Putting Parametric Derivatives Together
A parametric curve is described by two coordinate functions, \(x=x(t)\) and \(y=y(t)\). A complete curve-analysis problem may ask for the point at a particular parameter value, the tangent line there, whether the tangent is horizontal or vertical, and whether the curve is concave up or down nearby.
The key is to keep the two coordinate derivatives distinct. The value of \(x'(t)\) describes how the horizontal coordinate changes, while \(y'(t)\) describes how the vertical coordinate changes. Their ratio gives the slope when that ratio is defined. The second derivative requires one additional step: the slope must be differentiated with respect to \(x\), not directly with respect to \(t\).
A Reliable Order of Work
For each requested parameter value, first evaluate both coordinates. Next evaluate \(x'\) and \(y'\) to identify the tangent type or compute its slope. Only then use the second-derivative formula, if \(x'\ne 0\), to test concavity. Keeping these tasks in order helps prevent a common mix-up: a vertical tangent is not a place where \(dy/dx=0\), and the second-derivative test is not available in its usual graph form when \(x'=0\).
Substitute the parameter into both coordinate functions to obtain \((x(t_0),y(t_0))\).
Compare \(x'(t_0)\) and \(y'(t_0)\). Use \(y'/x'\) for the slope only when \(x'(t_0)\ne 0\).
If \(x'\ne 0\), calculate \((x'y''-y'x'')/(x')^3\). A positive value indicates concavity up locally; a negative value indicates concavity down locally, provided the sign persists nearby.
Confirm that the tangent line passes through the calculated point and that the derivative calculations support the stated classification.
Why the Second-Derivative Formula Works
The slope formula follows from the chain rule: when \(x'(t)\ne 0\), \(y\) can be treated locally as a function of \(x\), and \(dy/dx=(dy/dt)/(dx/dt)=y'/x'\). To obtain the second derivative, differentiate that slope with respect to \(t\), then convert from change per unit \(t\) to change per unit \(x\) by dividing by \(dx/dt=x'\).
The middle equality uses the quotient rule. The condition \(x'\ne 0\) matters twice: it makes the slope ratio meaningful and permits division by \(x'\) when changing the differentiation variable from \(t\) to \(x\). Thus, at a vertical tangent, this formula cannot be used to claim ordinary graph concavity at that parameter value.
Worked Polynomial Example: A Horizontal Tangent and a Vertical Tangent
Worked Example: Analyze a Polynomial Parametrization
Let \(x=t^2+1\) and \(y=t^3-3t\). Find the point, tangent line, and concavity at \(t=1\). Also classify the tangent at \(t=0\).
At \(t=1\), the coordinates are
The point is \((2,-2)\). Differentiate both coordinate functions:
At \(t=1\), \(x'=2\) and \(y'=0\). Because \(x'\ne 0\), the slope is \(y'/x'=0/2=0\), so the tangent is horizontal. Through \((2,-2)\), its equation is
For concavity, substitute the derivatives into the second-derivative formula:
The curve is concave up near \(t=1\). Indeed, for \(t>0\) the formula simplifies to \(\frac{d^2y}{dx^2}=\frac{3(t^2+1)}{4t^3}>0\), so the sign remains positive in a neighborhood of \(1\).
Now consider \(t=0\). The point is \((x(0),y(0))=(1,0)\), and \(x'(0)=0\), while \(y'(0)=-3\ne 0\). Therefore the tangent is vertical, with equation \(x=1\). This classification can also be seen from the slope \(y'/x'=(3t^2-3)/(2t)\), whose magnitude grows without bound as \(t\) approaches zero from either side.
Check: The horizontal line \(y=-2\) contains \((2,-2)\), and its slope is \(0\), matching \(y'(1)/x'(1)=0\). The vertical line \(x=1\) contains \((1,0)\), and the coordinate derivatives there have \(x'=0\), \(y'\ne 0\), as required.
Worked Trigonometric Example: Tangent Types and Concavity
Worked Example: Analyze a Trigonometric Parametrization
Let \(x=\cos t\) and \(y=\sin(2t)\). Find the tangent lines at \(t=0\) and \(t=\pi/4\), and determine the concavity at \(t=\pi/4\).
At \(t=0\), the point is \((\cos 0,\sin 0)=(1,0)\). The derivatives are
Thus \(x'(0)=0\) and \(y'(0)=2\ne 0\). The tangent is vertical, so its equation through \((1,0)\) is \(x=1\).
