Concavity of Parametrized Curves

Concavity for a Parametrized Curve

A parametrized curve is described by two functions, \(x=x(t)\) and \(y=y(t)\). To determine its concavity, we want to know how the slope of the curve changes as \(x\) increases. That last phrase matters: the parameter \(t\) may move in the direction of increasing \(x\), or it may move in the opposite direction.

When \(x'(t)\ne 0\) throughout an interval, the curve can locally be viewed as a graph \(y=f(x)\) on the corresponding branch. The sign of \(d^2y/dx^2\) then gives the usual concavity information: a positive second derivative means concave up, and a negative second derivative means concave down. In parametric calculations, however, the second derivative must be computed with respect to \(x\), not merely with respect to \(t\).

Definition: On an interval where a curve is the graph of a twice-differentiable function \(y=f(x)\), the graph is concave up where \(f''(x)>0\), and concave down where \(f''(x)<0\). For a parametrization, this test applies on parameter intervals where \(x'(t)\ne 0\) and the branch is a graph over \(x\).

The Parametric Concavity Formula

Assume that \(x(t)\) and \(y(t)\) have continuous second derivatives on an interval and that \(x'(t)\ne 0\) there. The parametric slope formula is \(dy/dx=y'/x'\). To find the second derivative with respect to \(x\), differentiate that slope with respect to \(t\), then divide by \(dx/dt\). This gives

\frac{d^2y}{dx^2} = \frac{1}{x'(t)}\frac{d}{dt}\left(\frac{y'(t)}{x'(t)}\right) = \frac{x'(t)y''(t)-y'(t)x''(t)}{[x'(t)]^3}.

The denominator is the cube of \(x'(t)\), so its sign is the sign of \(x'(t)\). It is not always positive. In particular, if \(x'(t)<0\), the curve is moving toward decreasing \(x\) as \(t\) increases, and the sign of the numerator alone does not determine concavity.

Sign test: At parameter values where \(x'(t)\ne 0\), calculate \(x'y''-y'x''\) and divide by \((x')^3\). A positive result means concave up; a negative result means concave down. Keep the sign of \(x'\) in the denominator.

Why the Formula Works

The key point is that \(d/dx\) and \(d/dt\) are different operations. Along a parametrized branch, a small change in \(t\) produces a change in \(x\), so differentiating a quantity with respect to \(x\) requires dividing its \(t\)-derivative by \(x'(t)\).

Find the slope.
When \(x'(t)\ne 0\), the chain rule gives \(dy/dx=(dy/dt)/(dx/dt)=y'/x'\).

Differentiate the slope with respect to \(t\).
The quotient rule gives \(d(y'/x')/dt=(x'y''-y'x'')/(x')^2\).

Convert the derivative back to \(x\).
Divide by \(dx/dt=x'\), producing \((x'y''-y'x'')/(x')^3\).

This also explains the condition \(x'(t)\ne 0\): the conversion from differentiation with respect to \(t\) to differentiation with respect to \(x\) requires division by \(x'(t)\). On an interval where \(x'\) is continuous and never zero, it has one sign throughout. The mean value theorem then shows that \(x(t)\) is strictly monotone on that interval, so the parametrization traces a graph branch there.

Worked Examples

Worked Example: A Curve Moving Toward Increasing \(x\)

Let \(x=2t+1\) and \(y=t^3\). Determine the concavity on the parameter intervals \(t<0\) and \(t>0\).

First calculate the derivatives:

x'=2,\qquad x''=0,\qquad y'=3t^2,\qquad y''=6t.

Substitute into the second-derivative formula:

\frac{d^2y}{dx^2} = \frac{(2)(6t)-(3t^2)(0)}{2^3} = \frac{12t}{8} = \frac{3t}{2}.

For \(t<0\), this is negative, so the curve is concave down. For \(t>0\), it is positive, so the curve is concave up. Since \(x'=2>0\), \(x\) increases with \(t\). From \(x=2t+1\), the intervals correspond to \(x<1\) and \(x>1\), respectively.

Check the calculation by eliminating the parameter: \(t=(x-1)/2\), so \(y=(x-1)^3/8\). Differentiating gives \(dy/dx=3(x-1)^2/8\) and \(d^2y/dx^2=3(x-1)/4\). Since \(3(x-1)/4=3t/2\), this agrees with the parametric result.

