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Tangent Vectors from Parametric Derivatives

Use the vector derivative to identify a curve’s instantaneous direction, and understand when it defines a nonzero tangent vector.

Beginner 9 min read

What You'll Learn

  • Define the tangent vector of a parametrized curve at a parameter value
  • Find a curve’s vector derivative by differentiating its components
  • Interpret the derivative using nearby secant vectors
  • Write a tangent line when the derivative is nonzero
  • Recognize why a zero derivative does not supply a tangent direction

From a Parametrized Curve to a Tangent Vector

A parametrized curve describes points in space as a parameter varies. For a differentiable curve \(\mathbf{r}(t)\), its derivative \(\mathbf{r}'(t)\) gives a vector that records the curve’s instantaneous direction at the parameter value \(t\). This is the vector-valued version of the familiar idea that the derivative of a scalar function describes its local change.

The word direction matters. A tangent vector is not the point on the curve, and it is not necessarily a unit-length vector. It is the derivative vector itself. When that vector is nonzero, it points along the tangent direction, oriented according to increasing parameter.

Definition: Let \(\mathbf{r}:I\to\mathbb{R}^n\) be a vector-valued function defined on an interval \(I\), and let \(t_0\) be an interior point of \(I\). If the vector derivative \(\mathbf{r}'(t_0)\) exists, it is the tangent vector to the parametrized curve at parameter \(t_0\). If \(\mathbf{r}'(t_0)\ne\mathbf{0}\), it gives a nonzero tangent direction at the point \(\mathbf{r}(t_0)\).

For a curve in three-dimensional space, write \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\). Its derivative is computed componentwise:

$$ \mathbf{r}'(t) = \left\langle x'(t),y'(t),z'(t) \right\rangle. $$

The same definition applies to a curve with two components or with more than three components. The derivative exists when the derivatives of all component functions exist, and its components are those scalar derivatives.

Why the Derivative Gives the Tangent Direction

At a parameter value \(t_0\), the point \(\mathbf{r}(t_0)\) lies on the curve. For a nearby parameter \(t_0+h\), the displacement from that point to the nearby curve point is \(\mathbf{r}(t_0+h)-\mathbf{r}(t_0)\). Dividing this displacement by \(h\) gives a vector difference quotient. As \(h\) approaches zero, that quotient approaches the derivative, when the derivative exists.

$$ \mathbf{r}'(t_0) = \lim_{h\to 0} \frac{\mathbf{r}(t_0+h)-\mathbf{r}(t_0)}{h}. $$

This limit explains the geometric interpretation. If \(\mathbf{r}'(t_0)=\mathbf{v}\), the definition of the limit says that the difference quotient is \(\mathbf{v}\) plus an error that tends to \(\mathbf{0}\). Multiplying by \(h\) shows that the displacement is approximately \(h\mathbf{v}\) for small \(h\). Thus, near \(t_0\), the curve’s displacement is approximated by a scalar multiple of \(\mathbf{v}\).

$$ \frac{\mathbf{r}(t_0+h)-\mathbf{r}(t_0)}{h} =\mathbf{v}+\boldsymbol{\varepsilon}(h), \qquad \lim_{h\to 0}\boldsymbol{\varepsilon}(h)=\mathbf{0}, $$ $$ \mathbf{r}(t_0+h)-\mathbf{r}(t_0) =h\mathbf{v}+h\boldsymbol{\varepsilon}(h). $$

When \(\mathbf{v}\ne\mathbf{0}\), the leading displacement \(h\mathbf{v}\) lies along the line in direction \(\mathbf{v}\). The remaining term becomes negligible relative to \(h\) as \(h\) approaches zero. This is why the nonzero derivative gives the curve’s instantaneous tangent direction. For positive \(h\), the leading displacement points along \(\mathbf{v}\); for negative \(h\), it points the other way, as the parameter moves toward smaller values.

Theorem: If \(\mathbf{r}\) is differentiable at \(t_0\) and \(\mathbf{r}'(t_0)\ne\mathbf{0}\), the tangent line at \(\mathbf{r}(t_0)\) has the parametrization $$ \boldsymbol{\ell}(s)=\mathbf{r}(t_0)+s\mathbf{r}'(t_0), \qquad s\in\mathbb{R}. $$ The point \(\mathbf{r}(t_0)\) gives the line’s location, and \(\mathbf{r}'(t_0)\) gives its direction.

