Tangent Lines at Regular Parameter Values

From a Parametric Curve to a Tangent Line

A parametric curve gives its coordinates as functions of a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). To write a tangent line at a specified parameter value, we need two pieces of information: the point on the curve and the line’s slope there.

The point is found by evaluating both coordinate functions at the same parameter value. The slope comes from comparing how quickly \(y\) and \(x\) change as the parameter changes. When \(dx/dt\) is nonzero, this comparison gives a finite slope and a point-slope equation for the tangent line.

Definition: Let \(x(t)\) and \(y(t)\) be differentiable near \(t_0\), and suppose \(x'(t_0)\ne 0\). The point of the curve at \(t_0\) is \((x(t_0),y(t_0))\), and its tangent slope is \(\displaystyle m=\frac{y'(t_0)}{x'(t_0)}\). The tangent line at that point is \(\displaystyle y-y(t_0)=m\bigl(x-x(t_0)\bigr)\).

Here, \(x'(t)\) and \(y'(t)\) are the derivatives of the coordinate functions with respect to \(t\). The ratio \(y'(t_0)/x'(t_0)\) is not generally the derivative \(y'(t_0)\) by itself: both coordinate rates are measured with respect to the parameter, and their ratio gives the rate of change of \(y\) with respect to \(x\).

Why the Slope Is a Ratio of Derivatives

A tangent line can be understood as the limiting position of secant lines through two nearby points on a curve. Fix \(t_0\), and compare the point at \(t_0\) with the point at \(t_0+h\), where \(h\) is a small nonzero change in the parameter. The secant slope is the change in \(y\) divided by the change in \(x\).

\frac{y(t_0+h)-y(t_0)}{x(t_0+h)-x(t_0)} = \frac{\dfrac{y(t_0+h)-y(t_0)}{h}} {\dfrac{x(t_0+h)-x(t_0)}{h}}.

As \(h\) tends to zero, the numerator on the right tends to \(y'(t_0)\), and the denominator tends to \(x'(t_0)\). Since \(x'(t_0)\ne0\), that denominator has a nonzero limit. In particular, for sufficiently small nonzero \(h\), the change in \(x\) is nonzero, so the secant slope is defined.

\lim_{h\to0} \frac{y(t_0+h)-y(t_0)}{x(t_0+h)-x(t_0)} = \frac{y'(t_0)}{x'(t_0)}.

This limit is the tangent slope. The line with that slope through the curve point \((x(t_0),y(t_0))\) has the point-slope equation stated above. This also explains why the parameter’s rate of change does not itself determine the tangent slope: the same parameter increment changes both coordinates, and the ratio compares those changes directly.

Theorem: If \(x\) and \(y\) are differentiable at \(t_0\) and \(x'(t_0)\ne0\), then the tangent line at the parametrized point \((x(t_0),y(t_0))\) has slope \(y'(t_0)/x'(t_0)\). The condition \(x'(t_0)\ne0\) is essential for this finite-slope formula.

A Reliable Calculation Process

Keeping the point evaluation separate from the derivative calculation helps prevent errors. A useful order is:

Evaluate the point.
Find \(x(t_0)\) and \(y(t_0)\) using the specified parameter value.

Differentiate both coordinates.
Compute \(x'(t)\) and \(y'(t)\) with respect to \(t\).

Evaluate and form the ratio.
Find \(x'(t_0)\) and \(y'(t_0)\), confirm \(x'(t_0)\ne0\), and calculate \(m=y'(t_0)/x'(t_0)\).

Write and check the line.
Use \(y-y(t_0)=m(x-x(t_0))\), then verify the point and slope.

Worked Examples

Worked Example: A Polynomial Curve at \(t=2\)

Find the tangent line for \(x(t)=t^2+1\) and \(y(t)=t^3-2t\) at \(t_0=2\).

First evaluate the coordinates:

x(2)=2^2+1=4+1=5, \qquad y(2)=2^3-2(2)=8-4=4.

The point is \((5,4)\). Now differentiate each coordinate function and evaluate at \(2\):

x'(t)=2t, \qquad y'(t)=3t^2-2, $$ $$ x'(2)=2(2)=4, \qquad y'(2)=3(2^2)-2=12-2=10.

Since \(x'(2)=4\ne0\), the slope is defined:

m=\frac{y'(2)}{x'(2)}=\frac{10}{4}=\frac{5}{2}.

Using the point \((5,4)\), the tangent line is

y-4=\frac{5}{2}(x-5).

For a check, substituting \(x=5\) gives \(y-4=0\), so the line passes through \((5,4)\). Its slope is \(5/2\), matching the derivative ratio \(10/4\).

