A Worked Parametric Curve Analysis

Introduction

A parametrization gives the coordinates of a moving point in terms of a parameter $t$: $x=x(t)$ and $y=y(t)$. A complete curve analysis connects these coordinate functions to the geometric path: which points are traced, in what order, how quickly the point moves, and how the curve behaves near its tangencies.

We will analyze one curve throughout this tutorial:

x(t)=t^2+1,\qquad y(t)=t^3-3t,\qquad -2\le t\le 2.

The finite parameter interval makes the endpoints part of the analysis. The example also has a vertical tangent, horizontal tangencies, and branches with different concavity. These features make it useful for seeing how the various calculations fit together.

Coordinate Derivatives and the Curve’s Trace

The trace is the set of points in the plane reached as $t$ ranges over its domain. First eliminate $t$ to describe those points. From $x=t^2+1$, we have $t^2=x-1$, so $x$ ranges from $1$ to $5$. Also, $y=t(t^2-3)=t(x-4)$. Squaring this equation gives a relation involving only $x$ and $y$:

y^2=t^2(x-4)^2=(x-1)(x-4)^2,\qquad 1\le x\le 5.

This relation describes the trace for the given parameter interval. Indeed, for every $x$ in $[1,5]$, the values $t=\pm\sqrt{x-1}$ are in $[-2,2]$, and the parametrization supplies the corresponding points on the relation. At $x=1$, both choices give the same parameter $t=0$.

The trace alone does not tell us which point is visited first. For orientation, calculate the coordinate derivatives:

x'(t)=2t,\qquad y'(t)=3t^2-3=3(t^2-1).

For $t<0$, $x'(t)<0$, so the point moves toward smaller $x$-values; for $t>0$, $x'(t)>0$, so it moves toward larger $x$-values. Meanwhile $y'(t)>0$ when $|t|>1$, and $y'(t)<0$ when $|t|<1$. The parameter values $-1,0,1$ therefore divide the motion into intervals with different coordinate behavior.

Definition: The speed of a parametrized curve is the magnitude of the derivative pair: $$ v(t)=\sqrt{(x'(t))^2+(y'(t))^2}. $$ It is defined for every parameter where the coordinate derivatives exist.

Here the speed is

v(t)=\sqrt{4t^2+9(t^2-1)^2}.

The derivative pair $(x'(t),y'(t))$ also gives the direction of motion. A point is regular when this pair is not $(0,0)$; at a regular point, it supplies a nonzero tangent direction. We will use these facts to analyze tangents as well as speed.

Worked Example: Trace and Orientation

Find the endpoints and turning locations in the coordinate values, then describe the order in which the curve is traversed.

At the endpoints,

\begin{aligned} t=-2 &: (x,y)=(4+1,\,-8+6)=(5,-2),\\ t=2 &: (x,y)=(4+1,\,8-6)=(5,2). \end{aligned}

At the derivative zeros inside the interval,

\begin{aligned} t=-1 &: (x,y)=(2,\, -1+3)=(2,2),\\ t=0 &: (x,y)=(1,0),\\ t=1 &: (x,y)=(2,\,1-3)=(2,-2). \end{aligned}

As $t$ increases from $-2$ to $0$, $x$ decreases from $5$ to $1$; from $0$ to $2$, $x$ increases from $1$ to $5$. The $y$-coordinate increases on $(-2,-1)$, decreases on $(-1,1)$, and increases on $(1,2)$, as follows from the sign of $y'$. Thus the path starts at $(5,-2)$, passes through $(2,2)$ and $(1,0)$, then through $(2,-2)$, and ends at $(5,2)$. The endpoint coordinates and the displayed intermediate values follow directly by substitution into the parametrization.

The same point can be reached at two different parameter values. For example, $y=0$ when $t=0$ or $t=\pm\sqrt{3}$. The latter two values give $(x,y)=(4,0)$, so the path passes through that point twice. The trace relation describes the set of points, while the parameter order describes the traversal, including repeated visits.

Worked Example: Speed and Tangent Lines

Find the speed and tangent line at $t=-1$, $t=0$, and $t=1$.

At $t=-1$, the point is $(2,2)$, and the derivative pair is

(x'(-1),y'(-1))=(-2,0),\qquad v(-1)=\sqrt{(-2)^2+0^2}=2.

The tangent is horizontal because the derivative pair has zero $y$-component and nonzero $x$-component. Its equation through $(2,2)$ is $y=2$. Substitution of the point gives $2=2$, and the derivative pair confirms the horizontal direction.

At $t=1$, the point is $(2,-2)$, and

(x'(1),y'(1))=(2,0),\qquad v(1)=\sqrt{2^2+0^2}=2.

The tangent is horizontal, with equation $y=-2$. Substituting $(2,-2)$ verifies that the point lies on this line.

At $t=0$, the point is $(1,0)$, and

(x'(0),y'(0))=(0,-3),\qquad v(0)=\sqrt{0^2+(-3)^2}=3.

The nonzero derivative pair points vertically, so the tangent line is $x=1$. The point $(1,0)$ lies on that line. All three points are regular because their derivative pairs are nonzero.

Finding Tangencies from the Derivative Pair

At a regular parameter value, a horizontal tangent occurs when $y'(t)=0$ and $x'(t)\ne 0$. A vertical tangent occurs when $x'(t)=0$ and $y'(t)\ne 0$. Requiring the other component to be nonzero matters: if both derivatives vanish, the derivative pair does not directly give a tangent direction, and the point needs separate analysis.

For this curve, $y'(t)=3(t^2-1)$ vanishes at $t=\pm1$. At both values $x'(t)=2t\ne0$, so they give the horizontal tangencies already computed. The equation $x'(t)=2t=0$ gives $t=0$, where $y'(0)=-3\ne0$, so this is a vertical tangency. This checks all horizontal and vertical tangent parameters in the stated domain.

