Parametric differentiation — dy - dx, d²y - dx²: Worked examples

This page is the exercise gym for parametric differentiation. The parent note built the two formulas; here we drive them through every kind of situation a problem can present — good slopes, vertical tangents, degenerate points, negative signs, a real-world word problem, and an exam twist.

Recall The two tools we keep reusing (from the parent)

First derivative — the speed ratio (vertical speed divided by horizontal speed): $\frac{d y}{d x} = \frac{y ˙}{x ˙} (\overset{x}{˙} \neq = 0)$ Second derivative — quotient rule plus one extra $\overset{x}{˙}$ underneath: $\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}$ Here a dot means "rate of change with respect to the parameter $t$ ": $\overset{x}{˙} = d x / d t$ , $\overset{x}{¨} = d^{2} x / d t^{2}$ , and likewise for $y$ . If any of that feels new, re-read the parent — this page uses it, it doesn't re-derive it. The engine underneath is the Chain rule and the Quotient rule. Throughout this page, we call these two displayed equations the first-derivative formula and the second-derivative formula.

The scenario matrix

Every parametric-differentiation problem falls into one of these cells. Our worked examples below are labelled with the cell they hit, so by the end no cell is empty.

#	Cell (case class)	What makes it tricky	Example
A	Plain "nice" point, $\overset{x}{˙} \neq = 0$	none — warm-up	Ex 1
B	$\overset{x}{˙} = 0$ → vertical tangent	slope undefined, not an error	Ex 2
C	$\overset{y}{˙} = 0$ → horizontal tangent	slope exactly $0$	Ex 3
D	Sign of $d^{2} y / d x^{2}$ → concavity	up vs down, and sign of $\overset{x}{˙}$ matters	Ex 4
E	Degenerate point: $\overset{x}{˙} = \overset{y}{˙} = 0$ (cusp)	$0/0$ , need a limit	Ex 5
F	Real-world word problem (projectile)	units, choose the right $t$	Ex 6
G	Exam twist: find $t$ giving a required slope	solve backwards	Ex 7
H	Limiting behaviour as $t \to$ special value	asymptotic slope	Ex 8

Each cell shows a different way the same two formulas behave. Watch especially the sign of $\overset{x}{˙}$ in cells B, D, H — it is the quiet troublemaker.

Cell A — the warm-up (nice point)

Worked example 1 — Slope and concavity of

$x = t^{2}, y = t^{3}$ at $t = 2$ Forecast: Guess: is the slope positive or negative at $t = 2$ ? Is the curve bending up or down there? Jot a guess before reading.

Differentiate each coordinate w.r.t. $t$ . Why this step? the formulas are built entirely from $\overset{x}{˙}, \overset{y}{˙}, \overset{x}{¨}, \overset{y}{¨}$ . $\overset{x}{˙} = 2 t, \overset{y}{˙} = 3 t^{2}, \overset{x}{¨} = 2, \overset{y}{¨} = 6 t$
First derivative $= \overset{y}{˙} / \overset{x}{˙}$ . Why this step? slope is the ratio of vertical to horizontal speed. $\frac{d y}{d x} = \frac{3 t ^{2}}{2 t} = \frac{3 t}{2} \Rightarrow at t = 2 : \frac{3 ( 2 )}{2} = 3$
Second derivative via the second-derivative formula recalled at the top of this page. Why this step? we must go through $t$ , not divide accelerations — so we feed $\overset{x}{˙}, \overset{y}{˙}, \overset{x}{¨}, \overset{y}{¨}$ into $\frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}$ . $\frac{d ^{2} y}{d x ^{2}} = \frac{( 6 t ) ( 2 t ) - ( 3 t ^{2} ) ( 2 )}{( 2 t ) ^{3}} = \frac{12 t ^{2} - 6 t ^{2}}{8 t ^{3}} = \frac{6 t ^{2}}{8 t ^{3}} = \frac{3}{4 t}$ At $t = 2$ : $\frac{3}{8}$ .