At \(t=\pi/4\), the coordinates are
At this parameter, \(x'=-\sqrt{2}/2\) and \(y'=2\cos(\pi/2)=0\). Since \(x'\ne 0\), the tangent is horizontal. Its equation is \(y=1\).
For concavity, the second derivatives are \(x''=-\cos t\) and \(y''=-4\sin(2t)\). At \(t=\pi/4\), \(x''=-\sqrt{2}/2\) and \(y''=-4\). Therefore,
The value is negative, so the curve is concave down near this parameter. The sign persists locally because the derivatives in the formula are continuous and \(x'\) is nonzero near \(\pi/4\).
Check: The horizontal line \(y=1\) contains \((\sqrt{2}/2,1)\), and the derivative ratio there is \(0/(-\sqrt{2}/2)=0\). The vertical line \(x=1\) contains \((1,0)\), and at \(t=0\) the coordinate derivatives are \(0\) and \(2\). Both tangent classifications agree with the derivatives.
Worked Polynomial Example: One Point, Two Tangents
Worked Example: Compare Two Parameter Values at the Same Point
Let \(x=t^2-1\) and \(y=t^3-t\). Find the tangent lines and concavity at \(t=1\) and \(t=-1\). Explain what the results say about the point.
At \(t=1\), \(x=1-1=0\) and \(y=1-1=0\). At \(t=-1\), \(x=1-1=0\) and \(y=-1-(-1)=0\). Both parameter values trace the same point \((0,0)\).
The derivatives are
At \(t=1\), \(x'=2\) and \(y'=2\), so the slope is \(2/2=1\). The tangent line through the origin is \(y=x\). The second derivative is
This branch is concave up near \(t=1\). At \(t=-1\), \(x'=-2\) and \(y'=2\), so the slope is \(2/(-2)=-1\). The tangent line through the origin is \(y=-x\). Its second derivative is
This branch is concave down near \(t=-1\). In each case, \(x'\ne 0\), and the calculated second derivative is continuous and nonzero nearby, so its sign holds on a sufficiently small neighborhood of that parameter value.
Check: Both lines \(y=x\) and \(y=-x\) pass through \((0,0)\). Their slopes, \(1\) and \(-1\), match the derivative ratios at \(t=1\) and \(t=-1\), respectively. The second-derivative calculations give opposite signs, so the concavity differs on the two branches.
Conditions and Common Pitfalls
A horizontal tangent requires \(y'=0\) and \(x'\ne 0\). A vertical tangent requires \(x'=0\) and \(y'\ne 0\). If both derivatives are zero, neither test settles the tangent type; the point requires further analysis. These conditions apply at the parameter value being studied, not merely at the geometric point, because a single point can be traced more than once.
For concavity, the sign of \(d^2y/dx^2\) is the relevant test: positive means concave up, and negative means concave down, on an interval where the curve is locally a graph \(y\) as a function of \(x\). The formula requires \(x'\ne 0\). At a vertical tangent, do not substitute \(x'=0\) into the formula or interpret the resulting division by zero as a concavity classification.
A frequent mistake is to use \(x'/y'\) as the slope. For example, in the trigonometric example at \(t=\pi/4\), \(x'=-\sqrt{2}/2\) and \(y'=0\). Reversing the ratio gives \((-\sqrt{2}/2)/0\), which is undefined, and might lead to the incorrect conclusion that the tangent is vertical. The correct slope is \(y'/x'=0/(-\sqrt{2}/2)=0\), so the tangent is horizontal.
Another mistake is to report a concavity sign from one parameter value without checking the conditions. A single positive value establishes the sign at that point, but a local concavity claim also relies on the expression being defined nearby and retaining its sign. Checking that \(x'\) stays nonzero and the second-derivative expression stays positive or negative near the parameter justifies the local conclusion.
Check Your Understanding
For each question, show the coordinate and derivative values that support your conclusion.
- For \(x=t^2+1\), \(y=t^3-3t\), find the tangent line at \(t=-1\). Is it horizontal, vertical, or neither?
- For \(x=\cos t\), \(y=\sin(2t)\), classify the tangent at \(t=\pi/3\) and find its slope.
- State the condition that permits use of the parametric second-derivative formula.
- For \(x=t^2-1\), \(y=t^3-t\), explain why evaluating only one parameter value at \((0,0)\) misses information about the curve.
- If \(x'(t_0)=0\) and \(y'(t_0)=0\), can the horizontal- or vertical-tangent tests alone classify the tangent? Explain.
