Parametric differentiation — dy - dx, d²y - dx²: Question bank

True or false — justify

True or false: The slope

d y / d x

of a parametric curve depends on how fast you walk along it.

False. Both

d y

and

d x

carry a common factor

d t

that cancels; the slope is a pure ratio

\overset{y}{˙} / \overset{x}{˙}

, so doubling your speed changes nothing about the path's steepness.

True or false: If

\overset{x}{˙} \neq = 0

everywhere, the curve can still have a vertical tangent.

False. A vertical tangent needs the run to vanish, i.e.

\overset{x}{˙} = 0

; if

\overset{x}{˙}

is never zero the tangent is never vertical (though

\overset{y}{˙} = 0

can still give horizontal ones).

True or false:

\frac{d ^{2} y}{d x ^{2}} = \frac{y ¨}{x ¨}

because you "differentiate twice the same way."

False. The second derivative means differentiate

d y / d x

with respect to $x$ , not

t

; going back through

t

produces a quotient-rule expression over

\overset{x}{˙}^{3}

, not a simple ratio of accelerations.

True or false: The formula

\frac{y ¨ x ˙ - y ˙ x ¨}{x ˙ ^{3}}

has

\overset{x}{˙}^{3}

only because of the Quotient rule.

False. The quotient rule gives

\overset{x}{˙}^{2}

; the third power comes from one extra division by

\overset{x}{˙}

in the final step

\frac{d}{d x} (y^{'}) = \frac{d ( y ^{'} ) / d t}{d x / d t}

True or false: If both

\overset{x}{˙} = 0

and

\overset{y}{˙} = 0

at a point, the curve is smooth there.

False. That is a singular point (a possible cusp or corner, like a Cycloid at its base); the tangent direction is genuinely undefined and

\overset{y}{˙} / \overset{x}{˙}

gives the indeterminate

0/0

True or false: For

x = t, y = f (t)

, the parametric second derivative reduces to ordinary

f^{''} (x)

True. Then

\overset{x}{˙} = 1, \overset{x}{¨} = 0

, so the formula becomes

\overset{y}{¨} /1 = \overset{y}{¨} = f^{''} (t) = f^{''} (x)

— the parametric machinery correctly reproduces plain calculus.

True or false: The sign of

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

tells you concave up or down regardless of the direction

t

increases.

True. Concavity is a geometric property of the shape in the

x y

-plane; reversing

t

flips the sign of both

\overset{x}{˙}

and

\overset{y}{˙}

but leaves the concavity expression unchanged (see the "Why" section below).

True or false: A parametric curve where

\overset{x}{˙} < 0

over some interval must be drawn wrong.

False.

\overset{x}{˙} < 0

just means the point moves leftward there; the curve can double back over the same

x

-values, which is exactly why we use a parameter instead of

y = f (x)

Spot the error

Student writes

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{y ˙}{x ˙})

and stops. What's missing?

They differentiated w.r.t.

t

but never converted back to

x

; they must still divide by

\overset{x}{˙}

, giving the extra factor that turns

\overset{x}{˙}^{2}

into

\overset{x}{˙}^{3}

Student writes

\frac{d y}{d x} = \frac{x ˙}{y ˙}

. Why is this upside down?

The variable you solve for (here

y

) goes on top and the variable you differentiate against (

x

) on the bottom;

\overset{y}{˙} / \overset{x}{˙}

is rise-over-run, and flipping it gives run-over-rise.

Student computes the numerator as

\overset{y}{˙} \overset{x}{¨} - \overset{y}{¨} \overset{x}{˙}

. What went wrong?

The two terms are swapped in sign. The Quotient rule on

\overset{y}{˙} / \overset{x}{˙}

is (bottom · derivative-of-top − top · derivative-of-bottom), i.e.

\overset{y}{¨} \overset{x}{˙} - \overset{y}{˙} \overset{x}{¨}

; reversing it negates the whole second derivative.

Student cancels

d t

"because it's small" rather than "because it's a common factor." Fix the reasoning.

Smallness is irrelevant —

d t

cancels because it multiplies both

\overset{x}{˙}

and

\overset{y}{˙}

identically, so it divides out algebraically; the cancellation would hold for any nonzero

d t

Student claims

\overset{x}{˙}^{3}

in the denominator can be replaced by

∣ \overset{x}{˙} ∣^{3}

"to be safe." Is that right?

No. The sign of

\overset{x}{˙}

genuinely matters — it encodes whether the point moves right or left — and the derivation gives

\overset{x}{˙}^{3}

exactly; forcing absolute value would corrupt the sign of the concavity.

Student uses Chain rule as

\frac{d y}{d t} = \frac{d y}{d x} + \frac{d x}{d t}

. Spot the error.

The chain rule multiplies, it doesn't add:

\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}

. Solving that for

d y / d x

is exactly what gives the slope formula.

