Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: parametric curves (201-SN3-RE)

This sheet does not restate the formulas of the chapter: they fit in four lines. It deals with what decides the mark in 201-SN3-RE (formerly 201-NYB), that is, reading a curve that is not a graph, telling a vertical tangent line from a cusp, and the two formulas in which a factor is systematically lost.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

A parametric curve is not the graph of a function: it can cross itself, go back over its path, stop dead. Almost every lost mark comes from treating $t$ like an $x$ , when $t$ is a TIME and the curve is a path.

This chapter is part of 201-SN3-RE, Integral Calculus

The essentials

What a parametric curve can do, and a graph cannot

• A parametric curve $\left(x(t),y(t)\right)$ can CROSS ITSELF: two values of $t$ give the same point.
• It can have a CUSP: the moving point stops and leaves again in the opposite direction.
• It has an ORIENTATION, given by the direction of increasing $t$ , which must be shown with an arrow.
• The same curve has INFINITELY MANY parametrizations, which trace it at different speeds.
• The vertical line test does not apply: $x$ is not the variable, it is a function of time.

The curve

x=t^{2}-1

y=t^{3}-t

passes through the origin again for

t=-1

and

t=1

: no graph of a function does this, and the double point only shows on the parametrization.

Finding the values of $t$ that give the same point is a frequent question: solve the SYSTEM $x(t_{1})=x(t_{2})$ and $y(t_{1})=y(t_{2})$ .

The four formulas, and the factor that gets lost

• Slope: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y'(t)}{x'(t)}$ , a quotient of two derivatives with respect to $t$ .
• Second derivative: $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)}$ . The denominator $x'(t)$ is not optional.
• Arc length: $L=\int_{a}^{b}\sqrt{\left[x'(t)\right]^{2}+\left[y'(t)\right]^{2}}\,dt$ , with BOTH derivatives.
• Area under the curve: $\mathcal{A}=\int_{a}^{b}y(t)\,x'(t)\,dt$ , where the $x'(t)$ comes from $dx=x'(t)\,dt$ .
• Surface of revolution about the $x$ -axis: $S=2\pi\int\left|y(t)\right|\sqrt{\left[x'\right]^{2}+\left[y'\right]^{2}}\,dt$ .

At a point, the velocity vector has components

x'(t)

and

y'(t)

: the slope of the tangent line is their quotient,

\frac{dy}{dx}=\frac{y'(t)}{x'(t)}

All these formulas come from $dx=x'(t)\,dt$ and $dy=y'(t)\,dt$ . Rebuilding them from there is safer than memorizing them.

Special tangent lines and cusps

• HORIZONTAL tangent: $y'(t)=0$ AND $x'(t)\neq 0$ .
• VERTICAL tangent: $x'(t)=0$ AND $y'(t)\neq 0$ .
• Both zero at the same time: SINGULAR point, often a cusp, and the slope is found with a limit.
• Speed of the moving point: $\sqrt{\left[x'\right]^{2}+\left[y'\right]^{2}}$ . It vanishes exactly at the singular points.

x=t^{2}

y=t^{3}-3t

y'(t)=0

t=\pm 1

gives two horizontal tangent lines (green),

x'(t)=0

t=0

a vertical tangent line (orange).

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Treating the curve as the graph of a function

the whole question, when you try to fix a correct calculation

What not to write

“The curve passes through the origin twice, there is an error in my calculation.”

What to write

“ $t=-1$ and $t=1$ both give the point $(0, 0)$ : it is a double point, perfectly normal for a parametric curve.”

Why: A path can cross itself: a car doing another lap of a race track passes the same spot at a different time. The vertical line test, learned for functions, has no validity here.

2. Forgetting the denominator in the second derivative

the whole concavity question

What not to write

$\frac{d^{2}y}{dx^{2}}=\frac{d}{dt}\left(\frac{dy}{dx}\right)$

What to write

$\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)}$

Why: The second derivative is the derivative of the slope WITH RESPECT TO $x$ , not with respect to $t$ . So the same chain rule is applied a second time, dividing by $x'(t)$ . Without this factor, the sign can even be reversed where $x'(t)<0$ , and the whole concavity table is wrong.

