Why the Parameter Interval Matters
A parametric curve is described by coordinate functions \(x(t)\) and \(y(t)\), but the formulas alone do not specify which part of the curve is included. The parameter domain matters. Restricting \(t\) to an interval can remove curve points, make a point an endpoint, or leave a candidate tangent condition outside the domain altogether.
At an interior parameter value, nearby points on the curve can be approached using parameter values on both sides. At an included endpoint, the curve is available from only one side. This changes how we interpret tangent behavior: an endpoint can have a well-defined one-sided tangent even though there is no two-sided approach within the restricted domain.
Tangent Tests on a Restricted Domain
At an interior parameter value \(t_0\), when \(x'(t_0)\ne 0\), the slope of the tangent line is
This is the parametric slope formula. It requires \(x'(t_0)\ne 0\). If \(x'(t_0)=0\) while \(y'(t_0)\ne 0\), the tangent direction is vertical. If both derivatives vanish, this first-derivative test does not determine the tangent; further analysis is needed.
At an endpoint, use the same ideas with one-sided derivatives, provided they exist. For instance, at the left endpoint \(a\) of \([a,b]\), the derivatives are taken as \(t\to a^+\). A nonzero one-sided derivative vector gives a one-sided tangent line. If the vector is zero, examine the one-sided secants instead.
The test for a horizontal or vertical tangent also depends on the domain. At a regular interior point, a horizontal tangent requires \(y'=0\) and \(x'\ne 0\); a vertical tangent requires \(x'=0\) and \(y'\ne 0\). At a regular endpoint, the same derivative conditions give a one-sided tangent direction. Only values belonging to the restricted domain can be considered.
Why a Nonzero Endpoint Derivative Gives a Tangent
The endpoint theorem follows directly from the definition of a derivative. Consider the left endpoint \(t_0\), and let \(h>0\) approach zero through values that remain in the parameter domain. The secant vector from the endpoint to the nearby curve point is
Dividing this vector by \(h\) gives a vector whose components approach the one-sided derivatives. If that derivative vector is nonzero, the secant directions approach its direction, so the limiting secant line is the line through \((x(t_0),y(t_0))\) in that direction. At a right endpoint, the same reasoning uses \(h<0\); reversing a direction vector does not change the line. This establishes a one-sided tangent at either endpoint.
If \(x'(t_0)=0\) and \(y'(t_0)\ne 0\), the direction is vertical. Equivalently, nearby horizontal changes are small relative to vertical changes, so the secant lines approach a vertical line. If both derivatives are zero, the displayed limit gives the zero vector and does not establish a direction. In that case, the secant lines must be checked directly or by a justified limiting slope.
Worked Example: One-Sided Tangents at Both Ends of a Truncated Circle
Worked Example: One-Sided Tangents at Both Ends of a Truncated Circle
Consider \(x=\cos t\), \(y=\sin t\) for \(0\le t\le \pi/2\). Find the tangent behavior at both endpoints and describe the interior slope.
Differentiate the coordinates:
At \(t=0\), the curve is at \((1,0)\), and the one-sided derivative vector is
This is nonzero and vertical, so the endpoint tangent line is \(x=1\). At interior values \(0<t<\pi/2\), the slope is
As \(t\to 0^+\), this slope becomes unbounded in magnitude, consistent with a vertical endpoint tangent. At \(t=\pi/2\), the point is \((0,1)\), and the one-sided derivative vector is
This gives a horizontal tangent line, \(y=1\). The endpoint checks agree with the limiting curve behavior from within the allowed interval.
Check: The line \(x=1\) passes through \((1,0)\), and the line \(y=1\) passes through \((0,1)\). The derivative vectors \((0,1)\) and \((-1,0)\) have, respectively, vertical and horizontal directions.
Worked Example: An Endpoint Horizontal Tangent
Worked Example: An Endpoint Horizontal Tangent
Let \(x=t\), \(y=t^2\), with \(0\le t\le 2\). Determine the tangent at the left endpoint and decide whether there are any horizontal tangents at interior parameter values.
The coordinate derivatives are
At the endpoint \(t=0\), the point is \((0,0)\), and the derivative vector is \((1,0)\). It is nonzero, so the endpoint has a one-sided horizontal tangent. The slope formula confirms this:
For interior values \(0<t<2\), \(2t>0\), so there is no interior horizontal tangent. The only solution of \(y'(t)=0\) is \(t=0\), which is an endpoint rather than an interior value.
Check: At \(t=0\), the curve point is \((0,0)\), and the horizontal tangent line is \(y=0\). Substituting \(t=0\) into the slope \(2t\) gives zero, as required.