The tangent line is a useful geometric representation, but it is not a replacement for the curve. It describes the line that matches the curve’s first-order direction at one point.

Worked Example: A Polynomial Curve

Worked Example: A Polynomial Curve

Consider \(\mathbf{r}(t)=\langle t,t^2,2t+1\rangle\). Find the derivative as a function of \(t\), then identify the point and tangent vector when \(t=1\).

Differentiate each component:

$$ \frac{d}{dt}(t)=1,\qquad \frac{d}{dt}(t^2)=2t,\qquad \frac{d}{dt}(2t+1)=2. $$

Therefore,

$$ \mathbf{r}'(t)=\langle 1,2t,2\rangle. $$

At \(t=1\), the curve’s point is

$$ \mathbf{r}(1)=\langle 1,1^2,2(1)+1\rangle =\langle 1,1,3\rangle. $$

The tangent vector is

$$ \mathbf{r}'(1)=\langle 1,2(1),2\rangle =\langle 1,2,2\rangle. $$

To check the derivative directly, compare the curve point at \(1+h\) with the point at \(1\):

$$ \mathbf{r}(1+h) = \langle 1+h,(1+h)^2,2(1+h)+1\rangle = \langle 1+h,1+2h+h^2,3+2h\rangle, $$ $$ \frac{\mathbf{r}(1+h)-\mathbf{r}(1)}{h} = \frac{\langle h,2h+h^2,2h\rangle}{h} = \langle 1,2+h,2\rangle \longrightarrow \langle 1,2,2\rangle. $$

The limit agrees with \(\mathbf{r}'(1)\). The tangent line is \(\langle 1,1,3\rangle+s\langle 1,2,2\rangle\).

Reading the Components Geometrically

The components of the derivative describe the curve’s instantaneous change in each coordinate. For \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), the first component of \(\mathbf{r}'(t_0)\) is the instantaneous change in \(x\), the second is the instantaneous change in \(y\), and the third is the instantaneous change in \(z\). Together, these changes make one vector direction in space.

A component can be zero without the whole tangent vector being zero. For example, \(\langle 0,3,1\rangle\) is a nonzero vector: it indicates no first-coordinate change to first order at that parameter, while the other two coordinates change. The relevant condition for a nonzero tangent direction is that the entire derivative vector is not \(\mathbf{0}\).

Worked Example: A Curve with Trigonometric Components

Worked Example: A Curve with Trigonometric Components

Let \(\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle\). Find its derivative and determine the tangent vector and tangent line at \(t=\frac{\pi}{2}\).

Differentiate componentwise:

$$ \frac{d}{dt}(\cos t)=-\sin t,\qquad \frac{d}{dt}(\sin t)=\cos t,\qquad \frac{d}{dt}(t)=1. $$

Thus,

$$ \mathbf{r}'(t)=\langle-\sin t,\cos t,1\rangle. $$

First find the point on the curve:

$$ \mathbf{r}\left(\frac{\pi}{2}\right) = \left\langle \cos\frac{\pi}{2},\sin\frac{\pi}{2},\frac{\pi}{2} \right\rangle = \left\langle 0,1,\frac{\pi}{2}\right\rangle. $$

Then evaluate the derivative:

$$ \mathbf{r}'\left(\frac{\pi}{2}\right) = \left\langle -\sin\frac{\pi}{2},\cos\frac{\pi}{2},1 \right\rangle = \langle-1,0,1\rangle. $$

The vector is nonzero, so it supplies the tangent direction. The tangent line is

$$ \boldsymbol{\ell}(s) = \left\langle 0,1,\frac{\pi}{2}\right\rangle +s\langle-1,0,1\rangle. $$

For a direct derivative check, the difference quotient at \(t_0=\frac{\pi}{2}\) is

$$ \frac{\mathbf{r}\left(\frac{\pi}{2}+h\right)-\mathbf{r}\left(\frac{\pi}{2}\right)}{h} = \left\langle \frac{-\sin h}{h}, \frac{\cos h-1}{h}, 1 \right\rangle \longrightarrow \langle-1,0,1\rangle. $$

Here the standard scalar limits give the first two component limits, and the third component is constantly \(1\). The resulting vector agrees with the componentwise derivative.

