Introduction
A parametrized curve describes its coordinates as functions of a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). At a parameter value where the horizontal coordinate is momentarily not changing but the vertical coordinate is changing, the curve has a vertical tangent. The key is to check both coordinate derivatives: the slope formula alone may be undefined even when there is no vertical tangent.
The parametric slope formula from the earlier material on deriving the parametric slope formula is \(dy/dx=(dy/dt)/(dx/dt)\), when \(dx/dt\ne 0\). A zero denominator suggests that a vertical tangent may occur, but does not by itself verify one. We will make that distinction precise.
Definition and test for a vertical tangent
Suppose \(x(t)\) and \(y(t)\) are differentiable near an interior parameter value \(t_0\), and the curve passes through \(P=(x(t_0),y(t_0))\). A tangent line at \(P\) is vertical if it is the line \(x=x(t_0)\). The most direct test uses the two coordinate derivatives at \(t_0\).
Here \(x'(t_0)=0\) says that the \(x\)-coordinate has zero instantaneous rate of change. The condition \(y'(t_0)\ne 0\) says that the curve is changing vertically at a nonzero instantaneous rate. Together they give a vertical direction of motion.
Since \(x'(t_0)=0\), the expression \(y'(t_0)/x'(t_0)\) is not defined. Thus the test does not claim that the slope formula has a finite value at \(t_0\). Instead, it establishes the tangent direction directly. In cases where the slope \(dy/dx\) is defined for nearby parameter values, its magnitude often grows without bound near \(t_0\), but the derivative-pair test is the important condition at the parameter value itself.
Why the test works
A tangent direction can be understood by looking at the displacement from the point \(P\) to nearby points on the parametrized curve. Write the nearby parameter as \(t_0+h\). Differentiability gives the first-order approximations
The notation \(o(h)\) represents a quantity whose ratio to \(h\) tends to zero as \(h\) tends to zero. If \(x'(t_0)=0\) and \(y'(t_0)\ne0\), the horizontal displacement is smaller than first order in \(h\), while the vertical displacement has a nonzero first-order term. Consequently, the ratio of the horizontal displacement to the vertical displacement tends to zero.
This means the secant directions approach the vertical direction: their horizontal change is negligible compared with their vertical change. Therefore the tangent line at \(P\) is vertical. Because the line passes through \(P\) and is vertical, its equation is \(x=x(t_0)\).
This argument also explains why both conditions matter. If \(y'(t_0)=0\) as well, the displayed ratio has no such conclusion: both coordinate changes may be smaller than first order, and their relative sizes must be examined separately.
A practical procedure
For a curve given by \(x=x(t)\) and \(y=y(t)\), use the following steps to locate and verify vertical tangents.
Compute \(x'(t)\) and \(y'(t)\) on the parameter domain.
Solve \(x'(t)=0\), retaining only parameter values in the domain.
At each candidate \(t_0\), evaluate \(y'(t_0)\). The test applies if and only if this value is nonzero.
Compute \(x(t_0)\) and \(y(t_0)\). The tangent line is \(x=x(t_0)\).
Solving \(x'(t)=0\) produces candidates, not automatic conclusions. The check of \(y'(t_0)\) is essential, and a candidate at which both derivatives vanish requires separate analysis.
Worked examples
Worked Example: A single verified vertical tangent
Consider \(x(t)=t^2-1\) and \(y(t)=2t+3\), with \(t\in\mathbb{R}\). Find any vertical tangent.
Differentiate the coordinates:
Set \(x'(t)=0\): \(2t=0\), so the only candidate is \(t_0=0\). At that value, \(y'(0)=2\ne0\), so the vertical-tangent test applies. The point is
Therefore the tangent line is \(x=-1\). For a further check, eliminate the parameter using \(t=(y-3)/2\): the curve satisfies \(x=(y-3)^2/4-1\). At \(y=3\), this gives \(x=-1\), and \(dx/dy=(y-3)/2=0\). Thus near the point the horizontal coordinate changes to second order in the vertical displacement, confirming the vertical tangent. Substituting \(t=0\) into the original parametrization also returns \((-1,3)\), the point on the line \(x=-1\).
Worked Example: Two vertical tangents on one parametrization
Consider \(x(t)=t^3-3t\) and \(y(t)=2t\), with \(t\in\mathbb{R}\). Find and verify all vertical tangents.
The derivatives are
The equation \(x'(t)=0\) has solutions \(t=1\) and \(t=-1\). At both values, \(y'(t)=2\ne0\), so both give vertical tangents.
At \(t=1\), the point is
The tangent line is \(x=-2\). To check the behavior near this point, factor the coordinate difference:
As \(t\) tends to \(1\), the horizontal displacement has a squared factor in \(t-1\), while the vertical displacement has a first-power factor. Their ratio tends to zero, as required for a vertical tangent.