3. Confusing a vertical tangent line with a cusp

1 to 2 marks, and a wrong sketch at the most visible point of the curve

What not to write

“ $x'(t)=0$ , so the tangent line is vertical.”

What to write

“ $x'(t)=0$ AND $y'(t)\neq 0$ gives a vertical tangent line; if $y'(t)=0$ as well, it is a singular point and you must compute $\lim\frac{y'}{x'}$ .”

On the astroid

x=\cos^{3}t

y=\sin^{3}t

, both derivatives vanish at the four sharp points: these are not vertical tangent lines but CUSPS, where the curve forms a point.

Why: On the astroid, both derivatives vanish at the four sharp points: the curve forms a point there, not a vertical tangent line. Checking the other derivative before concluding takes five seconds and keeps you from smoothing out a cusp on the sketch.

4. Eliminating the parameter without carrying over the restrictions

1 to 2 marks

What not to write

“ $x=\cos t$ , $y=\sin t$ , so $x^{2}+y^{2}=1$ : it is the whole circle.”

What to write

“If $t\in\left[0,\pi\right]$ , only the upper HALF-circle is traced, with $y\ge 0$ .”

x^{2}+y^{2}=1

describes the whole circle, but

t\in\left[0,\pi\right]

traces only its upper half (green): the restriction carries over.

Why: Eliminating the parameter gives a Cartesian equation, but it loses the information about the domain. The parametrization traces a PIECE of that curve, and you must say which one by carrying the bounds on $t$ over to $x$ and $y$ .

5. Forgetting one of the two derivatives in the arc length

the whole question

What not to write

$L=\int\sqrt{1+\left[y'(t)\right]^{2}}\,dt$

What to write

$L=\int\sqrt{\left[x'(t)\right]^{2}+\left[y'(t)\right]^{2}}\,dt$

Why: The formula with $1+\left[f'\right]^{2}$ holds for a graph $y=f(x)$ , where $x$ moves forward at speed $1$ . Here $x$ moves at speed $x'(t)$ , which has no reason to equal $1$ : the $1$ is replaced by $\left[x'(t)\right]^{2}$ .

6. Forgetting $x'(t)$ in the area under the curve

the whole question

What not to write

$\mathcal{A}=\int_{a}^{b}y(t)\,dt$

What to write

$\mathcal{A}=\int_{a}^{b}y(t)\,x'(t)\,dt$ , since $dx=x'(t)\,dt$

Why: The area is $\int y\,dx$ , and the change of variable $x=x(t)$ requires translating the $dx$ as well. It is exactly the same omission as forgetting the $dx$ in an ordinary substitution, and it gives a number without the right units.

7. Not showing the orientation

1 mark, the one for the complete sketch

What not to write

“Here is the curve.”

What to write

“Here is the curve, traced in the direction of increasing $t$ , shown by the arrows; it starts at $(3, 0)$ for $t=0$ .”

The ellipse

x=3\cos t

y=2\sin t

starts at

(3, 0)

t=0

and turns counterclockwise: the arrows are part of the answer.

Why: Two parametrizations can give the same geometric curve while tracing it in opposite directions, and questions about speed or position at a given instant depend on the direction. An arrow and the starting point are enough, and the marking scheme expects them.

8. Believing that a parametrization is unique

1 interpretation mark

What not to write

“ $x=t$ , $y=t^{2}$ is THE parametrization of the parabola.”

What to write

“It is A parametrization. $x=2u$ , $y=4u^{2}$ describes the same parabola, twice as fast.”

Why: The geometric curve and the motion that traces it are two distinct objects. The arc length does not depend on the parametrization; the speed does. Knowing which of the two the question asks for keeps you from computing the wrong quantity.

