Why the Direction of the Parameter Matters
A parametric curve is described by two functions, \(x=x(t)\) and \(y=y(t)\). To determine its concavity, we need to know how the slope changes as \(x\) changes. But the parameter \(t\) does not always move in the same direction as \(x\): as \(t\) increases, \(x\) might increase or decrease.
This distinction matters because concavity is defined with respect to increasing \(x\), not increasing \(t\). The parametric second derivative provides the needed information, but interpreting its sign requires checking the sign of \(dx/dt\).
Concavity for a Parametric Curve
Suppose \(x=x(t)\) and \(y=y(t)\) are twice differentiable on a parameter interval, and \(dx/dt\ne 0\) throughout that interval. Then \(x\) changes strictly in one direction locally, so the curve can be treated there as a graph \(y\) as a function of \(x\). Its slope is
The second derivative with respect to \(x\) measures how this slope changes as \(x\) changes. It—not merely the change in slope as \(t\) changes—determines concavity.
For the parametric equations, write \(x'=dx/dt\), \(y'=dy/dt\), \(x''=d^2x/dt^2\), and \(y''=d^2y/dt^2\). When \(x'\ne 0\), the parametric second derivative formula is
The formula converts parameter derivatives into a derivative with respect to \(x\). In particular, its sign—not the sign of \(d(dy/dx)/dt\) alone—gives the concavity.
Why the Formula Gives the Right Concavity
The chain rule explains why differentiating with respect to \(t\) is not the same as differentiating with respect to \(x\). If \(g(t)\) is a differentiable quantity along the curve and \(x'(t)\ne 0\), then
Apply this to the slope \(g(t)=dy/dx=y'/x'\). The quotient rule gives
Dividing by \(dx/dt=x'\), as required to differentiate with respect to \(x\), yields
This also identifies the direction issue. The rate of change of slope with respect to the parameter is \(d(dy/dx)/dt\). To get the rate of change of slope with respect to \(x\), that quantity must be divided by \(dx/dt\). If \(x'<0\), the division reverses its sign.
A Sign-Based Procedure
To classify concavity on a parameter interval, first check that \(x'\ne 0\) there. Then find \(dy/dx\), differentiate it with respect to \(t\), and divide by \(x'\). Alternatively, calculate the numerator and denominator in the parametric second derivative formula directly.
Confirm that \(x'(t)\ne 0\) throughout the interval being analyzed.
Use \(d^2y/dx^2=(1/x')\,d(dy/dx)/dt\), or the equivalent formula involving \(x'\), \(y'\), \(x''\), and \(y''\).
Include the sign of \(x'\) in the calculation; do not classify from the sign of the parameter derivative of the slope alone.
A positive second derivative means concave up; a negative second derivative means concave down, with respect to increasing \(x\).
Worked Examples
Worked Example: Concavity When \(x\) Increases
Let \(x=t^2\) and \(y=t^3\) for \(1<t<3\). Determine the concavity on this parameter interval.
Differentiate each coordinate:
Since \(1<t<3\), we have \(x'=2t>0\), so the formula applies throughout the interval. First find the slope:
Differentiate this slope with respect to \(t\), then divide by \(x'=2t\):
Therefore the curve is concave up on \(1<t<3\). As a check, at \(t=2\), the point is \((x,y)=(4,8)\), the slope is \(3\), and the second derivative is \(3/8>0\), consistent with the formula \(3/(4t)\).
Worked Example: Concavity When \(x\) Decreases
Let \(x=4-t\) and \(y=t^2\) for \(0<t<2\). Determine the concavity. This example illustrates why the sign of \(dx/dt\) must be included.
Differentiate the coordinate functions:
The slope and its parameter derivative are
To differentiate the slope with respect to \(x\), divide by \(x'=-1\):
Thus the curve is concave up, even though the slope decreases as \(t\) increases. The parameter moves toward decreasing \(x\), so the slope increases when viewed as \(x\) increases. As a check, \(t=1\) gives \((x,y)=(3,1)\), slope \(-2\), and second derivative \(2\). Eliminating \(t\) gives \(y=(4-x)^2\), whose second derivative with respect to \(x\) is also \(2\).
Worked Example: Classifying Two Parameter Intervals
Let \(x=t\) and \(y=t^3-3t\) on \(-2<t<2\), excluding \(t=0\). Classify concavity on \((-2,0)\) and \((0,2)\).
Here \(x'=1\), \(y'=3t^2-3\), and \(x''=0\). Since \(x'=1\ne 0\),
On \((-2,0)\), \(6t<0\), so the curve is concave down. On \((0,2)\), \(6t>0\), so the curve is concave up. For a check, at \(t=-1\) the second derivative is \(-6\), while at \(t=1\) it is \(6\); these values agree with the signs on the respective intervals. Because \(x=t\), these parameter values are also the corresponding \(x\)-values.
A Common Sign Error
A frequent mistake is to classify concavity using only \(d(dy/dx)/dt\). That quantity tells how the slope changes as \(t\) increases. Concavity, however, depends on how the slope changes as \(x\) increases.
Worked Example: Correcting a Direction Mistake
For \(x=4-t\), \(y=t^2\), the slope is \(dy/dx=-2t\). A mistaken calculation might say: “The slope has parameter derivative \(-2<0\), so the curve is concave down.” This conclusion is wrong because \(x'=-1\).
The correction is to convert the parameter derivative into a derivative with respect to \(x\):
The curve is concave up. The mistaken answer treated increasing \(t\) as increasing \(x\), although \(dx/dt=-1\) shows that \(x\) decreases as \(t\) increases.
Conditions and Interpretation
The formula requires \(x'(t)\ne 0\) at the parameter values where it is used. If \(x'=0\), the displayed quotient formula is undefined there, so it cannot by itself classify concavity at that parameter value. When an interval contains a value where \(x'=0\), analyze the portions on which \(x'\ne 0\) separately rather than extending a sign conclusion across the undefined point.
For an interval on which \(x'\) is continuous and never zero, its sign is constant: it cannot change from positive to negative without being zero somewhere. This makes the direction of \(x\) consistent throughout that interval. Nevertheless, calculate the sign of the full second derivative, because its numerator can also vary with \(t\).
Concavity statements should identify the interval on which the sign was established. A zero of \(d^2y/dx^2\) does not, by itself, tell you the concavity on either side; the sign must be examined on intervals. Likewise, the formula's sign is meaningful only where it is defined.
Check Your Understanding
For each question, show the derivative signs that support your conclusion.
- For \(x=t^2\), \(y=t^3\), and \(1<t<2\), calculate \(d^2y/dx^2\) and state the concavity.
- For \(x=5-t\), \(y=t^2\), and \(0<t<1\), why is the sign of \(d(dy/dx)/dt\) alone not enough to classify concavity?
- State the condition on \(dx/dt\) needed to apply the parametric second derivative formula at a parameter value.
- If \(dx/dt<0\) and \(d(dy/dx)/dt>0\), what is the sign of \(d^2y/dx^2\)? Explain.
- For \(x=t\) and \(y=t^3-3t\), classify concavity on \((-1,0)\) and \((0,1)\).