When the Derivative Is Zero

The derivative vector can be \(\mathbf{0}\) at a parameter value. In that case, it still correctly represents the derivative, but it does not provide a nonzero direction for the tangent line formula. The first-order displacement argument says that the displacement is smaller than a nonzero constant multiple of \(h\); it does not identify a direction from the zero vector.

This limitation must be stated carefully: a zero derivative does not prove that the curve has no geometric tangent. It means only that the derivative at that parameter does not itself supply a nonzero tangent vector. Further analysis may be needed to describe the curve’s local geometry.

Worked Example: A Zero Derivative at One Parameter

Consider \(\mathbf{r}(t)=\langle t^2,t^3,4\rangle\) at \(t=0\). Differentiate the components:

$$ \mathbf{r}'(t)=\langle 2t,3t^2,0\rangle. $$

At the parameter value, the point and derivative are

$$ \mathbf{r}(0)=\langle 0,0,4\rangle,\qquad \mathbf{r}'(0)=\langle 0,0,0\rangle. $$

Check the derivative using the difference quotient:

$$ \frac{\mathbf{r}(0+h)-\mathbf{r}(0)}{h} = \frac{\langle h^2,h^3,0\rangle}{h} = \langle h,h^2,0\rangle \longrightarrow \langle 0,0,0\rangle. $$

The derivative is indeed zero. Consequently, this derivative does not give a nonzero tangent direction at \(t=0\). The calculation does not justify claiming that the curve has no tangent; it identifies exactly what the first derivative tells us and where that information stops.

A Common Mistake: Using the Point as the Direction

A frequent error is to use \(\mathbf{r}(t_0)\) as the tangent vector. The position vector locates a point relative to the origin; the derivative describes the curve’s local change. They are different quantities and can point in different directions.

For example, let \(\mathbf{r}(t)=\langle t^2,t\rangle\) and take \(t=1\). A mistaken calculation might say, “The tangent vector is \(\mathbf{r}(1)=\langle1,1\rangle\).” But this is the position, not the derivative. The correction is:

$$ \mathbf{r}'(t)=\langle 2t,1\rangle, \qquad \mathbf{r}'(1)=\langle 2,1\rangle. $$

Thus the point is \(\langle1,1\rangle\), while the tangent vector is \(\langle2,1\rangle\). The tangent line uses both, in their separate roles:

$$ \boldsymbol{\ell}(s)=\langle1,1\rangle+s\langle2,1\rangle. $$

Substituting \(s=0\) gives the point \(\langle1,1\rangle\); changing \(s\) moves along the direction \(\langle2,1\rangle\). This verifies the distinction between location and direction.

How to Interpret a Tangent Vector

When a curve is given by component functions, the reliable procedure is short: differentiate the components, evaluate that vector at the parameter of interest, and distinguish the derivative from the curve point. If the derivative is nonzero, it gives the tangent direction. If it is zero, do not use it as though it were a direction; state that the derivative supplies no nonzero tangent vector there.

The derivative’s direction also depends on the parameter’s orientation. The vector points in the direction of increasing parameter because it is the limit of difference quotients formed by dividing the displacement by a positive increment \(h\). The tangent line itself has no preferred orientation, but the tangent vector does. This distinction is useful when interpreting a parametrized curve: the same geometric line can have tangent vectors pointing in opposite directions under different descriptions.

Key takeaway: The vector derivative \(\mathbf{r}'(t_0)\) is the curve’s instantaneous direction at \(t_0\). It is a nonzero tangent vector only when \(\mathbf{r}'(t_0)\ne\mathbf{0}\). The curve point \(\mathbf{r}(t_0)\) locates the tangent, while the derivative determines its direction.

Check Your Understanding

Use the definitions and examples above to answer these questions.

  1. For \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), how is \(\mathbf{r}'(t)\) computed?
  2. Why does the difference quotient help explain the tangent direction?
  3. For \(\mathbf{r}(t)=\langle t,t^2\rangle\), find the tangent vector at \(t=2\). Is it nonzero?
  4. What information does \(\mathbf{r}(t_0)\) provide, and what information does \(\mathbf{r}'(t_0)\) provide?
  5. If \(\mathbf{r}'(t_0)=\mathbf{0}\), what can you conclude about the derivative’s ability to give a tangent direction? What can you not conclude about the curve’s geometric tangent?

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