How a Parametric Curve Has a Vertical Tangent
A parametric curve describes its coordinates as functions of a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). At a particular parameter value, the derivatives \(dx/dt\) and \(dy/dt\) describe the coordinate changes as \(t\) changes.
A vertical tangent occurs when the curve has a vertical direction at a point. For a regular parametrization, the velocity vector in the plane is \(\left(dx/dt,dy/dt\right)\). If its horizontal component is zero while its vertical component is not, the vector points vertically. This gives a direct test for vertical tangents.
The nonzero condition on \(dy/dt\) matters: it ensures that the parametrization is moving in a nonzero vertical direction at \(t_0\). The point must also be calculated from both coordinate functions. A value of \(t\) by itself is not a point on the curve.
Why the Conditions Give a Vertical Tangent
For a small parameter change \(h\), the change in the curve’s position is \(\bigl(x(t_0+h)-x(t_0),\,y(t_0+h)-y(t_0)\bigr)\). Differentiability means that, after dividing this change by \(h\), the result approaches the derivative vector. Under the conditions above, that vector has zero horizontal component and a nonzero vertical component.
Because \(dy/dt(t_0)\ne0\), the limiting direction is vertical, not the zero vector. Equivalently, for nearby parameter values where the denominator is nonzero, compare the horizontal and vertical changes:
Thus the horizontal change is negligible compared with the vertical change near the point. The curve’s tangent direction is vertical, so the tangent line through \(\bigl(x(t_0),y(t_0)\bigr)\) is \(x=x(t_0)\). This also explains why the usual parametric slope formula cannot be used by direct division at this parameter value: \(dy/dx=(dy/dt)/(dx/dt)\) has a zero denominator there.
A Reliable Method
To find vertical tangents, first solve \(dx/dt=0\). Then check each solution in \(dy/dt\). Keep only parameter values where \(dy/dt\ne0\), calculate the corresponding point, and write the vertical line through it.
Find \(dx/dt\) and \(dy/dt\).
Solve \(dx/dt=0\) on the parameter interval being considered.
Evaluate \(dy/dt\) at each candidate; a candidate meets the vertical-tangent condition only if this value is nonzero.
Substitute the parameter value into both \(x(t)\) and \(y(t)\), then give the tangent line \(x=x(t_0)\).
The parameter interval is part of the problem. A solution of \(dx/dt=0\) outside the specified interval does not describe a point on the portion of the curve under consideration. If a curve passes through the same point at different parameter values, test each parameter value separately.
Worked Examples
Worked Example: A Parabola with a Vertical Tangent
Find the vertical tangent for \(x=1+t^2\), \(y=-2+3t\) at \(t=0\).
Differentiate each coordinate:
At \(t=0\), \(dx/dt=2(0)=0\), while \(dy/dt=3\ne0\). The vertical-tangent condition is satisfied. Calculate the point:
Therefore the point is \((1,-2)\), and the tangent line is \(x=1\). To check the direction directly, note that \(y+2=3t\), so \(t=(y+2)/3\). Substituting into \(x\) gives
The horizontal change relative to the vertical change tends to zero at \((1,-2)\), confirming the vertical tangent \(x=1\).
Worked Example: Two Vertical Tangents
Find all vertical tangents for \(x=t^3-3t\), \(y=t\) on \(-2\le t\le2\).
Differentiate:
Setting \(dx/dt=0\) gives \(t=1\) or \(t=-1\). Both values lie in the interval, and \(dy/dt=1\ne0\) at each. At \(t=1\),
This gives the point \((-2,1)\) and the tangent line \(x=-2\). At \(t=-1\),
This gives \((2,-1)\) and the tangent line \(x=2\). The tangent directions can also be checked by writing \(t=1+h\) and \(t=-1+h\), respectively. For \(t=1+h\), the coordinate changes from \((-2,1)\) are \(3h^2+h^3\) horizontally and \(h\) vertically, so their ratio is \(3h+h^2\to0\). For \(t=-1+h\), the changes from \((2,-1)\) are \(-3h^2+h^3\) and \(h\), so the ratio is \(-3h+h^2\to0\). Both tangents are vertical, as found by the derivative test.
Worked Example: Vertical Tangents on a Circle
Find the vertical tangents for \(x=\cos t\), \(y=\sin t\) on \(0\le t\le2\pi\).
Differentiate and solve for the candidate parameters:
The equation \(-\sin t=0\) gives \(t=0,\pi,2\pi\) on this interval. At \(t=0\), \(dy/dt=1\ne0\); at \(t=\pi\), \(dy/dt=-1\ne0\); and at \(t=2\pi\), \(dy/dt=1\ne0\). The associated points are
The tangent lines are \(x=1\) at \((1,0)\) and \(x=-1\) at \((-1,0)\). The parameter values \(0\) and \(2\pi\) give the same point and the same tangent line. To check the direction near \((1,0)\), the circle equation \(x^2+y^2=1\) gives \(x=\sqrt{1-y^2}\) near that point. Hence
Near \((-1,0)\), \(x=-\sqrt{1-y^2}\), and
Both limiting checks confirm the vertical tangent lines. The endpoint parameter values represent the same point on the circle; the tangent line there is still \(x=1\).
Common Mistake: Checking Only \(dx/dt=0\)
A frequent error is to treat every solution of \(dx/dt=0\) as a vertical tangent. That equation alone does not establish the tangent condition; \(dy/dt\) must also be checked.
Worked Example: A Zero Horizontal Derivative Is Not Enough
Consider \(x=t^2\), \(y=t^2\) at \(t=0\). A mistaken argument might say, “\(dx/dt=2t=0\), so the tangent is vertical.” This conclusion skips the other derivative and is incorrect.
In fact,
The required nonzero condition on \(dy/dt\) fails, so the vertical-tangent test does not apply. For every parameter value, \(x=y\), and the curve lies on the line \(y=x\), not on a vertical line. The correction is to check both coordinate derivatives before identifying a vertical tangent. When both derivatives vanish, this test alone gives no conclusion.
What to Report
A complete answer should identify the parameter value, verify the derivative conditions, give the point, and state the tangent line. A compact response might say: “At \(t=t_0\), \(dx/dt=0\) and \(dy/dt\ne0\); the point is \((x(t_0),y(t_0))=(a,b)\), so the vertical tangent is \(x=a\).” This makes clear both why the tangent is vertical and where it is located.
The derivative test is useful because it translates a geometric question into equations in the parameter. Its main limitation is equally important: if both coordinate derivatives are zero, one cannot decide the tangent direction using this test alone. Do not divide by \(dx/dt=0\), and do not label the point vertical without further analysis.
Check Your Understanding
For each question, use the derivative conditions and include the point when a vertical tangent is identified.
- For \(x=t^2+2\), \(y=4t-1\), find the vertical tangent, if one occurs.
- For \(x=t^3-3t\), \(y=2t+1\), find all parameter values giving vertical tangents and write their tangent lines.
- Why must \(dy/dt\ne0\) be checked after solving \(dx/dt=0\)?
- For \(x=\cos t\), \(y=\sin t\), what point and tangent line correspond to \(t=\pi\)?
- If both \(dx/dt\) and \(dy/dt\) are zero at a candidate parameter, what can the vertical-tangent test conclude?
