From Polynomial Coordinates to Tangents
A parametrized curve is given by coordinate functions \(x=x(t)\) and \(y=y(t)\). When those functions are polynomials, their derivatives and values can be calculated directly. The main task is to organize those calculations so that a parameter value gives the correct point, tangent direction, and—when defined—second derivative.
At a specified parameter \(t=a\), first find the point \((x(a),y(a))\). Then use \(x'(a)\) and \(y'(a)\) to determine the tangent. If \(x'(a)\ne 0\), the slope is \(y'(a)/x'(a)\). If \(x'(a)=0\) but \(y'(a)\ne 0\), the tangent is vertical. The distinction matters: a zero denominator in the slope formula does not produce an ordinary numerical slope.
Polynomial coordinate functions can be differentiated using the ordinary rules for polynomials. After differentiating, evaluate each derivative at the parameter of interest. Keep the parameter value separate from the point’s coordinates: \(a\) is the input, while \((x(a),y(a))\) is the point on the curve.
Finding a Tangent Line at a Regular Parameter
When \(x'(a)\ne0\), calculate the slope \(m=y'(a)/x'(a)\). The point-slope equation of the tangent is then \(y-y(a)=m(x-x(a))\). This equation is a line in the coordinate plane, so its point must use the coordinate values at \(a\), not the parameter itself.
Worked Example: A Horizontal Tangent and Its Second Derivative
Consider \(x(t)=t^2+1\) and \(y(t)=t^3-3t\). Find the tangent line and the second derivative at \(t=1\).
First evaluate the coordinates:
The point is \((2,-2)\). Differentiate the coordinate functions:
At \(t=1\), \(x'(1)=2\) and \(y'(1)=0\). Since \(x'(1)\ne0\), the slope is \(0/2=0\), so the tangent is horizontal. Its equation through \((2,-2)\) is
To find the second derivative, differentiate again: \(x''(t)=2\) and \(y''(t)=6t\). The parametric second-derivative formula gives
At \(t=1\), the numerator is \(2(6)-0(2)=12\), and the denominator is \(2^3=8\). Therefore \(\frac{d^2y}{dx^2}=12/8=3/2\).
Check the tangent: it contains the point because \(y=-2\) at \((2,-2)\), and its slope is \(0=y'(1)/x'(1)\). For a second-derivative check, the first derivative as a function of \(t\) is \(\frac{dy}{dx}=(3t^2-3)/(2t)\). Its derivative with respect to \(t\) is \(\frac{3}{2}(1+1/t^2)\); dividing by \(dx/dt=2t\) gives \(\frac{3(1+1/t^2)}{4t}\), which is \(3/2\) at \(t=1\), as above.
Horizontal and Vertical Tangencies
For a regular parameter value, the derivative pair \((x'(a),y'(a))\) gives the tangent direction. A horizontal tangent occurs when \(y'(a)=0\) and \(x'(a)\ne0\). A vertical tangent occurs when \(x'(a)=0\) and \(y'(a)\ne0\). Both conditions include a nonzero derivative: if both coordinate derivatives vanish, the parameter value is not regular, and these tests alone do not classify the tangent.
With polynomial coordinates, finding these tangencies means solving polynomial equations. For horizontal tangencies, find the roots of \(y'(t)=0\) and retain only those where \(x'(t)\ne0\). For vertical tangencies, solve \(x'(t)=0\) and retain only those where \(y'(t)\ne0\). For each retained value, evaluate both original coordinate functions to state the location of the tangent.
Worked Example: Locating Horizontal and Vertical Tangencies
Consider \(x(t)=t^3-3t\) and \(y(t)=t^2+1\). Find the horizontal and vertical tangencies for \(-2\le t\le2\).
Differentiate:
For a horizontal tangent, solve \(y'(t)=0\). This gives \(2t=0\), so \(t=0\). At that value, \(x'(0)=-3\ne0\), so the point is regular and the tangent is horizontal. Its point is \(x(0)=0\), \(y(0)=1\), hence \((0,1)\), and its tangent line is \(y=1\).
For a vertical tangent, solve \(x'(t)=0\). The roots are \(t=1\) and \(t=-1\). At each, \(y'(1)=2\ne0\) and \(y'(-1)=-2\ne0\), so both are regular vertical tangencies. At \(t=1\), the point is \(x(1)=1-3=-2\), \(y(1)=1+1=2\), giving the vertical line \(x=-2\). At \(t=-1\), the point is \(x(-1)=-1+3=2\), \(y(-1)=1+1=2\), giving the vertical line \(x=2\).
Check the classifications: \(y'(0)=0\) while \(x'(0)=-3\), and \(x'(\pm1)=0\) while \(y'(\pm1)=\pm2\). The stated points also satisfy the original coordinate equations at their respective parameter values. Thus the horizontal line passes through \((0,1)\), and each vertical line passes through its corresponding point.
