From a Parametrized Curve to a Tangent Line
A parametrized curve records points by assigning a position vector to each parameter value. Near a particular parameter value, the curve may bend, but its local direction can often be described by a straight line. The derivative of the position function provides that direction whenever the derivative is nonzero.
The distinction between a direction and a line matters. A direction alone does not specify where the tangent line lies. To write the line, we need both the position on the curve and the derivative at the parameter value.
Definitions and Tangent-Line Formula
Let \(\mathbf{r}:I\to\mathbb{R}^n\) be a differentiable vector-valued function, where \(I\) is an interval and \(t_0\) is an interior point of \(I\). Its position at \(t_0\) is \(\mathbf{r}(t_0)\). Its derivative is defined by the vector limit
provided this limit exists. The difference quotient is a vector: for a plane curve, it has two components; for a space curve, it has three.
The parameter \(s\) in this line equation is a new line parameter. It does not have to equal the curve parameter \(t\). Any nonzero scalar multiple of \(\mathbf{r}'(t_0)\) is also a valid direction vector, since it gives the same line.
For a plane curve \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), the formula can be written componentwise as
This form works whether or not the tangent is vertical. It also generalizes directly to curves in three-dimensional space.
Why the Derivative Gives the Tangent Direction
Points on the curve at parameters \(t_0\) and \(t_0+h\) determine a secant line. The vector from the first point to the second is \(\mathbf{r}(t_0+h)-\mathbf{r}(t_0)\). Dividing that vector by \(h\) gives the difference quotient in the derivative definition. Since \(h\) is a nonzero scalar, the quotient points in the same line direction as the secant, up to reversal.
As \(h\) approaches zero, the secant joins points closer and closer to \(\mathbf{r}(t_0)\). If the derivative exists and is nonzero, these secant directions approach the derivative direction. The following argument makes the connection precise.
For \(h\ne 0\), define \(\mathbf{q}_h=\dfrac{\mathbf{r}(t_0+h)-\mathbf{r}(t_0)}{h}\). Its direction is the direction of the secant line, with a possible reversal when \(h\) changes sign.
Differentiability gives \(\mathbf{q}_h\to\mathbf{r}'(t_0)\) as \(h\to 0\). If \(\mathbf{r}'(t_0)\ne\mathbf{0}\), then \(\mathbf{q}_h\ne\mathbf{0}\) for all sufficiently small \(h\), and its normalized direction approaches \(\dfrac{\mathbf{r}'(t_0)}{\|\mathbf{r}'(t_0)\|}\).
The secant lines pass through \(\mathbf{r}(t_0)\), and their directions approach the line direction \(\mathbf{r}'(t_0)\). Thus the tangent line is the line through \(\mathbf{r}(t_0)\) with that direction.
The normalized-direction limit in Step 2 follows because the norm is continuous: if \(\mathbf{q}_h\to\mathbf{v}\ne\mathbf{0}\), then \(\|\mathbf{q}_h\|\to\|\mathbf{v}\|>0\), so division by the norms is valid for sufficiently small \(h\). Reversal of direction does not change the underlying line. This is why a nonzero derivative gives the tangent line.
Worked Examples
Worked Example: Tangent Line to a Plane Curve
Let \(\mathbf{r}(t)=\langle t^2,2t+1\rangle\). Find the tangent line at \(t_0=1\).
First find the position at the specified parameter:
Differentiate each component and evaluate at \(t=1\):
This direction is nonzero, so the tangent line is
In component form, \(x=1+2s\) and \(y=3+2s\). Subtracting the first coordinate equation from the second gives \(y-x=2\), or \(y=x+2\). Check: at \(s=0\), the line gives \((1,3)=\mathbf{r}(1)\); its direction \(\langle 2,2\rangle\) is exactly \(\mathbf{r}'(1)\). The Cartesian equation also contains the point because \(3=1+2\).
Worked Example: Tangent Line to a Space Curve
Let \(\mathbf{r}(t)=\langle t,t^2,1-t\rangle\). Find the tangent line at \(t_0=-1\).