Worked Example: A Polynomial Curve at \(t=4\)

Find the tangent line for \(x(t)=t^2-3t+2\) and \(y(t)=2t^2+t-1\) at \(t_0=4\).

Evaluate the coordinates at the specified parameter:

$$ x(4)=4^2-3(4)+2=16-12+2=6, $$ $$ y(4)=2(4^2)+4-1=2(16)+4-1=32+4-1=35. $$

Thus, the point is \((6,35)\). Differentiate, then evaluate both derivatives:

x'(t)=2t-3, \qquad y'(t)=4t+1, $$ $$ x'(4)=2(4)-3=8-3=5, \qquad y'(4)=4(4)+1=16+1=17.

The \(x\)-derivative is nonzero, and the tangent slope is

m=\frac{17}{5}.

The line through \((6,35)\) with this slope is

y-35=\frac{17}{5}(x-6).

Checking the point, \(x=6\) makes the right-hand side zero, so \(y=35\). The slope in the equation is \(17/5\), equal to \(y'(4)/x'(4)=17/5\); both required pieces agree.

Worked Example: A Trigonometric Curve at \(t=\pi/2\)

Find the tangent line for \(x(t)=t+\sin t\) and \(y(t)=t^2+\cos t\) at \(t_0=\pi/2\).

Using \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\), the point is

x\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+1, \qquad y\left(\frac{\pi}{2}\right)=\left(\frac{\pi}{2}\right)^2+0 =\frac{\pi^2}{4}.

Differentiate the coordinate functions:

x'(t)=1+\cos t, \qquad y'(t)=2t-\sin t.

At \(t=\pi/2\), these derivatives are

x'\left(\frac{\pi}{2}\right)=1+0=1, \qquad y'\left(\frac{\pi}{2}\right)=2\left(\frac{\pi}{2}\right)-1=\pi-1.

Therefore, the slope is \(m=(\pi-1)/1=\pi-1\). The tangent line through the evaluated point is

y-\frac{\pi^2}{4} = (\pi-1)\left(x-\left(\frac{\pi}{2}+1\right)\right).

Substituting \(x=\pi/2+1\) makes the right-hand side zero, recovering \(y=\pi^2/4\). Also, the line’s slope \(\pi-1\) agrees with the derivative ratio at \(t=\pi/2\). The calculation uses exact values, so no rounding is needed.

A Common Mistake: Using Only \(dy/dt\)

A frequent error is to treat the derivative of \(y\) with respect to \(t\) as the slope in the \(xy\)-plane. In the first example, that mistake would give the following line:

$$ y-4=10(x-5). $$

This is wrong. The value \(10\) is \(dy/dt\) at \(t=2\), not the rate of change of \(y\) with respect to \(x\). At that parameter, \(x\) changes at rate \(dx/dt=4\), so the correct slope is \(10/4=5/2\). The corrected line is

y-4=\frac{5}{2}(x-5).

Both lines pass through \((5,4)\), so checking the point alone would not reveal the error. The slope must also agree with the derivative ratio. A second common source of mistakes is using \(t_0\) as if it were the curve point: the point must be \((x(t_0),y(t_0))\), not \((t_0,t_0)\).

What the Condition Does—and Does Not—Say

The condition \(x'(t_0)\ne0\) ensures that the quotient \(y'(t_0)/x'(t_0)\) is defined and that the tangent has a finite slope in the \(xy\)-plane. It is the condition needed for the method in this tutorial. If \(x'(t_0)=0\), this quotient cannot be used; the calculation must not proceed by dividing by zero.

When \(x'(t_0)\ne0\), the value of \(y'(t_0)\) may be zero, in which case the formula gives slope zero. The point-slope equation still applies. In all cases covered here, the essential habits are to evaluate both coordinates at the same parameter, differentiate both functions with respect to that parameter, and divide in the order \(y'(t_0)/x'(t_0)\).

Key takeaway: At a parameter value \(t_0\) with \(x'(t_0)\ne0\), evaluate the point \((x(t_0),y(t_0))\), calculate \(m=y'(t_0)/x'(t_0)\), and write \(y-y(t_0)=m(x-x(t_0))\). Check both that the point lies on the line and that the line’s slope matches the derivative ratio.

Check Your Understanding

For each question, keep the point evaluation distinct from the derivative calculation.

For \(x(t)=t^2+2\) and \(y(t)=t^3\), find the tangent-line equation at \(t_0=1\).
For \(x(t)=3t-1\) and \(y(t)=t^2+4\), find the curve point and tangent slope at \(t_0=2\).
State the condition on \(x'(t_0)\) that permits the finite-slope formula, and explain why the denominator matters.
In the slope formula, why is \(dy/dt\) divided by \(dx/dt\), rather than used by itself?
What two checks can you make on a proposed tangent-line equation?