Introduction
A trigonometric parametrization describes a curve by coordinate functions involving sine or cosine. Its tangent behavior depends on how both coordinates change as the parameter changes. Differentiating the coordinates reveals the slope where it is defined, as well as the parameter values that give horizontal or vertical tangents.
Periodicity adds a useful structure: when the coordinates repeat after a shared parameter period, their derivatives and tangent behavior repeat as well. The key is to track both coordinate derivatives, rather than infer tangent behavior from one coordinate alone.
Differentiate Both Coordinates
For a parametrized curve \(x=x(t)\), \(y=y(t)\), the coordinate derivatives \(x'(t)\) and \(y'(t)\) describe the changes in the horizontal and vertical coordinates. For trigonometric coordinates, apply the chain rule: the derivative of a sine or cosine includes the derivative of its inside expression.
The nonzero condition matters. If \(x'(t_0)=0\), the quotient for slope is undefined there; if \(y'(t_0)=0\) too, it is not enough to label the tangent horizontal. The tests identify horizontal and vertical tangents when exactly one coordinate derivative vanishes.
Why the Parametric Slope Formula Works
Suppose \(x'(t_0)\ne 0\). Near \(t_0\), the horizontal coordinate changes at a nonzero rate, so the curve's slope can be found by comparing the changes in \(y\) and \(x\). For a small parameter change \(h\), differentiability gives the local approximations below.
Here \(o(h)\) denotes a quantity whose ratio to \(h\) tends to zero as \(h\) tends to zero. Dividing the changes in \(y\) by the changes in \(x\), the common factor \(h\) cancels. Since \(x'(t_0)\ne 0\), the resulting limit is the quotient of the derivatives:
This explains both the slope formula and its condition: the denominator derivative must be nonzero. When \(x'(t_0)=0\) but \(y'(t_0)\ne 0\), the tangent direction is vertical instead.
Worked Example: Tangents to an Ellipse
Worked Example: Tangents to an Ellipse
Consider \(x=2\cos t\), \(y=3\sin t\). Find the slope at \(t=\frac{\pi}{3}\), and identify all horizontal and vertical tangent parameters.
Differentiate each coordinate:
At \(t=\frac{\pi}{3}\), \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) and \(\cos(\frac{\pi}{3})=\frac12\), so
Because \(x'\ne 0\), the slope is
For horizontal tangents, solve \(y'(t)=0\): \(3\cos t=0\), so \(t=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb Z\). At these values, \(\sin t=(-1)^k\), hence \(x'(t)=-2(-1)^k\ne 0\). They are horizontal tangents. For vertical tangents, solve \(x'(t)=0\): \(-2\sin t=0\), so \(t=k\pi\). There \(\cos t=(-1)^k\), and \(y'(t)=3(-1)^k\ne 0\), so they are vertical tangents.
Check: Substituting \(t=\frac{\pi}{3}\) into the derivative quotient gives \((3/2)/(-\sqrt{3})=-\sqrt{3}/2\), as computed. The nonzero derivative checks also confirm both tangent classifications.
Periodic Tangent Results
A shared period makes the tangent analysis repeat. Suppose \(P>0\) satisfies \(x(t+P)=x(t)\) and \(y(t+P)=y(t)\) for every relevant \(t\). Differentiating these identities gives \(x'(t+P)=x'(t)\) and \(y'(t+P)=y'(t)\). Thus the coordinates, derivative values, slope tests, and tangent classifications at \(t\) recur at \(t+P\).
For example, \(\sin(2t)\) and \(\cos(2t)\) both have period \(\pi\), since adding \(\pi\) to \(t\) adds \(2\pi\) to their angles. When the coordinates use different frequencies, check for a period shared by both; do not assume that the period of one coordinate is a period of the whole parametrization.
Worked Example: A Parametrization with Shared Period \(\pi\)
Let \(x=1+2\sin(2t)\) and \(y=-1+3\cos(2t)\). Find the horizontal and vertical tangent parameters, and describe the repeating results.
Differentiate using the chain rule:
For a horizontal tangent, set \(y'(t)=0\). Then \(\sin(2t)=0\), so \(2t=k\pi\), or \(t=\frac{k\pi}{2}\). At these parameters, \(\cos(2t)=(-1)^k\), giving \(x'(t)=4(-1)^k\ne 0\). The coordinates are \(x=1\) and \(y=-1+3(-1)^k\): for even \(k\), the point is \((1,2)\); for odd \(k\), it is \((1,-4)\).
For a vertical tangent, set \(x'(t)=0\). Then \(\cos(2t)=0\), so \(2t=\frac{\pi}{2}+k\pi\), or \(t=\frac{\pi}{4}+\frac{k\pi}{2}\). At these parameters, \(\sin(2t)=(-1)^k\), giving \(y'(t)=-6(-1)^k\ne 0\). The coordinates are \(x=1+2(-1)^k\) and \(y=-1\): for even \(k\), the point is \((3,-1)\); for odd \(k\), it is \((-1,-1)\).
