$x'(t)=4$ and $y'(t)=8-16t$ , so $\frac{dy}{dx}=2-4t$ . At $t=0$ : slope $2$ , tangent line $y=2x$ , which passes through $(2, 4)=P_{1}$ ✓. At $t=1$ : slope $-2$ , tangent line $y=-2(x-4)$ , which also passes through $(2, 4)$ ✓. This is the property that makes Bézier curves so practical: the control points fix the tangent lines at the endpoints.

d)

$y'(t)=0$ for $t=\frac{1}{2}$ : the highest point is $(2, 2)$ . Its height is HALF that of $P_{1}$ : the curve is pulled toward the control point without ever reaching it.

e)

$x$ is increasing, so the area is $\int y\,dx=\int_{0}^{1}8t(1-t)\cdot 4\,dt=32\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{16}{3}\approx 5.33$ . The triangle has area $\frac{4\times 4}{2}=8$ . Ratio: $\frac{16/3}{8}=\frac{2}{3}$ , Archimedes' result on the parabolic segment, twenty-two centuries old.

Exercise 14: Problem: two drones on crossing paths

Two drones fly at constant altitude. Their positions, in kilometres, are given as functions of the time $t$ in minutes: drone A at $(t, 2t)$ , drone B at $(4-t, 1+t)$ , for $0\leq t\leq 3$ .

a) Give the Cartesian equation of each path.
b) At which point do the two paths cross?
c) Do the drones collide? Justify using the times at which they pass.
d) Express the square of the distance between the drones in terms of $t$ , then find the minimum distance and the time at which it is reached.
e) By how many seconds would drone B's departure have to be moved earlier for the two drones to pass through the crossing point at the same time?

Show the solution

Answers

a) $y=2x$ and $y=5-x$
b) $\left(\frac{5}{3}, \frac{10}{3}\right)$
c) Passes at $\frac{5}{3}$ and $\frac{7}{3}$ min: no collision
d) $d^{2}=5t^{2}-18t+17$ ; $\approx 0.894$ km at $t=1.8$ min
e) $40$ s

a)

Drone A: $x=t$ , $y=2t$ , so $y=2x$ . Drone B: $x=4-t$ gives $t=4-x$ , and $y=1+t=5-x$ . Two lines.

b)

$2x=5-x$ gives $x=\frac{5}{3}$ and $y=\frac{10}{3}$ : the point $\left(\frac{5}{3}, \frac{10}{3}\right)$ .

c)

A passes there when $t=\frac{5}{3}$ min. B passes there when $4-t=\frac{5}{3}$ , that is $t=\frac{7}{3}$ min. The passes are $\frac{2}{3}$ min apart, that is $40$ s: the PATHS cross, the DRONES do not. The Cartesian equation loses time, exactly as it lost the orientation in exercise 1.

d)

$d^{2}(t)=\left(t-(4-t)\right)^{2}+\left(2t-(1+t)\right)^{2}=(2t-4)^{2}+(t-1)^{2}=5t^{2}-18t+17$ . This quadratic is minimal at $t=\frac{18}{10}=1.8$ min, where $d^{2}=5(3.24)-32.4+17=0.8$ . Minimum distance: $\sqrt{0.8}\approx 0.894$ km, reached BETWEEN the two passes through the crossing point: after A's, at about 1.67 min, and before B's, at about 2.33 min.

e)

B must reach the crossing point at $t=\frac{5}{3}$ instead of $\frac{7}{3}$ : its departure must be moved $\frac{2}{3}$ min earlier, that is $40$ s. In that case, $d^{2}$ would indeed vanish at $t=\frac{5}{3}$ : a collision avoidance system must therefore monitor the times of passage, not only the drawing of the paths. The figure in the solution shows the two paths and the positions at the moment of minimum distance.

Exercise 15: Problem: launching a projectile

A ball is thrown from the ground with a speed of $20$ m/s, at an angle of $30°$ above the horizontal. Air resistance is neglected and $g=9.8$ m/s $^{2}$ . The position is $x=v_{0}\cos\alpha\cdot t$ , $y=v_{0}\sin\alpha\cdot t-\frac{g}{2}t^{2}$ .

The figure below shows the situation; it is not to scale.

a) Write $x(t)$ and $y(t)$ with the numerical values.
b) Eliminate $t$ : what is the Cartesian equation of the path?
c) Compute the time of flight and the range.
d) Compute the maximum height and the time at which it is reached.
e) At what angle does the ball hit the ground? What range would you get by throwing at $45°$ ?

Show the solution

Answers

a) $x\approx 17.32t$ , $y=10t-4.9t^{2}$
b) $y\approx 0.5774x-0.01633x^{2}$
c) $\approx 2.041$ s, $\approx 35.35$ m
d) $\approx 5.10$ m at $\approx 1.020$ s
e) $-30°$ ; $40.82$ m at $45°$

a)

$\cos 30°=\frac{\sqrt{3}}{2}$ and $\sin 30°=\frac{1}{2}$ : $x=10\sqrt{3}\,t\approx 17.32\,t$ and $y=10t-4.9t^{2}$ .

b)

$t=\frac{x}{10\sqrt{3}}$ , so $y=\frac{x}{\sqrt{3}}-4.9\frac{x^{2}}{300}\approx 0.5774\,x-0.01633\,x^{2}$ : a parabola, opening downward.

c)

$y=0$ gives $t(10-4.9t)=0$ , that is $t=\frac{10}{4.9}\approx 2.041$ s. Range: $x=17.32\times 2.041\approx 35.35$ m.

d)

$y'(t)=10-9.8t=0$ for $t=\frac{10}{9.8}\approx 1.020$ s, half the time of flight. Maximum height: $10(1.0204)-4.9(1.0204)^{2}\approx 5.10$ m.

e)

At the ground, $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{10-9.8\times 2.041}{17.32}=\frac{-10}{17.32}\approx-0.577$ , that is an angle of $-30°$ : the ball comes down at the same angle, by the symmetry of the parabola, and with the same speed of $20$ m/s. At $45°$ , the range is $\frac{v_{0}^{2}\sin 90°}{g}=\frac{400}{9.8}\approx 40.82$ m, the maximum range. The figure in the solution compares the two paths.

Corrected exercises: parametric curves (201-SN3-RE)

Course recap

Calculations: slopes to compute

Exercise 0.1: Compute the slope of a parametric curve: ten direct calculations

Part A: Parametrization, tangent lines and shape of the curve (/50)

Exercise 1: From the parametrization to the Cartesian equation, and what gets lost on the way

Exercise 2: The slope, horizontal tangent lines and vertical tangent lines

Exercise 3: The second derivative, concavity and the formula that does not exist

Exercise 4: Double points, symmetries and reading the shape

Exercise 5: The cycloid: the parametrization you cannot do without

Part B: Area, length, surface and invariance under reparametrization (/50)

Exercise 6: The area enclosed by a parametric curve

Exercise 7: Arc length in parametric form

Exercise 8: The area of a surface of revolution

Exercise 9: Reparametrizing: the curve does not change, the integral sometimes does

Exercise 10: Synthesis problem: the astroid and its secret

Part C: The classics (/50)

Exercise 11: A parametrized hyperbola: branch, tangent lines and asymptotes

Exercise 12: The involute of a circle: unwinding a string

Exercise 13: The quadratic Bézier curve: three points are enough

Exercise 14: Problem: two drones on crossing paths

Exercise 15: Problem: launching a projectile

See also

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