Which method to choose

What happens at the point with parameter $t_{0}$

Compute $x'(t_{0})$ and $y'(t_{0})$ , then read the table. These two numbers decide everything.

If $x'\neq 0$ → tangent line with slope $\frac{y'}{x'}$ , the curve is locally the graph of a function

Example: cycloid at $t=\frac{\pi}{2}$ : slope $\frac{\sin t}{1-\cos t}=1$
If $y'=0$ and $x'\neq 0$ → HORIZONTAL tangent line

Example: cycloid at $t=\pi$ , at the top of the arch
If $x'=0$ and $y'\neq 0$ → VERTICAL tangent line

Example: circle $x=\cos t$ , $y=\sin t$ at $t=0$
If $x'=0$ AND $y'=0$ → SINGULAR point: compute $\lim_{t\to t_{0}}\frac{y'(t)}{x'(t)}$ to get the tangent line, and check whether there is a cusp

Example: astroid at its four sharp points, cycloid where it touches the ground

A singular point is not necessarily a cusp: it can also be an ordinary point passed through while slowing down. The limit of the slope decides.

How to eliminate the parameter

The goal is an equation in $x$ and $y$ alone. The method depends on the form of the two expressions.

If one of the two equations is easy to solve for t → isolate $t$ and substitute into the other

Example: $x=2t+1$ gives $t=\frac{x-1}{2}$ , to be substituted into $y$
If both contain $\cos t$ and $\sin t$ → isolate $\cos t$ and $\sin t$ , then use $\cos^{2}t+\sin^{2}t=1$

Example: $x=3\cos t$ , $y=2\sin t$ gives $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
If both contain powers of the same block → raise to the right power to bring out a common factor

Example: $x=\cos^{3}t$ , $y=\sin^{3}t$ gives $x^{2/3}+y^{2/3}=1$
If no simple elimination appears → do not insist: the study works very well on the parametrization, and the question often does not ask for it

Example: the cycloid has no elementary Cartesian equation

After eliminating, ALWAYS carry the bounds on $t$ over to $x$ and $y$ : the Cartesian equation often describes more than the parametrization.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Studying a parametric curve, written up for the grader

When to use it: A long question asking for tangent lines, special points and a sketch.

1 Give the domain of $t$ and compute $x'(t)$ and $y'(t)$ , factored.
2 Solve $x'(t)=0$ and $y'(t)=0$ separately, and draw up a SIGN table for the two derivatives.
3 Classify each critical value: horizontal tangent line, vertical tangent line, or singular point.
4 Compute the coordinates of each special point, not only its parameter.
5 Sketch following the direction of increasing $t$ , and draw an arrow.

Concluding sentence

At $t=\pi$ , $x'(\pi)=2\neq 0$ and $y'(\pi)=0$ : the tangent line is horizontal at the point $(\pi, 2)$ , the top of the arch.

The trap: Giving the special points by their PARAMETER instead of their coordinates. “Horizontal tangent line at $t=\pi$ ” locates nothing on the sketch; the pair $(x,y)$ is needed.

Marking: Usually 2 marks for the table of the two derivatives, 1 mark per correctly classified special point, 2 marks for the oriented sketch.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Check the slope on two neighbouring points

Compute the coordinates at $t_{0}$ and at $t_{0}+0.01$ , then the quotient of the changes: it must agree with $\frac{y'}{x'}$ .

Cycloid at $t=\frac{\pi}{2}$ : the formula gives $1$ , and the numerical quotient gives $0.995$ ✓.
Check the arc length against the chord

The length found must exceed the distance between the endpoints, and stay below the sum of the variations of $x$ and of $y$ .

An arch of the cycloid goes from $(0, 0)$ to $(2\pi, 0)$ , a chord of $6.28$ ; the length $8$ is plausible, a length of $5$ would not be.
Check that a double point really is double

Substitute the two values of $t$ found into $x$ AND into $y$ : both coordinates must agree, not just one of them.