At \(t=-1\), the point is
The tangent line is \(x=2\). Here the factorization gives
Again the ratio of horizontal to vertical displacement tends to zero as \(t\) tends to \(-1\). Finally, substituting each parameter into the original equations gives the stated point on its stated vertical line. Thus the two verified tangents are \(x=-2\) at \((-2,2)\) and \(x=2\) at \((2,-2)\).
Worked Example: An indeterminate slope that is not vertical
Consider \(x(t)=t^2\) and \(y(t)=2t^2\), with \(t\in\mathbb{R}\). At \(t=0\), determine what the slope formula does and whether the tangent is vertical.
The derivatives are \(x'(t)=2t\) and \(y'(t)=4t\). At \(t=0\), both derivatives are zero. The parametric slope formula there gives \(0/0\), which is indeterminate; the vertical-tangent test does not apply.
For \(t\ne0\), the slope is
In fact, the parametrization satisfies \(y=2x\) for every \(t\), since \(2t^2=2(t^2)\). The point at \(t=0\) is \((0,0)\), and the secant lines from that point to any other point of the trace have slope \(2\). The tangent direction is therefore not vertical; the tangent line is \(y=2x\). Substitution confirms that \((0,0)\) lies on this line. This example shows that a \(0/0\) slope expression does not establish a vertical tangent.
Worked Example: An indeterminate slope that does lead to a vertical tangent
Consider \(x(t)=t^3\) and \(y(t)=t^2\), with \(t\in\mathbb{R}\). At \(t=0\), both coordinate derivatives vanish. Check whether the curve nevertheless has a vertical tangent there.
We have \(x'(t)=3t^2\) and \(y'(t)=2t\), so both are zero at \(t=0\). The derivative pair alone cannot decide the tangent direction. For \(t\ne0\), compute the slope:
This slope has unbounded magnitude as \(t\) approaches zero. A direct check uses the displacement from the origin: \(x(t)=t^3\) and \(y(t)=t^2\), so
The secant directions therefore approach the vertical direction, and the tangent line at \((0,0)\) is \(x=0\). Substitution gives \(x(0)=0\) and \(y(0)=0\), so the point is on that line. This example, together with the previous one, confirms that when both derivatives vanish, the formula's \(0/0\) form is inconclusive: further analysis is needed.
Common pitfall: treating every zero denominator as a tangent
A common incorrect argument is: “At \(t=0\), \(dx/dt=0\), so the tangent is vertical.” This omits the check of \(dy/dt\). For example, with \(x=t^2\) and \(y=2t^2\), at \(t=0\) one gets \(dx/dt=0\), but also \(dy/dt=0\). Declaring a vertical tangent from the first equation alone is unjustified; for \(t\ne0\), the slope is \(dy/dx=4t/(2t)=2\), and the trace lies on \(y=2x\).
The correction is to distinguish three cases. If \(x'(t_0)=0\) and \(y'(t_0)\ne0\), the test verifies a vertical tangent. If \(x'(t_0)\ne0\), the usual slope formula gives a finite slope at that parameter value. If both derivatives vanish, the slope formula is indeterminate and the test gives no conclusion; inspect nearby slopes or secant directions instead. Do not divide by zero at \(t_0\), and do not treat \(0/0\) as a value.
Why locating vertical tangents matters
Vertical tangents identify points where the curve's direction is vertical, even though a local description of the form \(y=f(x)\) does not have a finite derivative there. In a parametric description, the two coordinate derivatives make it possible to test the direction without first solving for \(y\) as a function of \(x\). This can be especially useful when eliminating the parameter would be difficult or would obscure part of the curve.
The test is deliberately precise: solve \(x'(t)=0\), verify \(y'(t)\ne0\), and then report the point and the line \(x=x(t)\). When both derivatives vanish, do not force a conclusion from the slope formula. The examples show that further examination can reveal either a vertical or a nonvertical tangent.
Check Your Understanding
For each question, state the derivative conditions you use and show enough calculation to justify the conclusion.
- For \(x(t)=t^2+4\), \(y(t)=3t-1\), find any vertical tangent and its point.
- For \(x(t)=t^3-3t\), \(y(t)=t\), find all parameter values that pass the vertical-tangent test, and give the corresponding tangent lines.
- Why is \(dx/dt=0\) alone not enough to conclude that a tangent is vertical?
- For \(x(t)=t^2\), \(y(t)=5t^2\), examine the slope near \(t=0\). Is the tangent vertical?
- For \(x(t)=t^3\), \(y(t)=t^2\), explain why the \(0/0\) slope formula at \(t=0\) does not settle the tangent direction.