Worked Example: A Curve Moving Toward Decreasing \(x\)

Let \(x=-t\) and \(y=t^2\). Determine the concavity for all real \(t\).

The derivatives are \(x'=-1\), \(x''=0\), \(y'=2t\), and \(y''=2\). Therefore,

\frac{d^2y}{dx^2} = \frac{(-1)(2)-(2t)(0)}{(-1)^3} = \frac{-2}{-1} = 2.

The second derivative is positive, so the curve is concave up. Here \(x\) decreases as \(t\) increases. Eliminating the parameter confirms the result: \(t=-x\), so \(y=x^2\), and \(d^2y/dx^2=2\).

This example shows why the denominator's sign cannot be ignored: both the numerator and denominator are negative, and their quotient is positive.

Worked Example: Two Branches with Different Concavity

Let \(x=t^2\) and \(y=t^3\). Analyze the concavity separately for \(t>0\) and \(t<0\).

Compute the derivatives:

x'=2t,\qquad x''=2,\qquad y'=3t^2,\qquad y''=6t.

For \(t\ne 0\), substitution gives

\frac{d^2y}{dx^2} = \frac{(2t)(6t)-(3t^2)(2)}{(2t)^3} = \frac{12t^2-6t^2}{8t^3} = \frac{6t^2}{8t^3} = \frac{3}{4t}.

When \(t>0\), the second derivative is positive, so this branch is concave up. When \(t<0\), it is negative, so that branch is concave down. To check, \(x=t^2\) gives \(y=x^{3/2}\) for \(t>0\) and \(y=-x^{3/2}\) for \(t<0\). Their second derivatives are \(3/(4\sqrt{x})\) and \(-3/(4\sqrt{x})\), respectively, matching the formula because \(\sqrt{x}=|t|\).

The parameter value \(t=0\) is not included in either calculation: there \(x'=0\), and the formula cannot be used. The two intervals must be analyzed separately because \(x(t)=t^2\) does not move monotonically across both sides of zero.

A Common Sign Error

A tempting shortcut is to inspect only the numerator \(x'y''-y'x''\). That quantity is not, by itself, the second derivative. The denominator \((x')^3\) can reverse its sign.

Worked Example: Correcting a Sign Error

Use \(x=t^2\), \(y=t^3\) at \(t=-1\). The numerator is

x'y''-y'x'' = (2t)(6t)-(3t^2)(2) = 6t^2, \qquad 6(-1)^2=6.

A wrong conclusion would be: “The numerator is positive, so the curve is concave up.” But the denominator at \(t=-1\) is \((2t)^3=(-2)^3=-8\). The actual value is

\frac{d^2y}{dx^2} = \frac{6}{-8} = -\frac{3}{4}.

Thus this branch is concave down at \(t=-1\). The correction is to determine the sign of the entire quotient, including the cubic denominator.

Using the Test Carefully

To determine concavity on parameter intervals, first locate where \(x'(t)=0\), since those parameter values are excluded from the formula. Split the domain into intervals on which \(x'\ne 0\), compute the second derivative on each interval, and determine its sign. If the result is to be described using \(x\)-values, use the relation \(x=x(t)\) to identify the corresponding branch interval.

A positive value of \(d^2y/dx^2\) indicates concavity up.
A negative value of \(d^2y/dx^2\) indicates concavity down.
If \(x'<0\), the denominator \((x')^3\) is negative, so the numerator's sign is reversed in the quotient.
If \(x'=0\), this formula does not determine concavity at that parameter value. Analyze the neighboring valid intervals separately.

These conditions are part of the method, not optional technical details. Concavity describes how the slope changes with \(x\), and a parametrization may travel through the same curve in either direction. Keeping the derivative with respect to \(x\) distinct from the derivative with respect to \(t\) prevents that direction of travel from being mistaken for the curve's shape.

Check Your Understanding

For each question, keep track of both the numerator and the sign of \(x'(t)\).

What condition on \(x'(t)\) is needed to use the parametric formula for \(d^2y/dx^2\)?
For \(x=-t\), \(y=t^2\), what is \(d^2y/dx^2\), and why does the negative value of \(x'\) not make the curve concave down?
For \(x=t^2\), \(y=t^3\), on which of the intervals \(t<0\) and \(t>0\) is the curve concave up?
Why must the parameter domain be split when applying the formula to \(x=t^2\), \(y=t^3\)?
What information is missing if you know only the sign of \(x'y''-y'x''\)?