From Concavity to Inflection Points
A parametrized curve describes its coordinates using a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). The second derivative \(\frac{d^2y}{dx^2}\) tells us about the curve’s concavity when the curve is viewed as a graph of \(y\) against \(x\). An inflection point is a point where that concavity changes.
Finding a zero or an undefined value of the second derivative identifies a candidate, not automatically an inflection point. The decisive test is whether the concavity on one side differs from the concavity on the other. Once a candidate passes that test, substitute its parameter value into both coordinate functions to report the point.
Finding Candidate Parameter Values
Suppose \(x(t)\) and \(y(t)\) are twice differentiable near a parameter value, and \(x'(t)\ne 0\). The parametric second-derivative formula is
The formula is valid at parameter values where \(x'(t)\ne 0\). Its zeros are natural candidates for inflection points. Values where the expression is undefined also deserve attention, because the curve’s concavity may still change across such a value. But undefined does not mean inflection: the sign of the second derivative on either side must still be checked.
For a candidate to be an inflection point, the curve must be considered on both sides of it. A parameter-domain endpoint, for example, has no two-sided change of concavity within that domain. Also, when \(x'(t)=0\), do not use the displayed formula at that parameter value; instead, inspect the second derivative on either side, provided the curve branch can be viewed as a graph there.
Why the Sign Test Works
The sign of the second derivative determines concavity on intervals where that derivative exists: positive means concave up, while negative means concave down. Therefore, a change from positive to negative, or from negative to positive, on opposite sides of a point is precisely a change in concavity.
For a parametrization with \(x'(t)\ne 0\) near a candidate, \(x(t)\) is locally monotone, so parameters on opposite sides correspond to points on opposite sides of the candidate’s \(x\)-coordinate. The direction may be reversed if \(x'(t)<0\), but the two sides remain the two sides of the point. Thus, checking whether \(\frac{d^2y}{dx^2}\) changes sign as \(t\) passes through the candidate tests for a change in graph concavity.
Find zeros and undefined values of the parametric second derivative, and check whether the candidate lies inside the parameter domain.
Determine the sign of the second derivative immediately before and after the candidate, wherever it exists. If the formula is undefined at the candidate, use the signs on the neighboring intervals.
A change in concavity confirms an inflection point on a graph branch. Substitute the parameter value into \(x(t)\) and \(y(t)\) to find its coordinates.
Worked Examples
Worked Example: A Zero That Gives an Inflection Point
Consider \(x(t)=t\) and \(y(t)=t^3-3t\). Find any inflection point.
First compute the derivatives:
Since \(x'(t)=1\ne 0\), the second-derivative formula applies:
The second derivative is zero at \(t=0\). For \(t<0\), \(6t<0\), so the curve is concave down. For \(t>0\), \(6t>0\), so it is concave up. Concavity changes, confirming an inflection point. Its coordinates are
The inflection point is \((0,0)\). As a check, \(x=t\) means the curve is \(y=x^3-3x\), whose second derivative is \(6x\); it also changes sign at \(x=0\).
Worked Example: A Zero That Is Not an Inflection Point
Consider \(x(t)=t\) and \(y(t)=t^4\). Find whether the candidate at \(t=0\) is an inflection point.
Here \(x'(t)=1\), \(x''(t)=0\), \(y'(t)=4t^3\), and \(y''(t)=12t^2\). Thus
This second derivative is zero at \(t=0\), so \(t=0\) is a candidate. But for every \(t\ne 0\), \(12t^2>0\): the curve is concave up on both sides. There is no change in concavity, so the candidate is not an inflection point. Its coordinates, \((x(0),y(0))=(0,0)\), do not alter that conclusion. As a check, \(x=t\) gives \(y=x^4\), with second derivative \(12x^2\), which is positive on both sides of zero.
Worked Example: An Undefined Second Derivative with a Sign Change
Consider \(x(t)=t^3\) and \(y(t)=t\). Investigate the candidate \(t=0\).
For \(t\ne 0\), the derivatives are \(x'(t)=3t^2\), \(x''(t)=6t\), \(y'(t)=1\), and \(y''(t)=0\). Applying the formula away from zero gives
The expression is undefined at \(t=0\). For \(t<0\), \(t^5<0\), so \(-2/(9t^5)>0\). For \(t>0\), \(-2/(9t^5)<0\). The curve changes from concave up to concave down.
Here \(x=t^3\) is increasing through zero, and the curve is the graph \(y=\sqrt[3]{x}\). It is continuous at \(x=0\), and the opposite signs on either side confirm an inflection point. Its coordinates are \(x(0)=0\) and \(y(0)=0\), so the point is \((0,0)\). As a check, differentiating \(y=\sqrt[3]{x}\) for \(x\ne 0\) gives \(y''=-2/(9x^{5/3})\), positive for \(x<0\) and negative for \(x>0\), as found from the parametrization.
A Common Mistake: Treating Every Undefined Value as an Inflection
Worked Example: An Undefined Formula Without a Change in Concavity
A common error is to conclude that every parameter value where the parametric formula is undefined must be an inflection point. Consider \(x(t)=t^3\) and \(y(t)=t^6\), and examine \(t=0\).
The derivatives are \(x'=3t^2\), \(x''=6t\), \(y'=6t^5\), and \(y''=30t^4\). For \(t\ne 0\),
The original quotient is undefined at \(t=0\), but its value is \(2>0\) on both neighboring intervals. There is no change in concavity, so \(t=0\) is not an inflection point. Directly, \(y=t^6=(t^3)^2=x^2\), whose second derivative is \(2\), confirming the same conclusion.
The correction is to treat an undefined value only as a candidate. Check the signs on both sides and, where possible, examine the graph branch directly. In this example the parametrization makes the formula undefined at zero, but the graph itself has a well-defined second derivative there.
Reporting the Point and Its Coordinates
The parameter value is not itself the inflection point: it identifies a location along the parametrization. After confirming the concavity change at \(t=t_0\), calculate \(x(t_0)\) and \(y(t_0)\). Report the point as \((x(t_0),y(t_0))\), rather than reporting \(t_0\) alone.
Keep the candidate test and the confirmation distinct. A zero may fail because the second derivative has the same sign on both sides. An undefined value may also fail, either because the signs agree or because the expression has a removable gap. Conversely, an undefined value can mark a genuine inflection when concavity changes across it and the curve branch is a continuous graph there.
Check Your Understanding
For each question, distinguish a candidate from a confirmed inflection point.
- For \(x=t\), \(y=t^5\), the second derivative is \(20t^3\). Is \(t=0\) an inflection point? Explain using signs.
- For \(x=t\), \(y=t^4\), why is \(t=0\) a candidate but not an inflection point?
- What condition on \(x'(t)\) is required to apply the parametric second-derivative formula at a parameter value?
- If the second-derivative formula is undefined at \(t_0\), what must you check before calling the corresponding point an inflection point?
- If an inflection is confirmed at \(t=t_0\), how do you find its coordinates?