9.1 Vector Functions

INTRODUCTION

Recall that a curve C in the xy-plane is simply a set of ordered pairs (x, y). We say that C is a parametric curve if the x- and y-coordinates of a point on the curve are defined by a pair of functions x = f(t), y = g(t) that are continuous on some interval a ≤ t ≤ b. The notion of a parametric curve extends to 3-space as well. A parametric curve in space, or space curve, is a set of ordered triples (x, y, z) where

x = f(t), y = g(t), z = h(t), (1)

are continuous on an interval defined by a ≤ t ≤ b. In this section we combine the concepts of parametric curves with vectors.

Vector-Valued Functions

It is often convenient in science and engineering to introduce a vector r whose components are functions of a parameter t. We say that

r(t) = 〈f(t), g(t)〉 = f(t)i + g(t)j

and

r(t) = 〈f(t), g(t), h(t)〉 = f(t)i + g(t)j + h(t)k,

are vector-valued functions or simply vector functions. As shown in FIGURE 9.1.1, for a given value of the parameter, say t0, the vector r(t0) is the position vector of a point P on a curve C. In other words, as the parameter t varies, we can envision the curve C being traced out by the moving arrowhead of r(t).

2 graphs. Graph a. Caption. 2-space. A curve and a vector are graphed on an x y coordinate plane. The curve labeled C, enters the left side of the viewing window in the second quadrant, goes up and to the right, passes through the marked point (x(t subscript 0), y(t subscript 0)) in the first quadrant, reaches a high point, then goes down and to the right, and exits the bottom right of the viewing window. The vector labeled r(t subscript 0) = (x(t subscript 0), y(t subscript 0)), begins at the origin, goes up and to the right, and ends on the curve C at the marked point (x(t subscript 0), y(t subscript 0)). Graph b. Caption. 3-space. A curve and a vector are graphed on a three dimensional x y z coordinate system. The curve labeled C, enters the bottom of the viewing window in the fourth quadrant, goes up and to the right until a certain point in the first quadrant, changes direction and continues going up to the left, reaches a high point, then goes down to the left until the marked point (x(t subscript 0), y(t subscript 0), z(t subscript 0)) in the first quadrant, reaches a low point, then goes up and to the right, intersects itself, and exits the top right of the viewing window. The vector labeled r(t subscript 0) = (x(t subscript 0), y(t subscript 0), z(t subscript 0)), begins at the origin, goes up and to the right, and ends on the curve C at the marked point (x(t subscript 0), y(t subscript 0), z(t subscript 0)).

FIGURE 9.1.1 Curves defined by vector functions

We have already seen an example of parametric equations, as well as the vector function of a space curve, in Section 7.5, when we discussed the line in 3-space.

EXAMPLE 1 Circular Helix

Graph the curve traced by the vector function

r(t) = 2 cos ti + 2 sin tj + tk, t ≥ 0.

SOLUTION

The parametric equations of the curve are x = 2 cos t, y = 2 sin t, z = t. By eliminating the parameter t from the first two equations,

x2 + y2 = (2 cos t)2 + (2 sin t)2 = 22,

we see that a point on the curve lies on the circular cylinder x2 + y2 = 4. As seen in FIGURE 9.1.2 and the accompanying table, as the value of t increases, the curve winds upward in a spiral or circular helix.

A graph. A cylinder, a curve and 4 vectors are graphed on a three dimensional x y z coordinate system. The cylinder is labeled x^2 + y^2 = 4 and the center of the bottom base is at the origin. The bottom base passes through the marked point (2, 0, 0) and the top base passes through the marked point (0, 2, 9 pi over 2). The curve begins at the marked point (2, 0, 0), spirals up counter clockwise around the cylinder, passes through the marked points (0, 2, pi over 2), (negative 2, 0, pi), (0, negative 2, 3 pi over 2), (2, 0, 2 pi), (0, 2, 5 pi over 2), (negative 2, 0, 3 pi), (0, negative 2, 7 pi over 2), (2, 0, 4 pi), and ends at the marked point (0, 2, 9 pi over 2). The first vector begins at the origin, goes up and to the right, and ends at the marked point (0, 2, pi over 2). The second vector begins at the origin, goes up towards the back, and ends at the marked point (negative 2, 0, pi). The third vector begins at the origin, goes up and to the left, and ends at the marked point (0, negative 2, 3 pi over 2). The fourth vector begins at the origin, goes up and to the front, and ends at the marked point (2, 0, 2 pi).

