The Curve and Its Repeating Shape
Consider the parametrized curve
The parameter \(t\) can take any real value. Increasing \(t\) traces a sequence of arches. Since sine and cosine have period \(2\pi\), replacing \(t\) by \(t+2\pi\) increases \(x\) by \(2\pi\) and leaves \(y\) unchanged:
Thus each arch has the same shape and is shifted horizontally by \(2\pi\). We will examine the tangent behavior and concavity of the arches, paying particular attention to parameter values where the usual slope formula needs care.
Derivative Pair and Tangent Directions
Differentiating each coordinate with respect to \(t\) gives
At a parameter value where \(x'(t)\ne 0\), the curve has a locally defined slope as a function of \(x\), given by \(dy/dx=y'(t)/x'(t)\). If both derivatives vanish, the derivative pair is zero and this regular-point slope formula does not apply directly.
Here, \(x'(t)=0\) exactly when \(\cos t=1\), or \(t=2\pi k\) for an integer \(k\). At each such value, \(\sin t=0\) as well, so the derivative pair vanishes. These are stationary parameter values. In contrast, \(y'(t)=0\) also when \(t=(2k+1)\pi\). There, \(x'(t)=2\), so the point is regular and has a horizontal tangent.
For regular values, the slope simplifies using the half-angle identities:
This expression applies when \(t\ne 2\pi k\). It gives slope \(0\) at \(t=(2k+1)\pi\), consistent with the horizontal tangent there. At a stationary parameter, however, the expression is undefined; the tangent direction must be determined from the curve’s local behavior.
Worked Example: Finding a Regular Tangent Line
Find the tangent line at \(t=\pi/3\). First calculate the point:
The parameter is regular because
Therefore the slope is \((\sqrt{3}/2)/(1/2)=\sqrt{3}\). The point-slope equation is
Check: substituting the point makes both sides zero. The slope also agrees with \(\cot(\pi/6)=\sqrt{3}\).
Stationary Parameters Are Cusps
At \(t=2\pi k\), the point on the curve is \((2\pi k,0)\), and both coordinate derivatives are zero. To identify its tangent direction, write \(t=2\pi k+h\). Periodicity gives
As \(h\) approaches zero, the Taylor expansions \(\sin h=h-h^3/6+O(h^5)\) and \(\cos h=1-h^2/2+O(h^4)\) show that
Consequently, \((x(t)-2\pi k)/y(t)\) tends to zero. The displacement from the point therefore approaches the vertical direction, so the cusp has the vertical tangent line \(x=2\pi k\). The slope \(dy/dx\) becomes unbounded on either side; the zero derivative pair does not mean that the tangent direction is absent.
Worked Example: Tangent at the Cusp \(t=0\)
At \(t=0\), the point is \((0,0)\), and \(x'(0)=y'(0)=0\). For small \(h\),
The displacement ratio satisfies
Thus the secant direction approaches the vertical direction, and the tangent line is \(x=0\). This is consistent with the slope formula away from zero: \(\cot(h/2)\) tends to \(+\infty\) as \(h\to0^+\) and to \(-\infty\) as \(h\to0^-\). Both one-sided slopes describe the same vertical line. Substituting the point \((0,0)\) into \(x=0\) verifies that it lies on the tangent line.
Deriving the Concavity Formula
Where \(x'(t)\ne0\), the chain rule gives \(dy/dx=y'(t)/x'(t)\). Differentiating this slope with respect to \(t\), then dividing by \(dx/dt\), gives the second derivative with respect to \(x\):
This formula requires \(x'(t)\ne0\), so it applies on the open intervals between the cusp parameters. For this curve, \(x''(t)=\sin t\) and \(y''(t)=-\cos t\). Substitution and simplification yield
The denominator is positive wherever \(t\ne2\pi k\). Hence the sign of the second derivative is the opposite of the sign of \(1+2\cos t\). The curve is concave up when \(\cos t<-1/2\), and concave down when \(\cos t>-1/2\).
To see the interval boundaries, solve \(1+2\cos t=0\), which gives \(\cos t=-1/2\). Within one period these values occur at \(2\pi/3\) and \(4\pi/3\); translating by \(2\pi k\) gives the stated values in every period. The cosine is greater than \(-1/2\) near each period endpoint and less than \(-1/2\) between the two solutions. Since the second derivative changes sign at each solution, both points are inflection points.
Worked Example: Concavity and Inflection Points in One Period
Take the period from \(t=0\) to \(t=2\pi\). The second derivative is
At \(t=\pi/3\), \(\cos t=1/2\), so the numerator factor \(1+2\cos t=2\), while \((1-\cos t)^2=1/4\). Thus \(d^2y/dx^2=-2/(1/4)=-8\), and the curve is concave down there. At \(t=\pi\), the factor is \(-1\) and the denominator is \(4\), giving \(d^2y/dx^2=1/4\); the curve is concave up. At \(t=5\pi/3\), the cosine is again \(1/2\), so the second derivative is \(-8\), and the curve is concave down again.
The boundary parameters are \(2\pi/3\) and \(4\pi/3\). Their points are
At each boundary, \(1+2\cos t=0\), so the second derivative is zero. The signs on either side change from negative to positive at the first and from positive to negative at the second, confirming both are inflection points.
Horizontal Tangencies and a Common Pitfall
At \(t=(2k+1)\pi\), the point is \(((2k+1)\pi,2)\): indeed, \(\sin t=0\), \(\cos t=-1\), and \(x'(t)=2\). Its tangent is horizontal. The concavity formula gives \(d^2y/dx^2=1/4\) there, so this horizontal tangent occurs while the curve is concave up.
A common error is to compute the second derivative as \(y''(t)/x'(t)\). At \(t=\pi\), this would give \(1/2\), since \(y''(\pi)=1\) and \(x'(\pi)=2\). That is not the derivative of the slope: the slope itself changes with \(t\), and the quotient rule is needed. The correct calculation is
The distinction between cusp and horizontal tangent is equally important. At a cusp, both coordinate derivatives vanish and the regular slope formula fails. At a regular horizontal tangent, only \(y'(t)\) vanishes, while \(x'(t)\) remains nonzero. Checking the derivative pair before interpreting a slope prevents these two geometrically different situations from being confused.
Check Your Understanding
Use the derivative pair, tangent formula, and concavity sign to answer these questions.
- At which parameter values in one period does the derivative pair vanish, and what are the corresponding points?
- Find the tangent direction and tangent line at \(t=\pi\).
- Why does the regular parametric slope formula not apply at \(t=2\pi\), and what is the tangent line there?
- State the concavity intervals on the period \((0,2\pi)\).
- Find the coordinates of the inflection point corresponding to \(t=2\pi/3\).