Verify: slope $3 > 0$ (rising) and $d^{2} y / d x^{2} = 3/8 > 0$ (bending upward) — matches a curve climbing and curving up. Sanity: eliminate $t$ ( $x = t^{2}, y = t^{3} \Rightarrow y = x^{3/2}$ ), so $y^{'} = \frac{3}{2} x^{1/2}$ , and at $t = 2$ , $x = 4$ : $y^{'} = \frac{3}{2} \cdot 2 = 3$ . ✓

Cell B — vertical tangent ( $\overset{x}{˙} = 0$ )

The figure below shows the unit circle in lavender. The red dashed lines are the vertical tangents at the leftmost and rightmost points, with red arrows pointing straight up to stress that the motion there is purely vertical; the mint dotted lines mark the horizontal tangents (top and bottom) for contrast. As you read Example 2, find the two red points $(\pm 1, 0)$ on the picture.

Figure — Parametric differentiation — dy - dx, d²y - dx²

Figure 1 — Unit circle $x = cos t, y = sin t$ . Red dashed = vertical tangents where $\overset{x}{˙} = 0$ (points $(\pm 1, 0)$ ); mint dotted = horizontal tangents where $\overset{y}{˙} = 0$ (points $(0, \pm 1)$ ).

Worked example 2 — Where is the tangent vertical for

$x = cos t, y = sin t$ ? Take $t \in [0, 2 π)$ for the geometric answer, but remember $t$ can be any real number and the pattern repeats every $2 π$ . Forecast: the circle. Where on a circle does the tangent line stand straight up? Predict the points before computing.

Derivatives. Why this step? we need $\overset{x}{˙}$ to see where the run vanishes. $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t$
Set $\overset{x}{˙} = 0$ . Why this step? a vertical tangent means zero horizontal motion — pure up/down. $- sin t = 0 \Rightarrow sin t = 0 \Rightarrow t = nπ (n any integer)$ On the primary lap $[0, 2 π)$ this is $t = 0$ and $t = π$ ; all other solutions land on the same two points because the circle repeats.
Locate those points. At $t = 0$ : $(cos 0, sin 0) = (1, 0)$ . At $t = π$ : $(- 1, 0)$ . Why this step? to name the geometric spots.
Confirm slope blows up — and check its sign on each side. Why this step? an undefined slope is the definition of vertical here; but "which infinity" tells us how the curve leans. Near $t = 0$ , $\frac{d y}{d x} = \frac{c o s t}{- s i n t} = - cot t .$

As $t \to 0^{+}$ : $sin t > 0$ (small positive), $cos t \to 1 > 0$ , so $- cot t = - \frac{cos t}{sin t} \to - \infty$ .
As $t \to 0^{-}$ : $sin t < 0$ (small negative), $cos t \to 1 > 0$ , so $- cot t \to + \infty$ .

The sign of $sin t$ (the denominator) is what flips the infinity: the tangent line is vertical either way, but the curve leaves the point steeply downward on one side and steeply upward on the other. (The same one-sided flip happens at $t = π$ , with the roles of the signs swapped.)

Verify: $(\pm 1, 0)$ are the leftmost/rightmost points of the unit circle; look at the red vertical arrows in Figure 1 — the tangents there are exactly vertical. ✓ The formula is supposed to fail (give $\pm \infty$ ), and it does. See Tangents and normals.

Cell C — horizontal tangent ( $\overset{y}{˙} = 0$ )

Worked example 3 — Top and bottom of the same circle

Same curve $x = cos t, y = sin t$ , again take $t \in [0, 2 π)$ for the geometric answer (the pattern repeats every $2 π$ ). Forecast: where is the tangent perfectly flat (slope $0$ )? Guess the two points.

