Why Build a Sign Chart?
A parametric curve is described by \(x=x(t)\) and \(y=y(t)\). As the parameter changes, the point \((x(t),y(t))\) moves along the curve. The signs of \(x'(t)\), \(y'(t)\), and \(\frac{d^2y}{dx^2}\) summarize different aspects of that behavior.
A sign chart places the relevant parameter values in order and records the signs on each interval between them. It is a compact way to see whether the curve moves left or right, whether its tangent slope is positive or negative, and whether its graph is concave up or down. The chart is useful only when its entries are interpreted under the conditions of the derivative formulas.
What the Three Signs Tell You
If \(x'(t)>0\), the \(x\)-coordinate increases as \(t\) increases; if \(x'(t)<0\), it decreases. The corresponding statements for \(y'(t)\) describe vertical motion. These signs describe changes in the coordinates, not the slope by themselves.
At a parameter value where \(x'(t)\ne 0\), the parametric slope is \(\frac{dy}{dx}=\frac{y'(t)}{x'(t)}\). Thus its sign is positive when \(x'\) and \(y'\) have the same sign, and negative when their signs differ. If \(x'=0\), this quotient is not defined by the formula, so do not assign a slope sign from it at that parameter value.
Where the parametric second-derivative formula applies, a positive \(\frac{d^2y}{dx^2}\) indicates concavity up as the curve is viewed locally as a graph of \(y\) with respect to \(x\); a negative value indicates concavity down. A zero or undefined value is a breakpoint to investigate, not a sign to guess.
Deriving the Second-Derivative Formula
The second derivative measures how the slope changes with respect to \(x\), not with respect to \(t\). This distinction explains why the parametric formula divides by \(x'\). Suppose \(x'(t)\ne0\) on an interval and the coordinate functions have the derivatives required below. First, the chain rule gives \(\frac{dy}{dx}=\frac{y'}{x'}\). To differentiate this slope with respect to \(x\), use \(\frac{d}{dx}=\frac{1}{x'}\frac{d}{dt}\).
By the quotient rule, \(\frac{d}{dt}\left(\frac{y'}{x'}\right)=\frac{x'y''-y'x''}{(x')^2}\), where \(x'\ne0\).
Divide the \(t\)-derivative by \(x'\): \(\frac{d^2y}{dx^2}=\frac{1}{x'}\frac{d}{dt}\left(\frac{y'}{x'}\right)\).
Therefore, \(\frac{d^2y}{dx^2}=\frac{x'y''-y'x''}{(x')^3}\), provided \(x'\ne0\).
This derivation is also a useful check on sign-chart work: the denominator is \((x')^3\), so it has the same sign as \(x'\). Omitting the conversion from \(t\) to \(x\) would produce a different, incorrect result.
A Reliable Sign-Chart Procedure
Before choosing test values, find all parameter values in the stated domain where a sign might change or a formula might fail. These include zeros of \(x'\), zeros of \(y'\), and zeros or undefined values of the numerator of the second derivative. A zero of \(x'\) is especially important because it makes both displayed parametric formulas inapplicable there.
Compute \(x'\), \(y'\), and the parametric second derivative, noting the condition \(x'\ne0\).
Use relevant zeros, undefined values, and endpoints of the given parameter domain.
Choose a parameter value in each interval and evaluate the signs. A continuous expression that is nonzero throughout an interval keeps one sign there.
Use \(x'\) and \(y'\) for coordinate direction, their quotient for slope when \(x'\ne0\), and the second derivative for concavity where its formula applies.
Worked Example: A Curve with Two Breakpoints in the Slope Formula
Worked Example: A Curve with Two Breakpoints in the Slope Formula
Let \(x=t^3-3t\) and \(y=t^2\). Build a sign chart for all real \(t\), including the coordinate derivative signs and the second derivative sign wherever it is defined.
Differentiate the coordinate functions:
The slope formula applies when \(3(t^2-1)\ne0\), or \(t\ne -1,1\). The second derivative is
The breakpoints are \(t=-1,0,1\): \(x'\) vanishes at \(-1\) and \(1\), while \(y'\) vanishes at \(0\). Since \(t^2+1\) is always positive, the second derivative is negative outside \([-1,1]\) and positive between \(-1\) and \(1\), excluding the points where \(x'=0\). Testing the intervals gives:
| Parameter interval | \(x'\) | \(y'\) | \(\frac{d^2y}{dx^2}\) | Interpretation |
|---|---|---|---|---|
| \((-\infty,-1)\) | + | − | − | Right and down; negative slope; concave down |
| \((-1,0)\) | − | − | + | Left and down; positive slope; concave up |
| \((0,1)\) | − | + | + | Left and up; negative slope; concave up |
| \((1,\infty)\) | + | + | − | Right and up; positive slope; concave down |
Check representative values, including the second derivative values that can be easy to miscalculate. At \(t=-2\), \(x'=3(4-1)=9\), \(y'=-4\), and \(\frac{d^2y}{dx^2}=-6(5)/[27(3)^3]=-30/729=-10/243\). At \(t=-\frac12\), \(x'=3(\frac14-1)=-\frac94\), \(y'=-1\), and
At \(t=\frac12\), \(x'=-\frac94\), \(y'=1\), and the second derivative is again \(\frac{160}{243}\), because the expression depends only on \(t^2\). At \(t=2\), \(x'=9\), \(y'=4\), and \(\frac{d^2y}{dx^2}=-30/729=-10/243\). These values agree with the chart.