Both coordinates have period \(\pi\), so the full parameter results repeat when \(t\) is increased by \(\pi\). For instance, the horizontal parameter \(t=0\) recurs at \(t=\pi\); the vertical parameter \(t=\frac{\pi}{4}\) recurs at \(t=\frac{5\pi}{4}\).
Check: At a horizontal parameter, \(y'=0\) and \(x'=4(-1)^k\), which is nonzero. At a vertical parameter, \(x'=0\) and \(y'=-6(-1)^k\), also nonzero. The coordinates at the shifted parameters agree because both sine and cosine angles have increased by \(2\pi\).
Worked Example: Different Frequencies
Worked Example: Different Frequencies
Consider \(x=\cos t\), \(y=\sin(2t)\). Find the horizontal and vertical tangent parameters in one interval \(0\le t<2\pi\), and use periodicity to describe their recurrence.
The coordinate derivatives are
Horizontal tangents require \(\cos(2t)=0\). In the stated interval, this gives \(t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). At these values, \(\sin t\ne 0\), so \(x'(t)\ne 0\). The corresponding points are \((\frac{\sqrt{2}}{2},1)\), \((-\frac{\sqrt{2}}{2},-1)\), \((-\frac{\sqrt{2}}{2},1)\), and \((\frac{\sqrt{2}}{2},-1)\), respectively.
Vertical tangents require \(\sin t=0\), giving \(t=0,\pi\) in this interval. At either value, \(\cos(2t)=1\), so \(y'(t)=2\ne 0\). The points are \((1,0)\) at \(t=0\) and \((-1,0)\) at \(t=\pi\). Both coordinates have period \(2\pi\), so these results recur at parameters shifted by integer multiples of \(2\pi\).
Check: At all four horizontal parameters, \(y'=2\cos(2t)=0\) while \(-\sin t\ne 0\). At \(t=0,\pi\), \(x'=-\sin t=0\) while \(y'=2\cos(2t)=2\). Thus each classification satisfies its required nonzero condition.
A Common Chain-Rule Error
For a coordinate such as \(\sin(3t)\), the derivative is \(3\cos(3t)\), not just \(\cos(3t)\). Omitting the inner derivative can produce an incorrect slope when the coordinates have different frequencies.
Worked Example: Checking the Chain-Rule Factors
Let \(x=2\sin(3t)\) and \(y=\cos(2t)\). Find the slope at \(t=\frac{\pi}{4}\). Differentiating correctly gives
At \(t=\frac{\pi}{4}\), \(\cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}\) and \(\sin(\frac{\pi}{2})=1\). Therefore \(x'=-3\sqrt{2}\), \(y'=-2\), and
A common wrong calculation omits the inner derivatives and uses \(x'=2\cos(3t)\), \(y'=-\sin(2t)\). It produces the slope \((-1)/(-\sqrt{2})=\sqrt{2}/2\), which is not the correct result. The correction is to apply the chain rule to both coordinates before forming the quotient.
Check: Substituting the correct derivatives gives \((-2)/(-3\sqrt{2})=\sqrt{2}/3\). Substituting the omitted-factor derivatives gives a different value, confirming that the error changes the slope.
Check the Conditions Before Classifying
A zero in one derivative is not enough by itself: the other derivative must be nonzero for the horizontal or vertical test. If both vanish, the quotient test does not classify the tangent, and a separate analysis may be needed.
For instance, take \(x=\cos t\) and \(y=\sin^2 t\). At \(t=0\), \(x'= -\sin t=0\) and \(y'=2\sin t\cos t=0\). Calling this a horizontal tangent just because \(y'=0\) would ignore the required condition \(x'\ne0\). For nearby \(t\ne0\), the slope is \(\frac{2\sin t\cos t}{-\sin t}=-2\cos t\), which approaches \(-2\) as \(t\) approaches \(0\). The zero-derivative test alone therefore does not justify a horizontal classification.
Check Your Understanding
For each question, show the coordinate derivatives and check the nonzero condition before classifying a tangent.
- For \(x=3\cos t\), \(y=2\sin t\), find the slope at \(t=\frac{\pi}{4}\).
- For the same parametrization, find all horizontal and vertical tangent parameters.
- For \(x=1+\sin(2t)\), \(y=2+\cos(2t)\), find the horizontal tangent parameters in \(0\le t<2\pi\).
- For \(x=\sin(2t)\), \(y=\cos(3t)\), calculate the slope at \(t=\frac{\pi}{4}\), if it is defined.
- What must be checked if both coordinate derivatives are zero at a parameter value?