Recall the slope formula for this curve. Why this step? we need where the slope equals $0$ , so we first re-establish where the slope comes from. Using the first-derivative formula with $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t$ : $\frac{d y}{d x} = \frac{y ˙}{x ˙} = \frac{c o s t}{- s i n t} = - cot t .$
Set $\overset{y}{˙} = 0$ . Why this step? a flat tangent means zero vertical motion, which makes the numerator of the slope vanish. $cos t = 0 \Rightarrow t = \frac{π}{2} + nπ (n any integer)$ On the primary lap $[0, 2 π)$ this is $t = \frac{π}{2}$ and $t = \frac{3 π}{2}$ .
Points: $t = \frac{π}{2} \to (0, 1)$ ; $t = \frac{3 π}{2} \to (0, - 1)$ . Why this step? name the geometric spots.
Check slope is $0$ , not undefined. At $t = \frac{π}{2}$ : $\frac{d y}{d x} = - cot (\frac{π}{2}) = 0$ . Why this step? to be sure we hit "flat" and not "vertical" — here $\overset{x}{˙} = - sin (\frac{π}{2}) = - 1 \neq = 0$ , so the ratio is a clean $0$ (numerator zero, denominator non-zero).

Verify: $(0, \pm 1)$ are the top and bottom of the circle — tangents run horizontally there, slope $0$ . ✓

Cell D — concavity from the sign of $d^{2} y / d x^{2}$

Worked example 4 — Is

$x = cos t, y = sin t$ concave up or down at $t = \frac{π}{4}$ ? Forecast: you're on the upper-right of the circle. Does it curve up (like a valley) or down (like a hill)? Guess.

Derive the circle's second derivative from scratch. Why this step? we should not borrow the parabola's answer — the circle needs its own computation. Feed $\overset{x}{˙} = - sin t, \overset{y}{˙} = cos t, \overset{x}{¨} = - cos t, \overset{y}{¨} = - sin t$ into the second-derivative formula: $\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}} = \frac{( - s i n t ) ( - s i n t ) - ( c o s t ) ( - c o s t )}{( - s i n t ) ^{3}} = \frac{s i n ^{2} t + c o s ^{2} t}{- s i n ^{3} t} = \frac{- 1}{s i n ^{3} t},$ using $sin^{2} t + cos^{2} t = 1$ in the numerator.
Plug $t = \frac{π}{4}$ . Why this step? evaluate the sign at the required point. $sin \frac{π}{4} = \frac{1}{2} \Rightarrow sin^{3} \frac{π}{4} = \frac{1}{2 2} \Rightarrow \frac{d ^{2} y}{d x ^{2}} = \frac{- 1}{1/ ( 2 2 )} = - 22 \approx - 2.828$
Read the sign. Why this step? a negative second derivative means concave down (see Concavity and second derivative).

Verify: on the upper half of a circle the curve arches like a hill — concave down — so a negative value is exactly right. ✓ Contrast: on the lower half $sin t < 0$ , so $- 1/ sin^{3} t > 0$ → concave up (valley-shaped). The sign of $sin t$ (through $sin^{3} t$ ) flips concavity between halves — that's cell D in action.

Cell E — degenerate point (cusp, $\overset{x}{˙} = \overset{y}{˙} = 0$ )

The next figure shows the curve $x = t^{2}, y = t^{3}$ . The lavender branch is traced for $t \leq 0$ , the coral branch for $t \geq 0$ ; both meet at the slate dot at the origin, forming a sharp beak (cusp). The mint arrow shows the common horizontal tangent direction. As you read Example 5, notice that although the tangent is horizontal, the two branches leave the origin on the same line — that is what makes it a cusp rather than a smooth minimum.

Figure 2 — The cusp of $x = t^{2}, y = t^{3}$ at the origin. Both branches ( $t \leq 0$ lavender, $t \geq 0$ coral) meet with a horizontal limiting tangent (mint), yet the point is not smooth.

