The Derivative as a Tangent Vector

The derivative vector r′(t0) has an important geometric property: It points in the direction tangent to the path traced by r(t) at t = t0.

To understand why, consider the difference quotient, where Δr = r(t0 + h) − r(t0) and Δt = h with h ≠ 0:

Although it has been our convention to regard all vectors as based at the origin, the tangent vector r′(t) is an exception; we visualize it as a vector based at the terminal point of r(t). This makes sense because r′(t) then appears as a vector tangent to the curve (Figure 3).

The vector Δr points from the head of r(t0) to the head of r(t0 + h) as in Figure 2(A). The difference quotient Δr/Δt is a scalar multiple of Δr and therefore points in the same direction [Figure 2(B)].

The difference quotient points in the direction of Δr = r(t0 + h) − r(t0).

As h = Δt tends to zero, Δr also tends to zero but the quotient Δr/Δt approaches a vector r′(t0), which, if nonzero, points in the direction tangent to the curve. Figure 3 illustrates the limiting process. We refer to r′(t0) as the tangent vector or the velocity vector at r(t0).

The difference quotient converges to a vector r′ (t0), tangent to the curve.

The tangent vector r′(t0) (if it is nonzero) is a direction vector for the tangent line to the curve. Therefore, the tangent line has vector parametrization:

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EXAMPLE 5: Plotting Tangent Vectors

Plot r(t) = 〈cos t, sin t, 4 cos2 t〉 together with its tangent vectors at and . Find a parametrization of the tangent line at .

Solution The derivative is r′(t) = 〈− sin t, cos t, −8 cos t sin t〉, and thus the tangent vectors at and are

Figure 4 shows a plot of r(t) with based at and based at .

Tangent vectors to
r(t) = 〈cos t, sin t, 4 cos2 t〉 at .

At and thus the tangent line is parametrized by

There are some important differences between vector- and scalar-valued derivatives. The tangent line to a plane curve y = f(x) is horizontal at x0 if f′(x0) = 0. But in a vector parametrization, the tangent vector r′(t0) = 〈x′(t0), y′(t0)〉 is horizontal and nonzero if y′(t0) = 0 but x′(t0) ≠ 0.

EXAMPLE 6: Horizontal Tangent Vectors on the Cycloid

The function

r(t) = 〈t − sin t, 1 − cos t〉

traces a cycloid. Find the points where:

  • r′(t) is horizontal and nonzero.

  • r′(t) is the zero vector.

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Solution The tangent vector is r′(t) = 〈1 − cos t, sin t〉. The y-component of r′(t) is zero if sin t = 0—that is, if t = 0,π, 2π,…. We have

By periodicity, we conclude that r′(t) is nonzero and horizontal for t = π, 3π, 5π,…and r′(t) = 0 for t = 0, 2π, 4π,…(Figure 5).

Points on the cycloid r(t) = 〈t − sin t, 1 − cos t〉 where the tangent vector is horizontal.

CONCEPTUAL INSIGHT

The cycloid in Figure 5 has sharp points called cusps at points where x = 0, 2π, 4π,…. If we represent the cycloid as the graph of a function y = f(x), then f′(x) does not exist at these points. By contrast, the vector derivative r′(t) = 〈1 − cos t, sin t〉 exists for all t, but r′(t) = 0 at the cusps. In general, r′(t) is a direction vector for the tangent line whenever it exists, but we get no information about the tangent line (which may or may not exist) at points where r′(t) = 0.

The next example establishes an important property of vector-valued functions that will be used in Sections 13.4–13.6.

EXAMPLE 7: Orthogonality of r and r′ When r Has Constant Length

Prove that if r(t) has constant length, then r(t) is orthogonal to r′(t).

Solution By the Product Rule for Dot Products,

This derivative is zero because ∥r(t)∥ is constant. Therefore r(t) · r′(t) = 0, and r(t) is orthogonal to r′(t) [or r′(t) = 0].

GRAPHICAL INSIGHT

The result of Example 7 has a geometric explanation. A vector parametrization r(t) consisting of vectors of constant length R traces a curve on the surface of a sphere of radius R with center at the origin (Figure 6). Thus r′(t) is tangent to this sphere. But any line that is tangent to a sphere at a point P is orthogonal to the radial vector through P, and thus r(t) is orthogonal to r′(t).