When $x'(t)\ne0$, the slope of the tangent line is $dy/dx=y'/x'$. At $t=2$, for example, the point is $(5,2)$, and the slope is $9/4$. Thus its tangent line is $y-2=\frac94(x-5)$. Substitution of $(5,2)$ gives $0=0$, and the slope agrees with the ratio of the coordinate derivatives.

Takeaway: At a regular parameter value, use $(x'(t),y'(t))$ to identify tangent direction. Use the ratio $y'/x'$ for the slope only when $x'(t)\ne0$; when $x'(t)=0$ and $y'(t)\ne0$, the tangent is vertical.

Concavity on the Branches

Concavity as a graph $y=f(x)$ is described by the sign of $d^2y/dx^2$. This formula applies on a parameter interval where $x'(t)\ne0$, so that $x$ can be used locally as the independent variable. By the chain rule, differentiating with respect to $x$ on such an interval means applying $\frac{1}{x'(t)}\frac{d}{dt}$. Therefore,

\frac{dy}{dx}=\frac{y'}{x'},\qquad \frac{d^2y}{dx^2} =\frac{1}{x'}\frac{d}{dt}\left(\frac{y'}{x'}\right) =\frac{y''x'-y'x''}{(x')^3}.

The last equality follows by the quotient rule: the derivative with respect to $t$ of $y'/x'$ is $(y''x'-y'x'')/(x')^2$, and dividing by $x'$ gives the displayed expression. The condition $x'\ne0$ is essential: this calculation does not give the graph concavity at a vertical tangent.

Worked Example: Concavity on Each Branch

Compute the second derivative and determine its sign on the two intervals separated by the vertical tangent.

Here $x'=2t$, $x''=2$, $y'=3t^2-3$, and $y''=6t$. For $t\ne0$, substitute these values into the formula:

\begin{aligned} \frac{d^2y}{dx^2} &=\frac{(6t)(2t)-(3t^2-3)(2)}{(2t)^3}\\ &=\frac{12t^2-6t^2+6}{8t^3}\\ &=\frac{6(t^2+1)}{8t^3} =\frac{3(t^2+1)}{4t^3}. \end{aligned}

Since $t^2+1>0$, the sign is determined by $t^3$. Hence $d^2y/dx^2<0$ on $(-2,0)$, so the left-hand branch is concave down there; $d^2y/dx^2>0$ on $(0,2)$, so the right-hand branch is concave up there. At $t=0$, $x'=0$, and the formula is not defined. Concavity as a graph $y=f(x)$ is therefore not assigned there by this calculation.

As a numerical check at $t=2$, the formula gives $3(2^2+1)/(4\cdot2^3)=15/32$. Differentiating the slope directly gives the same value: $dy/dx=(3t^2-3)/(2t)=\frac32(t-1/t)$, so its $t$-derivative at $2$ is $15/8$; dividing by $x'(2)=4$ gives $15/32$.

A Common Calculation Error

A common mistake is to use $x''$ in place of $x'$ in the first product of the second-derivative numerator. For this curve, $x''=2$, but the first product must use $x'=2t$. At $t=2$, incorrectly substituting $2$ for $x'$ gives

(6t)(2)-(3t^2-3)(2)\bigg|_{t=2} =24-18=6,

and would falsely give $6/64=3/32$. The correct numerator uses $x'(2)=4$:

(6\cdot2)(4)-(3\cdot2^2-3)(2) =48-18=30,\qquad \frac{30}{(2\cdot2)^3}=\frac{30}{64}=\frac{15}{32}.

The distinction is simple but important: $x'$ appears in the formula’s first product and in its denominator, while $x''$ appears in the second product. Checking the derivative values separately before substitution prevents this error.

Putting the Analysis Together

The coordinate functions and their derivatives answer different questions. Eliminating the parameter describes the trace; signs of $x'$ and $y'$ describe how the point moves along it; the magnitude of the derivative pair gives speed; and the pair itself identifies tangent directions at regular points. Concavity requires one further step and an additional condition, $x'\ne0$, because it is expressed as a second derivative with respect to $x$.

For the example, the trace lies on $y^2=(x-1)(x-4)^2$ with $1\le x\le5$. The path starts at $(5,-2)$, reaches $(1,0)$, and finishes at $(5,2)$. It has horizontal tangents at $(2,2)$ and $(2,-2)$, and a vertical tangent at $(1,0)$. Its graph branches are concave down for negative parameters and concave up for positive parameters. The parameter $t=0$ separates those branches and is precisely where the $y$-as-a-function-of-$x$ second-derivative formula fails.

Check Your Understanding

Use the parametrization $x=t^2+1,\ y=t^3-3t$, with $-2\le t\le2$, to answer these questions.

What are the trace equation and the allowed range of $x$?
In which parameter intervals does the $x$-coordinate decrease and increase?
Find the speed at $t=0$ and at $t=1$.
Why are $t=-1$ and $t=1$ horizontal tangencies, while $t=0$ is a vertical tangency?
On which parameter intervals is the curve concave down or concave up as a graph of $y$ against $x$, and why is $t=0$ excluded from the second-derivative formula?

A Worked Parametric Curve Analysis

What You'll Learn

Introduction

Coordinate Derivatives and the Curve’s Trace

Worked Example: Trace and Orientation

Worked Example: Trace and Orientation

Worked Example: Speed and Tangent Lines

Worked Example: Speed and Tangent Lines

Finding Tangencies from the Derivative Pair

Concavity on the Branches

Worked Example: Concavity on Each Branch

Worked Example: Concavity on Each Branch

A Common Calculation Error

Putting the Analysis Together

Check Your Understanding

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