FIGURE 9.1.2 Circular helix in Example 1

≡

The curve in Example 1 is a special case of the vector function

r(t) = a cos ti + b sin tj + ctk,      a > 0, b > 0, c > 0,

which describes an elliptical helix. When a = b, the helix is circular. The pitch of a helix is defined to be the number 2πc. Problems 9 and 10 in Exercises 9.1 illustrate two other kinds of helixes.

EXAMPLE 2 Circle in a Plane

Graph the curve traced by the vector function

r(t) = 2 cos ti + 2 sin tj + 3k.

SOLUTION

The parametric equations of this curve are x = 2 cos t, y = 2 sin t, z = 3. As in Example 1, we see that a point on the curve must also lie on the cylinder x2 + y2 = 4. However, since the z-coordinate of any point has the constant value z = 3, the vector function r(t) traces out a circle 3 units above the xy-plane. See FIGURE 9.1.3. ≡

A graph. A circle and 4 vectors are graphed on a three dimensional x y z coordinate system. The circle is labeled x^2 + y^2 = 4, z = 3, parallel to the x y plane, and its center is on the vertical z axis. The first vector begins at the origin, goes up and to the right on the y z plane, and ends on the circle. The second vector begins at the origin, goes up towards the back on the x z plane, and ends on the circle. The third vector begins at the origin, goes up and to the left on the y z plane, and ends on the circle. The fourth vector begins at the origin, goes up and to the front on the x z plane, and ends on the circle.

FIGURE 9.1.3 Curve in Example 2

EXAMPLE 3 Curve of Intersection

Find the vector function that describes the curve C of intersection of the plane y = 2x and the paraboloid z = 9 – x2 – y2.

SOLUTION

We first parameterize the curve C of intersection by letting x = t. It follows that y = 2t and z = 9 – t2 – (2t)2 = 9 – 5t2. From the parametric equations x = t, y = 2t, z = 9 – 5t2, we see that a vector function describing the trace of the paraboloid in the plane y = 2x is given by r(t) = ti + 2tj + (9 − 5t2)k. See FIGURE 9.1.4. ≡

A graph. A paraboloid and a plane are graphed on a three dimensional x y z coordinate system. The paraboloid is labeled z = 9 minus x^2 minus y^2, and graphed around the vertical z axis, and reaches a high point on the vertical z axis. The circular base of the paraboloid labeled x^2 + y^2 = 9, is on the x y plane and its center is at the origin. The graphed plane is vertical, passes along the dashed line labeled y = 2x, and intersects the paraboloid along a curve labeled C. The curve labeled C begins in the quadrant at the back on the left in the x y plane, goes up along the graphed plane, reaches a high point on the vertical z axis at the high point of the paraboloid, goes down along the graphed plane, and ends in the quadrant at the front on the right in the x y plane.

FIGURE 9.1.4 Curve in Example 3

Limits, Continuity, and Derivatives

The fundamental notion of the limit of a vector function r(t) = 〈f(t), g(t), h(t)〉 is defined in terms of the limits of the component functions.

DEFINITION 9.1.1 Limit of a Vector Function

The notation t → a in Definition 9.1.1 can, of course, be replaced by t → a+, t → a−, t → ∞, or t → –∞.

As an immediate consequence of Definition 9.1.1, we have the following result.

THEOREM 9.1.1 Properties of Limits

If limt→a r1(t) = L1 and limt→a r2(t) = L2, then

(i) cr1(t) = cL1, c a scalar

(ii) [r1(t) + r2(t)] = L1 + L2

(iii) r1(t) · r2(t) = L1 · L2.

