What Makes a Tangent Horizontal?
A parametrized curve is described by coordinate functions \(x=x(t)\) and \(y=y(t)\), where \(t\) ranges over a specified parameter domain. As \(t\) changes, the point \((x(t),y(t))\) moves along the curve. At a parameter value \(t_0\), the derivatives \(x'(t_0)\) and \(y'(t_0)\) describe the rates of change of the horizontal and vertical coordinates.
A horizontal tangent has zero slope: as the curve passes through the point, its tangent direction is horizontal. For a regular parameter value, the parametric slope formula is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\), provided \(dx/dt\ne 0\). Thus, a horizontal tangent occurs when the vertical coordinate has zero rate of change while the horizontal coordinate has a nonzero rate of change.
The two conditions matter together. Solving \(y'(t)=0\) gives candidate parameter values, not automatically horizontal tangents. At each candidate, check that \(x'(t)\) is nonzero, then substitute the parameter into both coordinate functions to find the point. Different parameter values may describe the same point, so report the resulting points and tangent lines as well as the parameter values.
Why the Criterion Works
The slope formula follows from comparing changes in the two coordinates. For \(t\ne t_0\), the slope of the secant joining the points at \(t_0\) and \(t\) is \(\frac{y(t)-y(t_0)}{x(t)-x(t_0)}\), whenever the denominator is nonzero. Factor the changes using the parameter increment:
As \(t\) approaches \(t_0\), the numerator of the right-hand side approaches \(y'(t_0)\), and its denominator approaches \(x'(t_0)\). If \(x'(t_0)\ne 0\), the denominator has a nonzero limit, so the ratio tends to \(y'(t_0)/x'(t_0)\). In particular, if \(y'(t_0)=0\), the tangent slope is zero. Since \(x'(t_0)\ne0\), the tangent direction is not stationary: the curve is changing horizontally at that parameter value.
This argument also explains why the nonzero condition cannot be dropped. If \(x'(t_0)=0\), the quotient limit just used has a zero denominator in its limiting expression, and the parametric slope formula does not establish a horizontal tangent. A parameter value where both derivatives vanish needs separate analysis; the criterion here does not classify it.
A Reliable Procedure
For a parametrized curve on a stated parameter domain, use the following steps. Keep the domain in view: roots outside it do not describe points on the parametrized curve being considered.
Find \(x'(t)\) and \(y'(t)\).
Solve \(y'(t)=0\) within the parameter domain.
Evaluate \(x'(t)\). Keep a candidate as a horizontal tangent parameter only if \(x'(t)\ne0\).
Evaluate \(x(t)\) and \(y(t)\), then state the point and the horizontal line \(y=y(t)\). Check that the derivative pair satisfies the criterion.
When a parameter domain includes an endpoint, the same calculation applies if the parametrization extends differentiably to a neighborhood of that endpoint; otherwise, the tangent must be interpreted using the curve’s one-sided behavior. The examples below use parameter values where the derivatives are defined.
Polynomial Parametrizations
Worked Example: Two Horizontal Tangents on a Polynomial Curve
Find all horizontal tangents for \(x(t)=t^2+1\) and \(y(t)=t^3-3t\), with \(t\in\mathbb{R}\).
First differentiate each coordinate:
Solving \(y'(t)=0\) gives \(t=1\) or \(t=-1\). Check the other derivative at each candidate:
Both parameters give horizontal tangents. At \(t=1\), the point is \((x(1),y(1))=(1^2+1,1^3-3(1))=(2,-2)\). Its tangent line is \(y=-2\). At \(t=-1\), the point is \((x(-1),y(-1))=((-1)^2+1,(-1)^3-3(-1))=(2,2)\). Its tangent line is \(y=2\).
To verify the slopes, the parametric slope at either parameter is \(y'(t)/x'(t)\). At \(t=1\), it is \(0/2=0\); at \(t=-1\), it is \(0/(-2)=0\). The corresponding horizontal lines therefore have the claimed slopes and pass through their respective points.
This example also shows why it is useful to keep track of the parameter and the point separately. The two tangent points have the same \(x\)-coordinate, but different \(y\)-coordinates and different tangent lines.
