Why Polynomial Parametrizations Are Useful to Analyze
A parametrization describes a plane curve by giving both coordinates as functions of a parameter: \(x=x(t)\) and \(y=y(t)\). When these coordinate functions are polynomials, their derivatives are also polynomials. This often makes it possible to find tangent directions and concavity intervals exactly, using factorizations and sign charts.
The key is to analyze the coordinate derivatives together. A zero of \(y'(t)\) may indicate a horizontal tangent, but only if \(x'(t)\ne 0\). A zero of \(x'(t)\) may indicate a vertical tangent, but only if \(y'(t)\ne 0\). Concavity also depends on both coordinate functions, and its formula is valid only where \(x'(t)\ne 0\).
If both coordinate derivatives vanish, neither tangent classification follows from these tests alone. Such a parameter value requires separate analysis; do not label it horizontal or vertical just by substituting into the slope ratio.
From Coordinate Derivatives to Concavity
Concavity describes how the slope changes as \(x\) increases. For a parametrized curve, \(t\) is not necessarily the same as \(x\), so differentiating the slope with respect to \(t\) is not yet the second derivative with respect to \(x\). The chain rule supplies the required conversion.
Here, \(x''(t)\) and \(y''(t)\) mean second derivatives with respect to \(t\). The denominator is the cube of \(x'(t)\), not its square. Thus, when determining the sign, the sign of \(x'(t)\) matters.
Why the Second-Derivative Formula Works
At a point where \(x'(t)\ne 0\), the parametric slope is \(dy/dx=y'/x'\). Along the curve, a change with respect to \(x\) can be computed by first differentiating with respect to \(t\), then dividing by \(dx/dt\). Applying this chain-rule relationship to the slope gives the formula.
Where \(x'(t)\ne 0\), \(dy/dx=y'/x'\).
The quotient rule gives \(d(y'/x')/dt=(x'y''-y'x'')/(x')^2\).
Divide by \(dx/dt=x'\), obtaining \(d^2y/dx^2=(x'y''-y'x'')/(x')^3\).
This reasoning is complete wherever \(x'(t)\ne 0\). If \(x'(t)=0\), division by \(x'(t)\) is not permitted, so the formula does not classify concavity there.
A Reliable Analysis Process
For polynomial coordinate functions, organize the work around exact factors and critical parameter values. Keep the parameter \(t\) as the variable for your sign chart; translate to points on the curve only after the classification is clear.
Find \(x'(t)\), \(y'(t)\), and, when concavity is needed, \(x''(t)\), \(y''(t)\).
Check zeros of \(y'\) against \(x'\) for horizontal tangents, and zeros of \(x'\) against \(y'\) for vertical tangents.
Factor its numerator and denominator, and note every value where \(x'=0\).
Use the exact zeros of the numerator and denominator to divide the parameter line into intervals, then determine the sign on each interval.
Report concavity on parameter intervals where the formula applies, and give coordinates when identifying particular tangent points.
Worked Examples
Worked Example: Horizontal and Vertical Tangents with Concavity
Analyze \(x=t^2+1\), \(y=t^3-3t\). Find the horizontal and vertical tangents and the intervals of concavity.
Differentiate the coordinates:
Horizontal tangents require \(y'=0\) and \(x'\ne0\). The candidates are \(t=1\) and \(t=-1\); \(x'(1)=2\) and \(x'(-1)=-2\), so both are valid. Their points are \((x(1),y(1))=(2,-2)\) and \((x(-1),y(-1))=(2,2)\). The tangent lines are \(y=-2\) and \(y=2\).
A vertical tangent requires \(x'=0\) and \(y'\ne0\). At \(t=0\), \(x'=0\) and \(y'=-3\ne0\). The point is \((1,0)\), so the vertical tangent line is \(x=1\).
For concavity, substitute the derivatives into the formula:
Since \(t^2+1>0\) for every real \(t\), the sign is determined by \(t^3\). The curve is concave down for \(t<0\) and concave up for \(t>0\). The formula is undefined at \(t=0\), where the tangent is vertical. As a check, the first derivative is \((3t^2-3)/(2t)\); differentiating it with respect to \(t\) and dividing by \(2t\) gives the displayed second derivative.
Worked Example: A Sign Chart with Two Vertical Tangents
Analyze \(x=t^3-3t\), \(y=t^2\). Find all horizontal and vertical tangent parameters, then determine concavity.