DEFINITION 9.1.2 Continuity of a Vector Function

A vector function r is said to be continuous at t = a if

(i) r(a) is defined,

(ii) r(t) exists, and

(iii) r(t) = r(a).

Equivalently, r(t) is continuous at t = a if and only if the component functions f, g, and h are continuous there.

DEFINITION 9.1.3 Derivative of Vector Function

The derivative of a vector function r is

(2)

for all t for which the limit exists.

The derivative of r is also written dr/dt. The next theorem will show that on a practical level the derivative of a vector function is obtained by simply differentiating its component functions.

THEOREM 9.1.2 Differentiation of Components

If r(t) = 〈f(t), g(t), h(t)〉, where f, g, and h are differentiable, then

r′(t) = 〈f′(t), g′(t), h′(t)〉.

PROOF:

From (2) we have

Taking the limit of each component yields the desired result. ≡

Smooth Curves

When the component functions of a vector function r have continuous first derivatives and r′(t) ≠ 0 for all t in the open interval (a, b), then r is said to be a smooth function and the curve C traced by r is called a smooth curve.

Geometric Interpretation of r′(t)

If the vector r′(t) is not 0 at a point P, then it may be drawn tangent to the curve at P. As seen in FIGURE 9.1.5, the vectors

are parallel. If we assume limΔt→0 Δr/Δt exists, it seems reasonable to conclude that as Δt → 0, r(t) and r(t + Δt) become close and, as a consequence, the limiting position of the vector Δr/Δt is the tangent line at P. Indeed, the tangent line at P is defined as that line through P parallel to r′(t).

2 graphs. Graph a. A curve labeled C, 3 vectors and a line are graphed on a three dimensional x y z coordinate system. The curve C passes through a point labeled P. The first vector, labeled r(t), begins at the origin, goes up and to the right, and ends on the curve C at the point P. The second vector, labeled r(t + delta t), begins at the origin, goes up and to the right under the first vector, and ends at a second point on the curve C. The third vector, labeled delta r, begins on the curve C at the point P, goes down and to the right, and ends on the curve C at the end point of the second vector. An upward sloping line is tangent to the curve C at the point P. Graph b. A curve labeled C, 3 vectors and a line are graphed on a three dimensional x y z coordinate system. The curve C passes through a point labeled P. The first vector, labeled r(t), begins at the origin, goes up and to the right, and ends on the curve C at the point P, similar to the vector r(t) in graph a. The second vector, labeled r(t + delta t), begins at the origin, goes up and to the right under the first vector, and ends at a second point on the curve C, similar to the vector r(t + delta t) in graph a. The third vector, labeled (delta r) over (delta t), delta t > 0, begins on the curve C at the point P, goes down and to the right, and passes through the end point of the second vector. The third vector is parallel to the vector delta r in graph a. An upward sloping line is tangent to the curve C at the point P, similar to the tangent line in graph a.

FIGURE 9.1.5 Vector r′(t) is tangent to curve C at P

EXAMPLE 4 Tangent Vectors

Graph the curve C that is traced by a point P whose position is given by r(t) = cos 2ti + sin tj, 0 ≤ t ≤ 2π. Graph r′(0) and r′(π/6).

SOLUTION

By clearing the parameter from the parametric equations x = cos 2t, y = sin t, 0 ≤ t ≤ 2π, we find that C is the parabola x = 1 – 2y2, –1 ≤ x ≤ 1. From r′(t) = –2 sin 2ti + cos tj we find

r′(0) = j     and    

In FIGURE 9.1.6 these vectors are drawn tangent to the curve C at (1, 0) and (, ), respectively. ≡

A graph. A curve and 2 tangent vectors are graphed on an x y coordinate plane. The curve, labeled x = 1 minus 2 y^2, begins at the marked point (negative 1, negative 1), goes up and to the right with increasing steepness until the point (1, 0), then changes direction and continues going up to the left, passes through the marked point (1 over 2, 1 over 2), and ends at the marked point (negative 1, 1). The first tangent vector, labeled r prime(pi over 6), begins on the curve at the marked point (1 over 2, 1 over 2), and goes up to the left. The second tangent vector, labeled r prime(0), begins on the curve at the point (1, 0), and goes up vertically.