Trigonometric Parametrizations
Worked Example: Horizontal Tangents on a Trigonometric Curve
Find the horizontal tangents for \(x(t)=3\cos t\) and \(y(t)=2\sin t\), where \(0\le t<2\pi\).
Differentiate:
Set \(y'(t)=0\). Since \(2\cos t=0\) exactly when \(\cos t=0\), the candidates in the given interval are \(t=\frac{\pi}{2}\) and \(t=\frac{3\pi}{2}\). At these values,
Both candidates pass the test. At \(t=\frac{\pi}{2}\), the point is \((3\cos(\frac{\pi}{2}),2\sin(\frac{\pi}{2}))=(0,2)\), so the tangent line is \(y=2\). At \(t=\frac{3\pi}{2}\), the point is \((3\cos(\frac{3\pi}{2}),2\sin(\frac{3\pi}{2}))=(0,-2)\), so the tangent line is \(y=-2\).
For verification, the slope at the first parameter is \(y'/x'=0/(-3)=0\); at the second it is \(0/3=0\). Each line also passes through the point computed for its parameter.
When a Candidate Does Not Pass the Test
Worked Example: Checking Candidates Where Both Derivatives Vanish
For \(x(t)=t^3-3t\) and \(y(t)=t^4-2t^2\), find the parameter values that the horizontal-tangent criterion confirms.
Differentiate:
The candidates from \(y'(t)=0\) are \(t=-1,0,1\). Check \(x'\) at all three:
Only \(t=0\) has \(x'(t)\ne0\). Its point is \((x(0),y(0))=(0,0)\), and the slope is \(y'(0)/x'(0)=0/(-3)=0\). Thus the criterion confirms the horizontal tangent \(y=0\) at \((0,0)\).
At \(t=-1\) and \(t=1\), both derivatives are zero. It would be incorrect to substitute either into \(y'/x'\) and claim the result is zero: both quotients would be \(0/0\), which is undefined. This test alone does not determine the tangent at those parameters.
Common Pitfall: Stopping After Solving \(y'(t)=0\)
A common error is to treat every root of \(y'(t)\) as a horizontal tangent parameter. The last example gives a concrete countercheck. Solving \(4t(t-1)(t+1)=0\) produces three candidates, but at \(t=\pm1\) the denominator \(x'(t)\) is also zero. Therefore, the slope formula cannot be used there. The valid conclusion from this criterion is the horizontal tangent at \(t=0\), not three horizontal tangents.
Another avoidable error is to reverse the quotient: the slope is \(y'(t)/x'(t)\), not \(x'(t)/y'(t)\). The numerator measures change in height, and the denominator measures change in horizontal position, matching the ordinary definition of slope as rise over run. At a valid horizontal tangent, the rise rate is zero while the run rate is nonzero.
Why This Test Is Useful
Parametric equations often describe a curve more naturally than a single equation \(y=f(x)\). The parameter may trace a curve that bends back in the plane, or it may make different parameter values correspond to different points with the same horizontal coordinate. The derivative pair handles the motion directly: \(x'(t)\) records horizontal change, and \(y'(t)\) records vertical change.
For horizontal tangents, this turns a geometric question into a precise calculation: find where vertical change stops, then make sure horizontal change continues. The point and tangent line still need to be reported, because a parameter value alone does not specify the geometric answer. Keep the distinction clear between a candidate root, a parameter that passes the nonzero test, and the resulting point on the curve.
Check Your Understanding
For each question, use the condition \(y'(t)=0\) together with \(x'(t)\ne0\).
- For \(x(t)=t^2+2\) and \(y(t)=t^3-3t\), find the horizontal tangent parameter values, points, and tangent lines.
- For \(x(t)=2\cos t\) and \(y(t)=\sin t\), with \(0\le t<2\pi\), find and verify the horizontal tangents.
- Why does \(y'(t_0)=0\) alone not establish a horizontal tangent when \(x'(t_0)=0\)?
- For \(x(t)=t^3\) and \(y(t)=t^2\), what parameter value is a candidate from \(y'(t)=0\), and does the criterion confirm a horizontal tangent there?
- At a valid horizontal tangent, what is the equation of the tangent line in terms of the point’s \(y\)-coordinate?