The derivatives are
For a horizontal tangent, \(y'=0\) gives \(t=0\), and \(x'(0)=-3\ne0\). The point is \((0,0)\), with horizontal tangent \(y=0\). For vertical tangents, \(x'=0\) gives \(t=\pm1\). At both values \(y'=2t\ne0\). The points are \((x(1),y(1))=(-2,1)\) and \((x(-1),y(-1))=(2,1)\); the vertical tangent lines are \(x=-2\) and \(x=2\).
Now compute the second derivative:
The numerator is always negative. The denominator has the sign of \(t^2-1\), since \(27>0\) and cubing preserves sign. Thus the second derivative is positive for \(-1<t<1\) and negative for \(t<-1\) or \(t>1\). The curve is concave up on \((-1,1)\) and concave down on the two outer intervals. At \(t=\pm1\), \(x'=0\), so the formula is not defined. As a check, the first derivative is \(2t/[3(t^2-1)]\); differentiating by the quotient rule and dividing by \(x'\) gives the same expression.
Worked Example: Exact Concavity Boundaries
For \(x=t^3+t\), \(y=t^4-2t^2\), find horizontal tangents and concavity intervals, using exact parameter values.
Differentiate:
Since \(3t^2+1>0\) for every real \(t\), the slope is always defined and there are no vertical tangents. Horizontal tangents occur when \(y'=4t(t-1)(t+1)=0\), so \(t=0,\pm1\). These are valid because \(x'\) is positive at each. The corresponding points are \((0,0)\), \((2,-1)\), and \((-2,-1)\), respectively; each has horizontal tangent.
For concavity,
The denominator is positive. To find where the numerator is zero, set \(u=t^2\), so \(u\ge0\), and solve \(3u^2+6u-1=0\). The solutions are \(u=-1\pm 2\sqrt{3}/3\). Only \(u=(2\sqrt{3}-3)/3\) is nonnegative. Define \(a=\sqrt{(2\sqrt{3}-3)/3}\). The numerator is negative at \(t=0\), since \(3(0)^4+6(0)^2-1=-1\), and positive for sufficiently large \(|t|\). Its only real zeros are \(t=-a\) and \(t=a\). Therefore the curve is concave down on \((-a,a)\) and concave up on \((-\infty,-a)\) and \((a,\infty)\). The sign changes at each boundary. Since \(x'>0\) everywhere, the second-derivative formula applies throughout, including at those two parameter values.
A Common Sign Error
A frequent mistake is to use \((x')^2\) in the denominator of the second-derivative formula. The quotient rule first produces a denominator of \((x')^2\), but converting from differentiation with respect to \(t\) to differentiation with respect to \(x\) requires one more division by \(x'\). The final denominator is \((x')^3\).
Worked Example: Correcting the Denominator
For \(x=t^3-3t\), \(y=t^2\), an incorrect calculation would stop at
and call this \(d^2y/dx^2\). That is the derivative of the slope with respect to \(t\), not with respect to \(x\). The correction is to divide by \(x'=3t^2-3\):
This agrees with the formula derived above. The distinction matters for the sign: when \(x'<0\), cubing it gives a negative denominator, whereas squaring it would always give a positive one.
What to Report
A complete analysis should distinguish parameter values from points on the curve. A parameter value is an input \(t\); the corresponding point is \((x(t),y(t))\). When giving a tangent line, use that point as well as the tangent direction. For concavity, state the parameter intervals and identify any values where the formula is undefined because \(x'=0\).
Exact polynomial roots are especially useful: they give the exact boundaries for sign charts and avoid uncertainty from rounded approximations. A sign chart should be based on all real zeros that affect the expression, including zeros of \(x'\) in the denominator. Never assume a zero of one derivative is automatically a tangent classification or a concavity boundary without checking the other conditions.
Check Your Understanding
For each question, use exact parameter values and show the derivative or sign reasoning that supports your answer.
- For \(x=t^2\), \(y=t^3-t\), find the horizontal tangent parameters. Check whether any candidate also has \(x'=0\).
- For \(x=t^3+t\), \(y=t^2\), find all vertical and horizontal tangent parameters.
- State the parametric second-derivative formula and the condition required for using it.
- For \(x=t^3-3t\), \(y=t^2\), on which parameter intervals is the curve concave up?
- Explain why the sign of \(x'\) can affect the sign of \(d^2y/dx^2\).