FIGURE 9.1.6 Tangent vectors in Example 4

EXAMPLE 5 Tangent Line

Find parametric equations of the tangent line to the graph of the curve C whose parametric equations are x = t2, y = t2 – t, z = –7t at t = 3.

SOLUTION

The vector function that gives the position of a point P on the curve is given by r(t) = t2i + (t2 – t)j – 7tk. Now,

r′(t) = 2ti + (2t – 1)j – 7k and so r′(3) = 6i + 5j – 7k,

which is tangent to C at the point whose position vector is

r(3) = 9i + 6j – 21k;

that is, P(9, 6, –21). Using the components of r′(3), we see that parametric equations of the tangent line are x = 9 + 6t, y = 6 + 5t, z = −21 − 7t. ≡

Higher-Order Derivatives

Higher-order derivatives of a vector function are also obtained by differentiating its components. In the case of the second derivative, we have

r″(t) = 〈f″(t), g″(t), h″(t)〉 = f″(t)i + g″(t)j + h″(t)k.

EXAMPLE 6 Derivative of a Vector Function

If r(t) = (t3 – 2t2)i + 4tj + e−tk, then

r′(t) = (3t2 – 4t)i + 4j – e−tk     and    r″(t) = (6t – 4)i + e−tk. ≡

THEOREM 9.1.3 Chain Rule

If r is a differentiable vector function and s = u(t) is a differentiable scalar function, then the derivative of r(s) with respect to t is

EXAMPLE 7 Chain Rule

If r(s) = cos 2si + sin 2sj + e−3sk, where s = t4, then

≡

Details of the proof of the next theorem are left as exercises.

THEOREM 9.1.4 Rules of Differentiation

Let r1 and r2 be differentiable vector functions and u(t) a differentiable scalar function.

(i)

(ii)

(iii)

(iv)

Note of caution.

Since the cross product of two vectors is not commutative, the order in which r1 and r2 appear in part (iv) of Theorem 9.1.4 must be strictly observed.

Integrals of Vector Functions

If f, g, and h are integrable, then the indefinite and definite integrals of a vector function r(t) = f(t)i + g(t)j + h(t)k are defined, respectively, by

The indefinite integral of r is another vector function R + c such that R′(t) = r(t).

EXAMPLE 8 Integral of a Vector Function

If r(t) = 6t2i + 4e−2tj + 8 cos 4tk,

then

where c = c1i + c2j + c3k. ≡

Length of a Space Curve

If r(t) = f(t)i + g(t)j + h(t)k is a smooth function, then it can be shown that the length of the smooth curve traced by r is given by

(3)

Arc Length as a Parameter

A curve in the plane or in space can be parameterized in terms of the arc length s.

EXAMPLE 9 Example 1 Revisited

Consider the helix of Example 1. Since r′(t) = , it follows from (3) that the length of the curve from r(0) to an arbitrary point r(t) is

where we have used u as a dummy variable of integration. Using , we obtain a vector equation of the helix as a function of arc length:

(4)

Parametric equations of the helix are then

f(s) = 2 cos , g(s) = 2 sin , h(s) = . ≡

The derivative of a vector function r(t) with respect to the parameter t is a tangent vector to the curve traced by r. However, if the curve is parameterized in terms of arc length s, then r′(s) is a unit tangent vector. To see this, let a curve be described by r(s), where s is arc length. From (3), the length of the curve from r(0) to r(s) is . Differentiation of this last equation with respect to s then yields .

9.1 Exercises Answers to selected odd-numbered problems begin on page ANS-22.

In Problems 1–10, graph the curve traced by the given vector function.