Evaluate the position:
Differentiate the components, then evaluate:
The derivative is nonzero. Therefore, a vector equation for the tangent line is
Check the point by setting \(s=0\): the result is \(\langle -1,1,2\rangle\), the curve’s position at \(t=-1\). The line’s direction is \(\langle 1,-2,-1\rangle\), matching the derivative calculated at that parameter. Thus both the location and the direction are correct.
Worked Example: Tangent Line to a Parametrized Circle
Let \(\mathbf{r}(t)=\langle 3\cos t,2\sin t\rangle\). Find the tangent line at \(t_0=\dfrac{\pi}{2}\).
The position is
Differentiate, using the derivatives of sine and cosine:
The tangent line is
Here \(x=-3s\) and \(y=2\), so the line is \(y=2\). Check: \(s=0\) gives the curve point \((0,2)\), and the line has horizontal direction \(\langle -3,0\rangle\), which equals the evaluated derivative. The line therefore passes through the correct point and has the required direction.
What a Zero Derivative Does—and Does Not—Tell You
If \(\mathbf{r}'(t_0)=\mathbf{0}\), it cannot be used as a direction vector: the zero vector has no direction. Therefore, the tangent-line formula above does not apply at that parameter value. This conclusion concerns what the first derivative supplies; it does not establish that the curve has no tangent line.
For instance, consider \(\mathbf{r}(t)=\langle t^2,t^3\rangle\) at \(t_0=0\). Then \(\mathbf{r}(0)=\langle 0,0\rangle\) and \(\mathbf{r}'(t)=\langle 2t,3t^2\rangle\), so \(\mathbf{r}'(0)=\langle 0,0\rangle\). The derivative supplies no direction. Yet for \(t\ne 0\), the vector from the origin to the curve point is \(\langle t^2,t^3\rangle=t^2\langle 1,t\rangle\). Its direction, after normalization, approaches \(\langle 1,0\rangle\) as \(t\to 0\). In this example the nearby secants indicate the horizontal tangent line through the origin. Finding it requires information beyond the zero derivative.
At AP level, the key decision is to check whether the evaluated derivative is nonzero before using it as a direction. If it is zero, do not substitute it into the tangent-line formula and claim that the resulting expression is a tangent line.
A Common Mistake: Using the Direction as the Whole Line
A frequent error is to write the derivative vector as the tangent line itself, forgetting that a line must pass through the curve’s point. For the first worked example, one might incorrectly write \(\mathbf{L}(s)=s\langle 2,2\rangle\). This line passes through \(\langle 0,0\rangle\), not through \(\mathbf{r}(1)=\langle 1,3\rangle\), so it is not the tangent line at \(t=1\).
The correction is to start with the point and add a scalar multiple of the direction:
A reliable check is to set \(s=0\) and confirm that the line gives \(\mathbf{r}(t_0)\). Then confirm that its direction vector is a nonzero scalar multiple of \(\mathbf{r}'(t_0)\).
Using the Method Reliably
For each tangent-line problem, keep the roles of the two vectors separate: \(\mathbf{r}(t_0)\) locates the line, while \(\mathbf{r}'(t_0)\) directs it. A short sequence of calculations helps prevent errors:
- Evaluate the position function at the specified parameter.
- Differentiate the position function and evaluate its derivative at that same parameter.
- Check that the derivative vector is nonzero.
- Substitute the point and direction into \(\mathbf{L}(s)=\mathbf{r}(t_0)+s\mathbf{r}'(t_0)\).
- Verify the point and direction in the resulting equation.
This method applies in the plane and in space. The displayed line may look different in different dimensions, but the essential construction is unchanged: a line is determined by one point on it and a nonzero direction vector.
Check Your Understanding
For each question, identify the position and derivative roles before writing a tangent line.
- For \(\mathbf{r}(t)=\langle t+1,t^2\rangle\), find the tangent line at \(t_0=1\) in vector form.
- For \(\mathbf{r}(t)=\langle 2\cos t,\sin t\rangle\), find the position and tangent direction at \(t_0=0\).
- Why does a nonzero derivative determine a tangent direction from the nearby secant lines?
- If \(\mathbf{r}'(t_0)=\mathbf{0}\), what conclusion can you draw about using the derivative as a direction? What conclusion can you not draw from that fact alone?
- A student writes \(\mathbf{L}(s)=s\mathbf{r}'(t_0)\). What must be checked, and what term is generally missing?