Worked example 5 — The cusp of

$x = t^{2}, y = t^{3}$ at $t = 0$ Forecast: at $t = 0$ both $\overset{x}{˙}$ and $\overset{y}{˙}$ are $0$ — the formula gives $0/0$ . Does the curve have a slope there, or a sharp spike? Guess.

Check both speeds at $t = 0$ . Why this step? to detect the degeneracy before blindly dividing. $\overset{x}{˙} = 2 t = 0, \overset{y}{˙} = 3 t^{2} = 0 at t = 0$ Both zero — the point $(0, 0)$ is a stationary point of the parametrisation; direct substitution is the forbidden $0/0$ .
Take a limit instead. Why this step? $0/0$ is not "no answer" — it's "look at the trend nearby." From Ex 1, $\frac{d y}{d x} = \frac{3 t}{2}$ . $lim_{t \to 0} \frac{d y}{d x} = lim_{t \to 0} \frac{3 t}{2} = 0$
Interpret. Why this step? the limiting slope is $0$ (horizontal), but the curve actually reverses direction here (a cusp): approaching from $t < 0$ vs $t > 0$ the bug comes in and goes back out along the same tangent.

Verify: $y = x^{3/2}$ for $x \geq 0$ gives $y^{'} = \frac{3}{2} x^{1/2} \to 0$ as $x \to 0^{+}$ — matches the limiting slope $0$ . The picture (Figure 2) shows the sharp beak at the origin: the tangent is horizontal but the point is not smooth. ✓ This is why cell E always needs a limit, never a plug-in.

Cell F — real-world word problem (projectile)

Worked example 6 — A thrown ball

A ball is launched with horizontal speed $u = 3 m/s$ and initial upward speed $v = 4 m/s$ ; gravity $g = 10 m/s^{2}$ . Its path (with $t$ in seconds, positions in metres) is $x = 3 t, y = 4 t - 5 t^{2} .$ (a) What is the slope of the trajectory at $t = 0.2$ s? (b) At what time is the ball momentarily flat (peak)? Forecast: at launch the ball rises steeply; guess whether the slope at $t = 0.2$ is still positive.

Differentiate w.r.t. time. Why this step? $t$ is literally time here; $\overset{x}{˙}, \overset{y}{˙}$ are the velocity components (units m/s). $\overset{x}{˙} = 3, \overset{y}{˙} = 4 - 10 t$
Slope of path $= \overset{y}{˙} / \overset{x}{˙}$ . Why this step? slope on the map is the ratio of the two velocity components — the $d t$ (and the units s) cancel, leaving a pure number. $\frac{d y}{d x} = \frac{4 - 10 t}{3}$
(a) At $t = 0.2$ : $\frac{4 - 10 ( 0.2 )}{3} = \frac{4 - 2}{3} = \frac{2}{3} \approx 0.667$ . Why this step? evaluate at the asked time — still positive, still rising.
(b) Peak = horizontal tangent, $\overset{y}{˙} = 0$ . Why this step? at the top the vertical velocity is zero (cell C in disguise). $4 - 10 t = 0 \Rightarrow t = 0.4 s$

Verify: units check — $\overset{y}{˙} / \overset{x}{˙}$ is $(m/s) / (m/s)$ = dimensionless slope. ✓ Peak height: $y (0.4) = 4 (0.4) - 5 (0.4)^{2} = 1.6 - 0.8 = 0.8$ m, and $\overset{y}{˙} (0.4) = 0$ confirms flatness. ✓

Cell G — exam twist (solve backwards for $t$ )

Cell H — limiting behaviour of the slope

The final figure shows one arch of the cycloid in lavender, with the cusp at the origin. Three short tangent segments are drawn near the cusp — at $t = 0.6$ (mint), $t = 0.3$ (butter) and $t = 0.12$ (coral). Watch how they get steeper as $t$ shrinks toward $0$ : this is the visual meaning of the slope running off to $+ \infty$ in Example 8.

