Derivatives of Parametric Curves

A parametric derivative is interpreted at a particular parameter value. The same point can be visited twice with different tangent directions. A parameter can also momentarily stop while the underlying geometric path still has a well-defined tangent.

The chain rule gives the usual slope formula. The important additional work is identifying where its denominator vanishes and determining what that means geometrically.

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Derive the slope

If $x'(t)\ne0$ locally, the chain rule gives

\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt},\qquad\frac{dy}{dx}=\frac{y’}{x’}.

This compares changes in two coordinates; it is not the speed $\sqrt{x’^2+y’^2}$ . For physical time, the numerator and denominator are component velocities, while their ratio is geometric slope.

The nonzero condition permits a local graph over $x$ , not necessarily a single global function covering the entire curve.

Work with a looping curve

Let

x=t^2-1,\qquad y=t^3-3t.

Then $x’=2t$ and $y’=3t^2-3$ . For $t\ne0$ ,

\frac{dy}{dx}=\frac{3(t^2-1)}{2t}.

At $t=1$ , the point is $(0,-2)$ and the tangent is horizontal. At $t=-1$ , the point is $(0,2)$ with another horizontal tangent. These are distinct locations, not repeated descriptions of one point.

A loop and its local tangents

Equal axis units; unbounded branches continue beyond the frame.

Diagram explanation and exact data

The parameter distinguishes the two branches through (2,0). At t=0, the tangent at (−1,0) is vertical.

A loop and its local tangents: exact reference data
t	point	slope
−1	(0,2)	0
0	(−1,0)	vertical
1	(0,−2)	0
±√3	(2,0)	±√3

Classify horizontal and vertical tangents

At a regular point, a horizontal tangent has $y’=0$ and $x’\ne0$ . A vertical tangent has $x’=0$ and $y’\ne0$ . The companion nonzero condition prevents a conclusion based on an indeterminate $0/0$ .

At $t=0$ , the loop curve has point $(-1,0)$ , $x’=0$ and $y’=-3$ . Its tangent is the vertical line $x=-1$ .

A tangent-line equation avoiding division is

x'(t_0)(y-y_0)-y'(t_0)(x-x_0)=0.

It covers horizontal and vertical cases at regular points. If both derivative components vanish, it reduces to $0=0$ and does not specify a line.

Differentiate with respect to the correct variable

Differentiating the slope with respect to $t$ alone does not give the second derivative with respect to $x$ . Instead,

\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}(dy/dx)}{x’},\qquad x’\ne0.

For the loop,

\frac{dy}{dx}=\frac32\left(t-\frac1t\right),\qquad\frac{d^2y}{dx^2}=\frac{3(t^2+1)}{4t^3}.

The second derivative is positive for $t>0$ and negative for $t<0$ . These are concavity statements for local graphs over $x$ . At the vertical tangent the formula is unavailable.

The parametric-calculus reference supplies the chain-rule basis. A time derivative of the slope and an $x$ derivative answer different questions.

One coordinate point can have two tangent lines

At $t=\pm\sqrt3$ , the curve passes through $(2,0)$ . Its slopes are respectively $\pm\sqrt3$ , producing lines

y=\sqrt3(x-2),\qquad y=-\sqrt3(x-2).

Asking for the slope at that coordinate point without specifying a branch is ambiguous. The parameter identifies the visit. Eliminating it may describe the same geometric set but does not preserve this distinction automatically.

Examine stationary parameters separately

For $x=t^3$ , $y=t^3$ , both derivatives vanish at zero, but the image is the line $y=x$ with slope one. The parameter momentarily stops along a smooth geometric line.

For $x=t^2$ , $y=t^3$ , both derivatives again vanish at zero. The two-sided curve has a cusp. Its nearby slopes $3t/2$ tend to zero, establishing a horizontal limiting tangent, but the limiting direction of motion changes across the cusp.

The same $0/0$ formula issue can accompany different geometry. Inspect local coordinate expansions, ratios or tangent directions. A stationary parameter is not automatically a cusp; a tangent line is not automatically a regular parametrisation.

Distinguish ellipse slope from ellipse speed

For $x=3\cos t$ , $y=2\sin t$ ,

\frac{dy}{dx}=-\frac{2\cos t}{3\sin t},\qquad\sin t\ne0.

At $t=\pi/2$ the tangent is horizontal at $(0,2)$ , while at zero it is vertical at $(3,0)$ . Speed is

v=\sqrt{9\sin^2t+4\cos^2t},

which varies between two and three. Uniform angular change does not produce uniform speed around an ellipse.

The second derivative is $-2/(9\sin^3t)$ where defined. It is negative on the upper half and positive on the lower half. Reversing traversal does not change this geometric concavity.

Interpret reparameterisation and orientation

A regular monotone parameter change multiplies both coordinate derivatives by the same nonzero factor. Their quotient is unchanged. A negative factor reverses velocity direction but preserves the unoriented tangent line.

If the change-of-parameter derivative vanishes, the new velocity can be zero even where the original parametrisation is regular. Do not cancel that factor at its zero without examining the limit.

Separate geometric and temporal questions

Moving upward concerns $y'(t)$ ; rising as $x$ increases concerns $dy/dx$ . If $x'<0$ , a particle travels leftward, so those descriptions need not agree.

Similarly, $y”(t)$ is vertical acceleration, not graph concavity $d^2y/dx^2$ . A correct derivative with the wrong interpretation is still an incorrect answer. Name the differentiated variable and retain units in applications.

Practice questions

For $x=t^2$ , $y=t^3$ , find $dy/dx$ at one.
Find the point and tangent direction of the loop curve at zero.
If both coordinate derivatives vanish, must the geometric tangent be absent?

Check your answers

The slope is $3t/2$ for nonzero $t$ , giving $3/2$ at one.
The point is $(-1,0)$ and the tangent is vertical because $x’=0$ while $y’=-3\ne0$ .
No. The parametrisation $(t^3,t^3)$ has both derivatives zero at the origin but traces a straight line. Local analysis is necessary.

Sources and further reading

OpenStax, Calculus Volume 2: Calculus of Parametric Curves.