What Makes a Parametric Tangent Horizontal?
A parametric curve is described by two coordinate functions of a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). As \(t\) changes, the point \((x(t),y(t))\) moves along the curve. To find a horizontal tangent, we need to identify where the curve’s tangent direction is horizontal, not merely where one coordinate happens to stop changing.
At a parameter value \(t_0\), the coordinate derivatives \(x'(t_0)\) and \(y'(t_0)\) describe the components of the tangent vector. If the horizontal coordinate is changing while the vertical coordinate is momentarily not changing, the tangent vector points horizontally. This gives a precise test, with an important condition: the horizontal component must be nonzero.
The phrase “at \(t_0\)” matters: a curve may pass through the same point at different parameter values with different tangent directions. When a point is reached more than once, check each parameter value separately.
The Parametric Slope Formula
When \(x'(t_0)\ne 0\), the slope of the tangent at the point corresponding to \(t_0\) is given by the parametric derivative formula
For the usual AP Calculus functions, the formula follows by applying the chain rule after using \(x\) as a local variable. To justify that step carefully, suppose \(x\) is continuously differentiable near \(t_0\) and \(x'\) remains nonzero there. The inverse function theorem then gives a differentiable local inverse for \(x(t)\), so \(y\) can be expressed locally as a function of \(x\). The chain rule gives \(dy/dt=(dy/dx)(dx/dt)\), and division by the nonzero \(dx/dt\) yields the formula.
Even without making a local-graph claim, the tangent-vector interpretation gives the same horizontal-tangent test: the tangent vector is \((x'(t_0),y'(t_0))\). When \(x'(t_0)\ne 0\), this vector is nonzero, and if \(y'(t_0)=0\), its direction is horizontal. In particular, the slope formula has numerator zero and nonzero denominator, so the slope is \(0\).
A Reliable Procedure
For a specified parameter interval, use the following steps. The interval matters because a root of \(y'\) outside the allowed interval does not describe a point on the portion of the curve being considered.
Find \(x'(t)\) and \(y'(t)\).
Solve \(y'(t)=0\) on the specified parameter interval.
For each candidate, compute \(x'(t)\) and retain it only if \(x'(t)\ne 0\).
Substitute the parameter into \(x(t)\) and \(y(t)\). The horizontal tangent line is \(y=y(t)\).
Evaluate \(y'(t)/x'(t)\); it must equal \(0\) at every retained parameter value.
Worked Examples
Worked Example: A Polynomial Parametrization
Find the horizontal tangents to \(x=t^2+1\), \(y=t^3-3t\), for \(-2\le t\le 2\).
First differentiate each coordinate:
Set the vertical-coordinate derivative equal to zero and solve:
Check the horizontal-coordinate derivative at both candidates:
Both candidates give horizontal tangents. At \(t=-1\), the point is
so the tangent line is \(y=2\). Its slope check is \(y'(-1)/x'(-1)=0/(-2)=0\).
At \(t=1\), the point is
so the tangent line is \(y=-2\). Its slope check is \(y'(1)/x'(1)=0/2=0\). The two tangents have the same \(x\)-coordinate but different points and different tangent lines.
Worked Example: A Trigonometric Parametrization
Find the horizontal tangents for \(x=2\cos t\), \(y=\sin(2t)\), on \(0\le t<2\pi\).
Differentiate, using the chain rule for \(y(t)\):
Set \(y'(t)=0\):
On \(0\le t<2\pi\), this occurs at \(t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). At each value, \(\sin t\) is either \(\frac{\sqrt2}{2}\) or \(-\frac{\sqrt2}{2}\), so \(x'(t)=-2\sin t\ne0\). All four candidates qualify.
Substitute each value to find the points:
For example, at \(t=\frac{\pi}{4}\), \(x'=-\sqrt2\) and \(y'=2\cos(\frac{\pi}{2})=0\), giving slope \(0/(-\sqrt2)=0\). At \(t=\frac{3\pi}{4}\), \(x'=-\sqrt2\) and \(y'=2\cos(\frac{3\pi}{2})=0\), again giving slope zero. At \(t=\frac{5\pi}{4}\), \(x'=\sqrt2\) and \(y'=2\cos(\frac{5\pi}{2})=0\), so the slope is zero. At \(t=\frac{7\pi}{4}\), \(x'=\sqrt2\) and \(y'=2\cos(\frac{7\pi}{2})=0\), also giving slope zero. The four tangent lines are \(y=1\), \(y=-1\), \(y=1\), and \(y=-1\), respectively.
Worked Example: Rejecting a Candidate with Both Derivatives Zero
Consider \(x=\frac{t^3}{3}+\frac{t^2}{2}\) and \(y=\frac{t^3}{3}-\frac{t^2}{2}\), for \(-1\le t\le2\). Find any horizontal tangent that meets the stated conditions.
Differentiate:
The candidates from \(y'(t)=0\) are \(t=0\) and \(t=1\). At \(t=0\), both derivatives vanish: \(x'(0)=0\) and \(y'(0)=0\). Thus this value does not meet the horizontal-tangent condition \(x'\ne0\), and the slope formula cannot be used there to certify a horizontal tangent.
At \(t=1\), \(y'(1)=1-1=0\), while \(x'(1)=1+1=2\ne0\), so this candidate qualifies. Its coordinates are
The tangent line is \(y=-\frac16\). Checking the slope gives \(y'(1)/x'(1)=0/2=0\), as required.
Why the Nonzero Condition Matters
A common mistake is to solve \(y'(t)=0\) and declare every solution a horizontal tangent. For the third example, that would incorrectly label \(t=0\) as a horizontal tangent simply because \(y'(0)=0\). But \(x'(0)=0\) as well. The slope formula would require division by \(x'(0)\), which is division by zero:
and this expression does not give a slope. The correction is to check \(x'(t)\) at every candidate and to claim a horizontal tangent by this criterion only when \(x'(t)\ne0\). When both derivatives vanish, additional analysis is needed; the condition used in this tutorial alone does not settle the tangent direction.
A second useful distinction is between “the \(y\)-coordinate is momentarily not changing” and “the tangent is horizontal.” The equation \(y'(t_0)=0\) says the first. The nonzero condition \(x'(t_0)\ne0\) ensures that the curve is moving in a nonzero horizontal direction at that parameter value, which establishes the second.
Summary
For parametric equations \(x=x(t)\) and \(y=y(t)\), a horizontal tangent at an interior parameter value \(t_0\) requires \(y'(t_0)=0\) and \(x'(t_0)\ne0\). Once the candidate passes both tests, compute the point \((x(t_0),y(t_0))\); the horizontal tangent line is \(y=y(t_0)\). Always verify the slope using \(y'(t_0)/x'(t_0)\), and do not classify a parameter value where both coordinate derivatives vanish by this test.
Check Your Understanding
For each question, identify the candidate parameter values and apply both derivative conditions before reporting a tangent.
- For \(x=t^2\) and \(y=t^3-3t\), find all horizontal tangents on \(-2\le t\le2\). Give the point and line for each.
- For \(x=3\cos t\) and \(y=\sin t\), find the horizontal tangent on \(0\le t<2\pi\), if one exists. Verify the point and slope.
- Why is solving \(y'(t)=0\) alone insufficient to establish a horizontal tangent?
- For \(x=t^3\) and \(y=t^2\), explain why the parameter \(t=0\) does not satisfy the horizontal-tangent test given here.
- At a qualifying horizontal tangent, what is the slope according to the parametric slope formula, and what is the equation of the tangent line?
