Why an Inflection Point Needs a Test
A point where the second derivative is zero is a possible inflection point, not automatically an inflection point. Concavity changes only when the second derivative changes sign across the point. For a parametric curve, there is an additional step: the second derivative must be computed with respect to \(x\), not with respect to the parameter \(t\).
Thus the task has two parts. First locate parameter values where concavity might change. Then check the second derivative on both sides of each candidate and translate any confirmed parameter value into the point \((x(t),y(t))\) on the curve.
The Parametric Second Derivative
Write the curve as \(x=x(t)\), \(y=y(t)\). On an interval where \(x'(t)\ne 0\), the parametric slope is \(\frac{dy}{dx}=\frac{y'(t)}{x'(t)}\). Differentiating this slope with respect to \(t\), then dividing by \(\frac{dx}{dt}=x'(t)\), gives the second derivative with respect to \(x\):
Here \(x'\) and \(y'\) are derivatives with respect to \(t\), and \(x''\) and \(y''\) are second derivatives with respect to \(t\). The denominator matters: the numerator alone does not determine concavity. The formula requires \(x'(t)\ne 0\).
Why the Sign Test Works
When \(x'(t)\ne 0\) throughout a neighborhood of \(t_0\), \(x(t)\) is locally monotonic, so the parameter traces one graph branch there. If \(\frac{d^2y}{dx^2}>0\) on one side and \(\frac{d^2y}{dx^2}<0\) on the other, the graph changes from concave up to concave down or vice versa. That is precisely a change in concavity.
If \(x'(t)<0\), increasing \(t\) moves toward decreasing \(x\), so the left and right sides in \(x\) are reversed relative to the sides in \(t\). But reversal does not affect whether the signs differ: a positive sign on one parameter side and a negative sign on the other still means the concavity changes across the point.
Use \(\frac{d^2y}{dx^2}=\frac{x'y''-y'x''}{(x')^3}\) where \(x'\ne 0\).
Look for zeros of the second derivative and values where it is undefined. Check that each candidate is an interior parameter value.
Determine the sign of \(\frac{d^2y}{dx^2}\) immediately before and after the candidate. A sign change confirms an inflection; no sign change rejects it.
Substitute the confirmed parameter into both coordinate functions to give \((x(t_0),y(t_0))\).
Worked Examples
Worked Example: Confirming an Inflection at the Origin
Consider \(x(t)=t\), \(y(t)=t^3-3t\). Find and test the candidate inflection point.
First compute the derivatives with respect to \(t\):
Since \(x'=1\ne 0\), the parametric second derivative formula applies for every \(t\). Substitution gives
The candidate occurs at \(6t=0\), so \(t=0\). For \(t<0\), \(6t<0\), and for \(t>0\), \(6t>0\). The second derivative changes from negative to positive, so concavity changes across the candidate.
At \(t=0\), \(x(0)=0\) and \(y(0)=0^3-3(0)=0\). The inflection point is \((0,0)\). As a check, because \(x=t\), the curve is \(y=x^3-3x\); differentiating twice with respect to \(x\) gives \(6x\), which agrees with \(6t\).
Worked Example: Rejecting a Zero Without a Sign Change
Consider \(x(t)=t\), \(y(t)=t^4\). Determine whether the candidate at \(t=0\) is an inflection point.
The derivatives are
The denominator is \((x')^3=1\), so
The second derivative is zero at \(t=0\), making that value a candidate. But for every \(t\ne 0\), \(12t^2>0\). It is positive both immediately before and immediately after \(0\), so concavity does not change. There is no inflection point at this candidate.
The curve point is \((x(0),y(0))=(0,0)\), but it is not an inflection point. As a check, \(x=t\) makes the curve \(y=x^4\), whose second derivative is \(12x^2\), positive on both sides of \(x=0\).
Worked Example: Testing a Curve Traced with Decreasing \(x\)
Consider \(x(t)=-t\), \(y(t)=t^3\). Find and test the candidate inflection parameter.
Compute the parameter derivatives:
Because \(x'=-1\ne 0\), the formula applies. Keeping the cube in the denominator gives
The only candidate is \(t=0\). For \(t<0\), the second derivative is negative; for \(t>0\), it is positive. The sign changes, so this is an inflection parameter. The curve point is \(x(0)=-0=0\), \(y(0)=0^3=0\), namely \((0,0)\).
To check, eliminate the parameter using \(t=-x\). Then \(y=(-x)^3=-x^3\), so \(\frac{d^2y}{dx^2}=-6x\). Since \(x=-t\), this is \(6t\), exactly the result from the parametric formula. The concavity signs are reversed when described in terms of \(x\) rather than \(t\), but they still change across the point.
Common Pitfall: Using Only the Numerator
It is tempting to inspect \(x'y''-y'x''\) and treat its sign as the sign of \(\frac{d^2y}{dx^2}\). That is incorrect because the formula divides by \((x')^3\), which can be negative.
For a concrete example, let \(x=-t\) and \(y=t^4\). Then \(x'=-1\), \(x''=0\), \(y'=4t^3\), and \(y''=12t^2\). The numerator is
A wrong calculation would call this concave down because the numerator is negative for \(t\ne 0\). But the denominator is \((x')^3=(-1)^3=-1\), so the correct second derivative is
It is positive on both sides of \(t=0\), and therefore the candidate at \(0\) is not an inflection point. Always test the sign of the complete second derivative, including its denominator.
When the Formula Does Not Apply
If \(x'(t_0)=0\), the parametric second derivative formula is undefined at \(t_0\). That fact alone neither proves nor rules out an inflection point. The curve may have a vertical tangent, a cusp, or another local feature, and one must determine whether the relevant branch can be treated as a graph and whether its concavity changes. Do not substitute \(x'(t_0)=0\) into the formula and draw a conclusion from the resulting division by zero.
Likewise, an endpoint of the parameter interval has no parameter values on both sides within that interval. The two-sided sign-change test cannot confirm an inflection there. If a curve passes through the same geometric point at different parameter values, test each local branch separately; the branches need not have the same behavior.
Check Your Understanding
For each question, use the second derivative with respect to \(x\), test signs on both sides of any interior candidate, and report the curve point when an inflection is confirmed.
- For \(x=t\), \(y=t^3+2t\), find the parametric second derivative. Is there an inflection point at \(t=0\)? Explain.
- For \(x=t\), \(y=t^5\), test the candidate at \(t=0\). Does concavity change?
- For \(x=-t\), \(y=t^3-3t\), find and test the candidate inflection parameter. Give its curve point.
- Why is a zero of \(x'y''-y'x''\) only a candidate and not, by itself, proof of an inflection?
- What must be investigated separately when \(x'(t_0)=0\)?