$t=\pm 1$ gives $x=0$ and $y=0$ in both cases ✓. A single matching coordinate is not enough.
Compare the Cartesian equation with the parametrization

After eliminating, choose a value of $t$ , compute the point, and check that it satisfies the Cartesian equation obtained.

$x=3\cos t$ , $y=2\sin t$ at $t=\frac{\pi}{3}$ gives $(1.5, 1.73)$ , and $\frac{2.25}{9}+\frac{3}{4}=1$ ✓.

The typical problem, taken apart

One arch of the cycloid, from the tangent line to the length

The cycloid is given by $x(t)=t-\sin t$ and $y(t)=1-\cos t$ , for $t\in\left[0, 2\pi\right]$ .

Study the special tangent lines, then compute the length of one arch.

One arch of the cycloid, the path of a point on a wheel rolling without slipping: the curve touches the ground at

t=0

and

t=2\pi

, where the point is momentarily at rest.

Step 1

$x'(t)=1-\cos t$ and $y'(t)=\sin t$ . The first is POSITIVE or zero everywhere, and vanishes only at $t=0$ and $t=2\pi$ .

Why

Factoring and studying the sign of both derivatives is the first line: $1-\cos t\ge 0$ already says that the moving point never goes back horizontally.

Step 2

Horizontal tangent: $y'(t)=\sin t=0$ with $x'(t)\neq 0$ , so $t=\pi$ , at the point $\left(\pi, 2\right)$ , the top of the arch.

Why

Both conditions are checked. The point is given by its coordinates, not by its parameter: that is what lets you place it on the sketch.

Step 3

At $t=0$ and $t=2\pi$ : $x'=0$ AND $y'=0$ , SINGULAR points. The slope is found from $\lim_{t\to 0^{+}}\frac{\sin t}{1-\cos t}$ , of the form $\frac{0}{0}$ , which equals $\lim\frac{\cos t}{\sin t}=+\infty$ : the tangent line is vertical, and the point is a cusp.

Why

Both derivatives being zero rules out the formula $\frac{y'}{x'}$ : you must go through the limit, and L'Hôpital's rule gives it. Concluding that the tangent is vertical without this computation would be luck, not reasoning.

Step 4

Length: $L=\int_{0}^{2\pi}\sqrt{\left(1-\cos t\right)^{2}+\sin^{2}t}\,dt=\int_{0}^{2\pi}\sqrt{2-2\cos t}\,dt$ .

Why

Expanding gives $1-2\cos t+\cos^{2}t+\sin^{2}t=2-2\cos t$ , thanks to the Pythagorean identity. That is the simplification the formula expects, and it makes the radicand integrable.

Step 5

Half-angle identity: $1-\cos t=2\sin^{2}\frac{t}{2}$ , so $\sqrt{2-2\cos t}=2\left|\sin\frac{t}{2}\right|=2\sin\frac{t}{2}$ on $\left[0, 2\pi\right]$ , where the sine of the half-angle is positive.

Why

The absolute value is removed by justifying the SIGN on the interval: on $\left[0, 2\pi\right]$ , $\frac{t}{2}$ ranges over $\left[0,\pi\right]$ where the sine is positive. This justification is worth half a mark.

Step 6

$L=\int_{0}^{2\pi}2\sin\frac{t}{2}\,dt=\left[-4\cos\frac{t}{2}\right]_{0}^{2\pi}=-4(-1)+4(1)=8$ . Check: the chord is $2\pi\approx 6.28$ , and $8>6.28$ ✓.

Why

The result is an integer, a good sign of a correct calculation, and it exceeds the chord as every arc length must. Two checks for the price of one.

The conclusion, written out

The arch of the cycloid has a horizontal tangent line at $\left(\pi, 2\right)$ , two cusps with vertical tangent lines at its ends, and a length of $8$ units, exactly four times the diameter of the wheel.