Figure 3 — Cycloid $x = t - sin t, y = 1 - cos t$ . Tangent segments at $t = 0.6, 0.3, 0.12$ steepen toward vertical as the cusp at the origin is approached.

Worked example 8 — Slope of the cycloid

$x = t - sin t, y = 1 - cos t$ as $t \to 0^{+}$ Forecast: at $t = 0$ the cycloid touches the ground with a cusp. Does the tangent become vertical, horizontal, or something in between as we creep up from $t = 0$ ? Guess.

Slope (from parent Example 3). Why this step? reuse known work. $\frac{d y}{d x} = \frac{s i n t}{1 - c o s t}$
Try $t = 0$ directly: $\frac{sin 0}{1 - cos 0} = \frac{0}{0}$ . Why this step? to expose that it's degenerate — a limit is required (cell E's cousin, but now we want the limiting value).
Small- $t$ approximations. Why this step? for tiny $t$ , $sin t \approx t$ and $1 - cos t \approx \frac{t ^{2}}{2}$ — these turn the $0/0$ into something we can read. $\frac{d y}{d x} \approx \frac{t}{t ^{2} /2} = \frac{2}{t} t \to 0^{+} + \infty$
Interpret. Why this step? an infinite limiting slope means the cusp of the Cycloid rises with a vertical tangent.

Verify: numerically at $t = 0.01$ : $\frac{sin 0.01}{1 - cos 0.01} \approx \frac{0.0099998}{0.00005} \approx 200$ , matching $2/ t = 2/0.01 = 200$ . ✓ So the slope genuinely runs off to $+ \infty$ — the cusp is vertical, exactly as Figure 3's steepening tangent lines show.

Active recall

Which cell describes a point where

\overset{x}{˙} = 0

but

\overset{y}{˙} \neq = 0

Cell B — a vertical tangent, slope undefined.

Which cell needs a limit rather than a plug-in?

Cell E (cusp,

\overset{x}{˙} = \overset{y}{˙} = 0

, giving

0/0

) — and cell H for limiting slope.

For

x = t^{2}, y = t^{3}

, what is

d y / d x

and

d^{2} y / d x^{2}

\frac{3 t}{2}

and

\frac{3}{4 t}

At which

t

does

x = cos t, y = sin t

have vertical tangents?

t = nπ

(integer

n

), i.e. points

(\pm 1, 0)

At which

t

does

x = cos t, y = sin t

have horizontal tangents?

t = \frac{π}{2} + nπ

(integer

n

), i.e. points

(0, \pm 1)

Sign of

d^{2} y / d x^{2}

for the circle on its upper half?

Negative → concave down.

Projectile

x = 3 t, y = 4 t - 5 t^{2}

: when is the path flat?

When

\overset{y}{˙} = 4 - 10 t = 0

, i.e.

t = 0.4

s (the peak).

Connections

Parent: the two formulas — this page exercises them.
Tangents and normals — cells B and C are all about tangent lines.
Concavity and second derivative — cell D reads the sign of $d^{2} y / d x^{2}$ .
Cycloid, Parametric curves — cells E and H live here.
Chain rule, Quotient rule — the machinery inside every step.

Scenario map

The flowchart below is a decision tree you can walk for any parametric point. Start at the top with your $x (t), y (t)$ . First ask is $\overset{x}{˙}$ zero? If yes, you have a vertical tangent (Cell B). If no, ask is $\overset{y}{˙}$ zero? — yes gives a horizontal tangent of slope $0$ (Cell C). If neither is zero, ask are they both zero at once? — yes means a cusp needing a limit (Cell E, and its limiting-slope cousin Cell H); no means an ordinary point where you simply compute $\overset{y}{˙} / \overset{x}{˙}$ (Cell A). From that ordinary slope you can then branch to checking concavity (Cell D), interpreting real-world velocities (Cell F), or solving backwards for a required slope (Cell G). In words: zeros first, ordinary slope second, then everything else.