What a Tangent Line at a Parameter Value Means
A parametric curve is described by two coordinate functions of a parameter: \(x=x(t)\) and \(y=y(t)\). A specified parameter value \(t_0\) identifies a particular point on the curve, namely \((x(t_0),y(t_0))\). To find the tangent line there, we need both that point and the curve's slope at that parameter value.
The parameter is not necessarily the \(x\)-coordinate. In particular, substituting \(t_0\) directly for \(x\) or \(y\) is generally incorrect. First evaluate each coordinate function at \(t_0\); then use the coordinate derivatives to calculate the slope.
This slope ratio is the parametric slope formula. Its condition \(x'(t_0)\ne 0\) matters: it ensures the ratio is defined and gives a finite slope. The tangent-line equation is most directly written in point-slope form, \(y-y_0=m(x-x_0)\), with \(x_0=x(t_0)\) and \(y_0=y(t_0)\).
A Short Justification of the Slope Formula
The formula compares changes in the two coordinates as the parameter changes. For a small nonzero change \(h\), the slope of the secant line between the points at \(t_0\) and \(t_0+h\) is
When \(x'(t_0)\ne 0\), this quotient is defined for sufficiently small nonzero \(h\). Divide its numerator and denominator by \(h\), then take the limit as \(h\) approaches zero. Differentiability gives the limit of the numerator divided by \(h\) as \(y'(t_0)\), and the limit of the denominator divided by \(h\) as \(x'(t_0)\). Because that latter limit is nonzero, the quotient limit is the ratio of the limits.
Thus the tangent slope is the rate of change of \(y\) with respect to the parameter divided by the rate of change of \(x\) with respect to the parameter. This argument also explains why the condition \(x'(t_0)\ne 0\) cannot simply be ignored.
A Reliable Method
For a specified parameter value, organize the work so the point and slope remain distinct until both are known.
Find \(x_0=x(t_0)\) and \(y_0=y(t_0)\). Together they give the point of tangency \((x_0,y_0)\).
Compute \(x'(t)\) and \(y'(t)\), keeping each derivative attached to its coordinate.
Substitute \(t_0\) into each derivative. If \(x'(t_0)\ne 0\), calculate \(m=y'(t_0)/x'(t_0)\).
Use \(y-y_0=m(x-x_0)\). Confirm that the line passes through the curve point and has the calculated slope.
Worked Examples
Worked Example: A Polynomial Parametrization
Find the tangent line to \(x=t^2+1\), \(y=t^3-t\) at \(t=2\).
First evaluate both coordinates:
The point of tangency is \((5,6)\). Next differentiate each coordinate with respect to \(t\):
At \(t=2\), the coordinate derivatives are
Since \(x'(2)=4\ne 0\), the parametric slope formula applies:
Use point-slope form with \((x_0,y_0)=(5,6)\):
Check: when \(x=5\), the equation gives \(y-6=0\), so \(y=6\); the line passes through \((5,6)\). Its slope is \(11/4\), which matches the derivative ratio. The tangent line is therefore correct.
Worked Example: A Trigonometric Parametrization
Find the tangent line to \(x=2\cos t\), \(y=\sin(2t)\) at \(t=\pi/3\). Use exact values.
Evaluate the coordinates at the specified parameter:
The point is \(\left(1,\frac{\sqrt{3}}{2}\right)\). Differentiate with respect to \(t\). For the second coordinate, the chain rule gives a factor of \(2\):
Evaluate these derivatives:
Since \(x'(\pi/3)=-\sqrt{3}\ne 0\), the slope is
The tangent line in point-slope form is
Check: substituting \(x=1\) makes the right side zero, so the equation gives \(y=\sqrt{3}/2\), the curve point. The line's slope is \(\sqrt{3}/3\), equal to \(y'(\pi/3)/x'(\pi/3)\). Both checks agree.
Worked Example: Simplifying a Polynomial Tangent Equation
Find the tangent line to \(x=t^2-1\), \(y=t^3+2t\) at \(t=1\), and write the result in slope-intercept form as well as point-slope form.
Evaluate the coordinate functions:
The point is \((0,3)\). The derivatives are
At \(t=1\),
Because \(x'(1)=2\ne 0\), the slope is \(m=5/2\). Therefore,
Check: setting \(x=0\) gives \(y=3\), so the line contains the curve point. The coefficient of \(x\) is \(5/2\), which is the calculated tangent slope. The simplified equation passes both checks.
When the Formula Does Not Apply
The parametric slope formula requires \(x'(t_0)\ne 0\). If \(x'(t_0)=0\), the quotient \(y'(t_0)/x'(t_0)\) is undefined, so it cannot be used to claim a finite slope. That does not by itself settle what the tangent line is; the point needs separate analysis. If both coordinate derivatives vanish, their ratio is also undefined, and the first derivatives alone do not determine the tangent line.
For the standard point-slope calculation in this lesson, verify the denominator before dividing. A parameter value is given, but that does not guarantee the formula will produce a finite slope there.
Common Mistake: Reversing the Derivative Ratio
A frequent error is to divide the \(x\)-derivative by the \(y\)-derivative. In the first example, that incorrect calculation would be
This reverses the roles of the coordinates. Slope means change in \(y\) divided by change in \(x\), so the correct calculation is
The check catches the error: \(4/11\) does not equal the ratio of the vertical coordinate's rate of change, \(11\), to the horizontal coordinate's rate of change, \(4\). Keep the order \(y'\) over \(x'\), and then use that slope with the point \((x(t_0),y(t_0))\).
Why Careful Evaluation Matters
Finding a tangent line combines three different pieces of information: the parameter value, the point it produces, and the slope at that parameter. Confusing these roles can lead to an equation that uses the right slope at the wrong point, or the correct point with the wrong slope. Writing down \(x(t_0)\), \(y(t_0)\), \(x'(t_0)\), and \(y'(t_0)\) separately makes those errors easier to detect.
For exact trigonometric values, retain radicals and fractions through the calculation rather than rounding early. The resulting line is exact, and checking it against the exact curve point and derivative ratio provides a direct confirmation.
Check Your Understanding
For each question, identify the point of tangency and check whether the parametric slope formula applies before writing a line equation.
- For \(x=t^2+2\), \(y=t^3\), find the tangent line at \(t=1\).
- For \(x=3\cos t\), \(y=2\sin t\), find the tangent slope at \(t=\pi/3\).
- Why must \(x'(t_0)\ne 0\) for the finite-slope ratio \(y'(t_0)/x'(t_0)\) to apply?
- In the tangent-line method, what does the parameter value determine, and what do the coordinate derivatives determine?
- A learner computes \(x'(t_0)/y'(t_0)\). Explain the error and state the correct order of the ratio.