Student says at a horizontal tangent "

\overset{x}{˙} = 0

." Correct them.

A horizontal tangent needs zero rise, i.e.

\overset{y}{˙} = 0

(with

\overset{x}{˙} \neq = 0

). Zero run

\overset{x}{˙} = 0

is the condition for a vertical tangent — they've swapped the two.

Why questions

Why must

x

and

y

share the same parameter

t

for this method to work?

Because the slope is the ratio of their changes at the same instant; if they were driven by different clocks there'd be no common

d t

to cancel and no single moving point to trace the curve.

Why do we need the Quotient rule specifically for the second derivative but not the first?

The first derivative is a single division

\overset{y}{˙} / \overset{x}{˙}

; the second requires differentiating that whole fraction w.r.t.

t

, and differentiating a fraction is precisely what the quotient rule is for.

Why is Implicit differentiation a valid alternative when a curve isn't given parametrically?

When there's a relation

F (x, y) = 0

but no parameter, implicit differentiation gets

d y / d x

directly by treating

y

as an unknown function of

x

— it answers the same "slope without

y = f (x)

" question by a different route.

Why does the condition

\overset{x}{˙} \neq = 0

appear in the first derivative but

\overset{x}{˙}^{3} \neq = 0

(same thing) haunt the second?

Both fail exactly where

\overset{x}{˙} = 0

; the cube just makes the second derivative blow up faster, but the underlying geometry — a vertical tangent has no finite slope to differentiate — is identical.

Why can two very different parametrisations of the same curve give the same

d y / d x

at a shared point?

Because slope is an intrinsic property of the curve's shape, not of the parametrisation; reparametrising rescales

\overset{x}{˙}

and

\overset{y}{˙}

by the same chain-rule factor, which cancels in the ratio.

Why is the "cube on the bottom" a reliable fingerprint of a correct second derivative?

Any correct derivation ends with a division by

\overset{x}{˙}

after the quotient rule already produced

\overset{x}{˙}^{2}

; seeing

\overset{x}{˙}^{3}

confirms both steps happened, while

\overset{x}{˙}^{2}

betrays the forgotten final division.

Edge cases

At a point where

\overset{x}{˙} = 0

but

\overset{y}{˙} \neq = 0

, what does the curve look like and what is

d y / d x

A vertical tangent: the run is zero while the rise is nonzero, so

d y / d x = \overset{y}{˙} /0

is undefined (slope

\to \pm \infty

) — a legitimate feature, not a mistake.

At a point where

\overset{y}{˙} = 0

but

\overset{x}{˙} \neq = 0

, what is the tangent?

A horizontal tangent:

d y / d x = 0/ \overset{x}{˙} = 0

, so the curve is momentarily flat, like the top of a circle at

t = π /2

in the parent's Example 1.

What happens to

d y / d x

and

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

t

approaches a value where

\overset{x}{˙} \to 0^{+}

versus

\overset{x}{˙} \to 0^{-}

The slope diverges to

+ \infty

- \infty

depending on the sign of

\overset{y}{˙}

, and the concavity (with

\overset{x}{˙}^{3}

underneath) can flip sign across the vertical tangent — a genuine geometric transition, not a discontinuity error.

For the Cycloid

x = t - sin t, y = 1 - cos t

, why is the base point (

t = 2 π k

) a cusp and not a smooth min?

There both

\overset{x}{˙} = 1 - cos t = 0

and

\overset{y}{˙} = sin t = 0

, so

\overset{y}{˙} / \overset{x}{˙} = 0/0

is indeterminate; the point comes momentarily to rest and reverses direction, producing a sharp cusp rather than a rounded valley.

If a curve retraces its steps (same

(x, y)

at two different

t

values), can it have two different slopes there?

Yes. A self-intersection can have distinct tangent directions on each pass because

\overset{y}{˙} / \overset{x}{˙}

is evaluated at each

t

separately — this is precisely a case

y = f (x)

could never describe.

What is

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

for a straight-line path

x = a t, y = b t

with constants

a \neq = 0, b

Zero. Here

\overset{x}{¨} = \overset{y}{¨} = 0

, so the numerator

\overset{y}{¨} \overset{x}{˙} - \overset{y}{˙} \overset{x}{¨} = 0

; a line has no bending, exactly as concavity zero should say.

Connections

Chain rule — the reason $d y / d t = \frac{d y}{d x} \cdot \frac{d x}{d t}$ and the source of the final $\div \overset{x}{˙}$ .
Quotient rule — differentiates $\overset{y}{˙} / \overset{x}{˙}$ and supplies the $\overset{x}{˙}^{2}$ .
Implicit differentiation — the parameter-free alternative referenced above.
Tangents and normals — where vertical/horizontal tangent edge cases matter.
Concavity and second derivative — interprets the sign of $d^{2} y / d x^{2}$ .
Cycloid, Parametric curves — the cusp and self-intersection edge cases live here.