Worked Example: Both Coordinate Derivatives Vanish at an Endpoint
Worked Example: Both Coordinate Derivatives Vanish at an Endpoint
Consider \(x=t^2\), \(y=t^3\), on \(0\le t\le 1\). Find the one-sided tangent at \(t=0\), where the derivative vector is zero.
The endpoint corresponds to \((0,0)\). Differentiating gives
The first-derivative vector does not determine a direction. For \(t>0\), however, \(x'(t)\ne 0\), so the slope of the curve is
As \(t\to 0^+\), this slope approaches \(0\). The endpoint secant from \((0,0)\) to the point \((t^2,t^3)\) has slope
which also approaches \(0\). Thus the one-sided tangent line is horizontal: \(y=0\). The domain restriction is essential to the description: only the branch approached with \(t\ge 0\) is part of this curve.
Check: The secant slopes tend to zero, and the line \(y=0\) passes through the endpoint \((0,0)\). The zero derivative vector alone did not prove the tangent; the one-sided limit did.
Worked Example: Do Not Analyze Parameters Outside the Domain
Worked Example: Do Not Analyze Parameters Outside the Domain
Let \(x=t\), \(y=t^2\), but restrict the domain to \(1<t<3\). Find the interior tangent slopes and determine whether the curve has an included endpoint tangent.
Here \(x'=1\) and \(y'=2t\), so at every allowed parameter value the tangent slope is
Since \(1<t<3\), the slope is positive and lies strictly between \(2\) and \(6\). In particular, no allowed parameter value gives a horizontal tangent. The candidate \(t=0\), which would make \(2t=0\), is outside the domain and represents no point of this restricted curve.
The values \(t=1\) and \(t=3\) are also not in the domain. The curve approaches the corresponding points as \(t\) approaches those values, but those points are not included endpoints of this parametrized curve, so they are not endpoint tangent points for the given domain.
Check: For every allowed \(t\), \(1<t<3\) implies \(2<2t<6\). Thus the slope is never zero on the domain, and neither boundary value is included.
A Common Mistake: Treating an Endpoint as an Interior Point
A frequent error is to apply interior tangent language without checking whether the parameter value is interior. For \(x=t\), \(y=t^2\) on \([0,2]\), one might write, “There is no horizontal tangent at \(t=0\), because \(t=0\) is not an interior value.” That conclusion confuses the interior tangent test with endpoint analysis. The point \((0,0)\) is included, and its one-sided derivative vector \((1,0)\) gives a horizontal tangent.
The correction is to name the type of point precisely. At \(t=0\), there is a one-sided endpoint tangent, not an interior tangent. If a problem asks only for interior horizontal tangents, \(t=0\) is excluded from that list; if it asks for tangent behavior at the endpoint, the horizontal one-sided tangent must be reported.
A second error is to substitute into \(y'/x'\) when \(x'=0\), or to treat \(0/0\) as a slope. The formula requires a nonzero denominator. At a regular point with \(x'=0\) and \(y'\ne 0\), the tangent is vertical; when both derivatives vanish, as in the preceding example, the tangent must be decided by further analysis.
Why Restricted-Interval Analysis Matters
Parametric formulas can represent more than the portion of a curve named in a problem. A parameter value outside the stated domain may solve a derivative equation, but it does not identify a point on the restricted curve. Conversely, a value that was interior on a larger domain may become an endpoint after restriction, so only one-sided behavior is relevant there.
For careful analysis, first identify the domain and the point’s status within it. Then use the appropriate derivative test: the ordinary interior test, a one-sided endpoint test, or a secant limit when the derivative vector is zero. Keeping those cases distinct prevents both missing valid endpoint tangents and reporting tangents at points the restricted curve does not contain.
Check Your Understanding
For each question, pay attention to whether the parameter value is interior or an included endpoint.
- For \(x=t\), \(y=t^2\) on \([0,2]\), what is the tangent line at \(t=0\), and why is it a one-sided tangent?
- For \(x=\cos t\), \(y=\sin t\) on \([0,\pi/2]\), what is the tangent line at \(t=\pi/2\)?
- Why does \(x'(t_0)=0\), \(y'(t_0)\ne 0\) give a vertical tangent at a regular endpoint?
- For \(x=t^2\), \(y=t^3\) on \([0,1]\), why is the zero derivative vector at \(t=0\) insufficient by itself to determine the tangent?
- If a derivative equation has a solution outside the restricted parameter domain, should that solution be counted as a tangent point of the restricted curve? Explain.