  1. r(t) = 2 sin ti + 4 cos tj + tk; t ≥ 0
  2. r(t) = cos ti + tj + sin tk; t ≥ 0
  3. r(t) = ti + 2tj + cos tk; t ≥ 0
  4. r(t) = 4i + 2 cos tj + 3 sin tk
  5. r(t) = 〈et, e2t〉
  6. r(t) = cosh ti + 3 sinh tj
  7. r(t) = 〈〉; 0 ≤ t ≤ π/2
  8. r(t) = ti + t3j + tk
  9. r(t) = et cos ti + et sin tj + etk
  10. r(t) = 〈t cos t, t sin t, t2〉

In Problems 11–14, find the vector function that describes the curve C of intersection between the given surfaces. Sketch the curve C. Use the indicated parameter.

  1. z = x2 + y2, y = x; x = t
  2. x2 + y2 – z2 = 1, y = 2x; x = t
  3. x2 + y2 = 9, z = 9 – x2; x = 3 cos t
  4. z = x2 + y2, z = 1; x = sin t
  5. Given that r(t) = i + (t – 2)5j + t ln tk, find r(t).
  6. Given that lim t→a r1(t) = i – 2j + k and limt→a r2(t) = 2i + 5j + 7k, find:

    (a) [–4r1(t) + 3r2(t)]

    (b) r1(t) · r2(t).

In Problems 17–20, find r′(t) and r″(t) for the given vector function.

  1. r(t) = ln ti + j, t > 0
  2. r(t) = 〈t cos t – sin t, t + cos t〉
  3. r(t) = 〈te2t, t3, 4t2 – t〉
  4. r(t) = t2i + t3j + tan−1tk

In Problems 21–24, graph the curve C that is described by r and graph r′ at the indicated value of t.

  1. r(t) = 2 cos ti + 6 sin tj; t = π/6
  2. r(t) = t3i + t2j; t = −1
  3. r(t) = 2i + tj + k; t = 1
  4. r(t) = 3 cos ti + 3 sin tj + 2tk; t = π/4

In Problems 25 and 26, find parametric equations of the tangent line to the given curve at the indicated value of t.

  1. x = t, y = t2, z = t3; t = 2
  2. x = t3 – t, y = , z = (2t+ 1)2; t = 1

In Problems 27–32, find the indicated derivative. Assume that all vector functions are differentiable.

  1. [r(t) × r′(t)]
  2. [r(t) · (tr(t))]
  3. [r(t) · (r′(t) × r″(t))]
  4. [r1(t) × (r2(t) × r3(t))]

In Problems 33–36, evaluate the given integral.

In Problems 37–40, find a vector function r that satisfies the indicated conditions.

  1. r′(t) = 6i + 6tj + 3t2k; r(0) = i – 2j + k
  2. r′(t) = t sin t2i – cos 2tj; r(0) = i
  3. r″(t) = 12ti – 3t−1/2j + 2k; r′(1) = j, r (1) = 2i – k
  4. r″(t) = sec2ti + cos tj – sin tk;

    r′(0) = i + j + k, r(0) = −j + 5k

In Problems 41–44, find the length of the curve traced by the given vector function on the indicated interval.

  1. r(t) = a cos ti + a sin tj + ctk; 0 ≤ t ≤ 2π
  2. r(t) = ti + t cos tj + t sin tk; 0 ≤ t ≤ π
  3. r(t) = et cos 2ti + et sin 2tj + etk; 0 ≤ t ≤ 3π
  4. r(t) = 3ti + t2j + t3k; 0 ≤ t ≤ 1
  5. Express the vector equation of a circle r(t) = a cos ti + a sin tj as a function of arc length s. Verify that r′(s) is a unit vector.
  6. If r(s) is the vector function given in (4), verify that r′(s) is a unit vector.
  7. Suppose r is a differentiable vector function for which r(t) = c for all t. Show that the tangent vector r′(t) is perpendicular to the position vector r(t) for all t.
  8. In Problem 47, describe geometrically the kind of curve C for which r(t) = c.

Miscellaneous Problems

  1. Prove Theorem 9.1.4(ii).
  2. Prove Theorem 9.1.4(iii).
  3. Prove Theorem 9.1.4(iv).
  4. If v is a constant vector and r is integrable on [a, b], prove that .