Why the Second-Derivative Formula Works
The second derivative \(\frac{d^2y}{dx^2}\) describes how the slope \(\frac{dy}{dx}\) changes with \(x\). Along a parametrized curve, both slope and horizontal position depend on \(t\). When \(x'(t)\ne0\), the chain rule says that differentiating with respect to \(x\) is equivalent to differentiating with respect to \(t\) and dividing by \(dx/dt\).
Here is the derivation. First, the first derivative is \(\frac{dy}{dx}=\frac{y'}{x'}\). To differentiate that slope with respect to \(x\), apply the chain rule in the form \(\frac{d}{dx}=\frac{1}{x'}\frac{d}{dt}\). The quotient rule then gives:
The denominator has power three because one factor of \(x'\) comes from changing the differentiation variable from \(t\) to \(x\), while two more come from differentiating the quotient \(y'/x'\). This also explains why checking \(x'\ne0\) is essential before applying the formula.
Worked Example: A Sloping Tangent and Second Derivative
Let \(x(t)=t^2+2t\) and \(y(t)=t^3-t\). Find the tangent line and second derivative at \(t=1\).
The point is found from the original functions:
Differentiate to obtain \(x'(t)=2t+2\) and \(y'(t)=3t^2-1\). Thus \(x'(1)=4\) and \(y'(1)=2\). Since \(x'(1)\ne0\), the slope is \(2/4=1/2\). The tangent equation is
For the second derivative, \(x''(t)=2\) and \(y''(t)=6t\), so \(x''(1)=2\) and \(y''(1)=6\). Substitution into the formula gives
Check the tangent: at \((3,0)\), its equation gives \(0=\frac12(3)-\frac32\), and its slope is \(1/2=y'(1)/x'(1)\). For an independent second-derivative check, write the first derivative as a function of the parameter: \(\frac{dy}{dx}=(3t^2-1)/(2t+2)\). Its derivative at \(t=1\), by the quotient rule, is \(\frac{6(1)(4)-2(2)}{4^2}=20/16=5/4\). Dividing by \(x'(1)=4\) gives \(5/16\), agreeing with the formula.
A Common Pitfall: Dividing by Zero
At \(t=1\) in the preceding horizontal-and-vertical example, \(x'(1)=0\) and \(y'(1)=2\). A tempting but incorrect calculation is to write the slope as \(2/0\) and treat it as an ordinary number, or to conclude that no tangent exists because the quotient is undefined. Division by zero is undefined; the correct conclusion comes from the derivative pair. Since the point is regular and its horizontal derivative is zero, the tangent is vertical. The point is \((-2,2)\), so its tangent line is \(x=-2\).
A related error is to use the second-derivative formula at a parameter where \(x'=0\). Its denominator then vanishes, so the formula is not applicable. Check the required condition before substituting values; a vertical tangent is not an occasion to force a finite value of \(\frac{d^2y}{dx^2}\).
A Reliable Calculation Sequence
For a polynomial parametrization, the following sequence keeps the point, slope, and second derivative distinct and makes it easier to catch errors.
Substitute the specified parameter into \(x(t)\) and \(y(t)\) to obtain \((x(a),y(a))\).
Find \(x'(t)\) and \(y'(t)\), then evaluate both at \(a\).
If \(y'(a)=0\) and \(x'(a)\ne0\), it is horizontal. If \(x'(a)=0\) and \(y'(a)\ne0\), it is vertical. If both are nonzero, use their ratio for the slope.
Use the point and slope, or state the vertical line through the point. Verify that the line contains the point and has the calculated direction.
Evaluate \(x''(a)\) and \(y''(a)\), check \(x'(a)\ne0\), and substitute into the second-derivative formula.
Polynomial parametrizations are useful because the calculations reduce to polynomial differentiation, solving polynomial equations, and careful substitution. The same derivative tests distinguish tangent directions at individual parameter values, while the second-derivative formula measures how the tangent slope changes with horizontal position where that slope is defined.
Check Your Understanding
Use the derivative conditions and the coordinate functions carefully when answering.
- For \(x(t)=t^2+1\) and \(y(t)=t^3-3t\), what is the point and tangent slope at \(t=1\)?
- State the two derivative conditions for a horizontal tangent and the two conditions for a vertical tangent at a regular parameter value.
- For \(x(t)=t^3-3t\) and \(y(t)=t^2+1\), find the parameter values and points of all horizontal and vertical tangencies.
- Why must \(x'(a)\ne0\) before using the parametric second-derivative formula at \(t=a\)?
- For \(x(t)=t^2+2t\) and \(y(t)=t^3-t\), verify the tangent slope at \(t=1\) from the coordinate derivatives.