At \(t=-1\) and \(t=1\), the second-derivative formula is undefined because \(x'=0\); do not extend the interval signs to claim a value there. At \(t=0\), \(y'=0\), while \(x'=-3\), so the slope is \(0/(-3)=0\). The curve’s point is \((x(0),y(0))=(0,0)\). The interval signs summarize behavior on either side without assigning the interval signs as values at the breakpoints.
Worked Example: A Zero of the Vertical Coordinate Derivative
Worked Example: A Zero of the Vertical Coordinate Derivative
For \(x=t^2\), \(y=t^3-3t\), with parameter domain \(t>0\), build the sign chart and interpret the change at \(t=1\).
Differentiate:
Since \(t>0\), \(x'>0\), and the parametric second derivative is
The only breakpoint inside the domain is \(t=1\), where \(y'=0\). For \(0<t<1\), \(x'>0\), \(y'<0\), and the second derivative is positive: the point moves right and down, the slope is negative, and the curve is concave up. For \(t>1\), all three signs are positive: it moves right and up, has positive slope, and remains concave up.
Check one value in each interval. At \(t=\frac12\), \(x'=1\), \(y'=-\frac94\), and \(\frac{d^2y}{dx^2}=\frac{3(5/4)}{4(1/8)}=\frac{15/4}{1/2}=\frac{15}{2}\), which is positive. At \(t=2\), \(x'=4\), \(y'=9\), and \(\frac{d^2y}{dx^2}=\frac{3(5)}{4(8)}=\frac{15}{32}\), also positive. At the breakpoint \(t=1\), the slope is \(0/(2)=0\), and the second derivative is \(3(2)/4=3/2\). The chart’s interval signs agree with these checks.
Worked Example: A Concavity Sign Change
Worked Example: A Concavity Sign Change
Let \(x=t\) and \(y=t^3-3t\). Make a sign chart and identify what happens to the signs at the parameter values \(-1,0,1\).
Here \(x'=1\), \(y'=3t^2-3=3(t^2-1)\), \(x''=0\), and \(y''=6t\). Because \(x'\) is always positive, the slope is \(y'\), and
The signs on the four intervals are:
| Parameter interval | \(x'\) | \(y'\) | \(\frac{d^2y}{dx^2}\) |
|---|---|---|---|
| \((-\infty,-1)\) | + | + | − |
| \((-1,0)\) | + | − | − |
| \((0,1)\) | + | − | + |
| \((1,\infty)\) | + | + | + |
For a direct check, at \(t=-2\), \(y'=9\) and \(6t=-12\); at \(t=-\frac12\), \(y'=-\frac94\) and \(6t=-3\); at \(t=\frac12\), \(y'=-\frac94\) and \(6t=3\); at \(t=2\), \(y'=9\) and \(6t=12\). These values match every interval sign. At \(t=0\), the second derivative is zero and its sign changes from negative to positive. At \(t=-1\) and \(t=1\), the slope is zero, but the concavity signs do not change there.
A Common Sign-Chart Mistake
A frequent error is to treat the sign of \(y'\) as the sign of the slope. That works only if \(x'>0\). In the first example, on \((-1,0)\), both \(x'\) and \(y'\) are negative. A mistaken calculation might say, “\(y'<0\), so the slope is negative.” But the actual quotient has sign \((-)/(-)=+\), so the slope is positive.
Why the Chart Is Useful
A sign chart makes several pieces of information comparable on the same parameter intervals. It can reveal that a curve moves left while rising, or that its concavity changes even though its direction of motion does not. It also makes breakpoints visible, helping prevent formulas from being used at values where their denominators vanish.
The chart is a summary, not a substitute for checking the definitions and conditions. A positive \(x'\) means the horizontal coordinate increases with \(t\); it does not by itself prove that every feature of the full curve is captured by a single graph \(y=f(x)\) across the entire parameter domain. At points where \(x'=0\), or where the formulas are otherwise undefined, analyze those values separately. Within intervals where the formulas apply, the signs provide a precise and efficient account of local behavior.
Check Your Understanding
For each question, state any condition needed before applying a derivative formula.
- If \(x'>0\) and \(y'<0\) on an interval, what are the direction of horizontal motion and the sign of the slope?
- For \(x'<0\) and \(y'<0\), what is the sign of \(\frac{dy}{dx}\), assuming \(x'\ne0\)?
- Why must a sign chart be split at every zero of \(x'\) before using the parametric second-derivative formula?
- For \(x=t\), \(y=t^3-3t\), what are the signs of \(y'\) and \(\frac{d^2y}{dx^2}\) on \(0<t<1\)?
- In the first worked example, what is \(\frac{d^2y}{dx^2}\) at \(t=-\frac12\), and how can substitution verify its sign?