The classic mistake on this problem: Writing $\frac{dy}{dx}=\frac{\sin t}{1-\cos t}$ at $t=0$ and concluding “form $\frac{0}{0}$ , so no tangent line”. The indeterminate form rules nothing out: it calls for a limit, which exists here and is infinite, hence a vertical tangent line.

Learn by heart

• A parametric curve can cross itself and have cusps: the vertical line test does not apply.
• $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$ , a quotient of two derivatives with respect to $t$ .
• Second derivative: differentiate the SLOPE with respect to $t$ , then DIVIDE by $x'(t)$ .
• Horizontal tangent: $y'=0$ and $x'\neq 0$ . Vertical: $x'=0$ and $y'\neq 0$ .
• Both zero: singular point, the tangent line is read from $\lim\frac{y'}{x'}$ .
• Length: $\int\sqrt{\left[x'\right]^{2}+\left[y'\right]^{2}}\,dt$ , never $\sqrt{1+\left[y'\right]^{2}}$ .
• Area: $\int y(t)\,x'(t)\,dt$ , the $x'(t)$ coming from $dx=x'(t)\,dt$ .
• After eliminating the parameter, carry the bounds on $t$ over to $x$ and $y$ .

Frequently asked questions

Why can a parametric curve cross itself?

Because it describes a path over time, not the graph of a function. Two different instants can correspond to the same point of the plane, exactly as a car passes the same spot again after a lap of a race track. The vertical line test, valid for functions, does not apply here.

How do you compute the slope of a parametric curve?

Divide the derivative of y with respect to the parameter by the derivative of x with respect to the parameter. This comes from the chain rule: the two small dt cancel out. The formula only holds where the derivative of x does not vanish.

What is the difference between a vertical tangent line and a cusp?

A vertical tangent line corresponds to a zero derivative of x with a nonzero derivative of y: the moving point keeps going. A cusp corresponds to both derivatives being zero: the moving point stops and leaves again, forming a sharp point. So always check both derivatives before concluding.

Why does the arc length contain both derivatives?

Because the length of a small piece of path is the square root of the sum of the squares of the two displacements, by Pythagoras. On the graph of a function, x moves forward at speed one and the formula simplifies. Here both coordinates move at their own speed, and neither equals one.

Do you have to eliminate the parameter to study the curve?

No, and often it is not even possible. The cycloid has no elementary Cartesian equation, which prevents neither the computation of the tangent lines nor that of the length. When elimination is possible, remember to carry over the bounds on the parameter, otherwise the equation obtained describes more than the curve being studied.

Practise it

Corrected exercises: Parametric curves, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-parametric-curves. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: parametric curves (201-SN3-RE)

The essentials

What a parametric curve can do, and a graph cannot

The four formulas, and the factor that gets lost

Special tangent lines and cusps

The mistakes that cost marks

1. Treating the curve as the graph of a function

2. Forgetting the denominator in the second derivative

3. Confusing a vertical tangent line with a cusp

4. Eliminating the parameter without carrying over the restrictions

5. Forgetting one of the two derivatives in the arc length

6. Forgetting x′(t)x'(t)x′(t) in the area under the curve

7. Not showing the orientation

8. Believing that a parametrization is unique

Which method to choose

What happens at the point with parameter t0t_{0}t0​

How to eliminate the parameter

How the answer is expected to be written

Studying a parametric curve, written up for the grader

Check before you hand in

The typical problem, taken apart

One arch of the cycloid, from the tangent line to the length

Learn by heart

Frequently asked questions

Why can a parametric curve cross itself?

How do you compute the slope of a parametric curve?

What is the difference between a vertical tangent line and a cusp?

Why does the arc length contain both derivatives?

Do you have to eliminate the parameter to study the curve?

Corrected exercises: Parametric curves, 201-SN3-RE

See also

Looking for a 201-SN3-RE tutor in Montreal?

6. Forgetting $x'(t)$ in the area under the curve

What happens at the point with parameter $t_{0}$