Suppose we are tracking a particle's motion in space. If we run a vector r from the origin to the particle, then we can study the changes in r. If the particle's position coordinates are twice-differentiable functions of time, then so is r. Therefore we can find the particle's velocity and acceleration vectors at any time t by differentiating r.

DEFINITION: When a particle moves through space during a time interval I, the particle's coordinates are functions defined on I: x = f (t), y = g (t), z = h (t), t Î I.

The points (x, y, z) = (f (t), g (t), h (t)), t Î I make up the curve in space called the particle's path.

The vector r = OP = f (t) i + g (t) j + h (t) k from the origin to the particle's position P at time t is the particle's position vector. (See figure 1)

figure 1

The vector r is called a vector function of the real variable t on the interval I.

FACT:

A vector-valued function on a domain set D is a rule that assigns a vector in space to each element in D.

NOTE:

From now on, real-valued functions are now called scalar functions.

Limits of vector-valued functions are defined the same way as limits of scalar functions.

DEFINITION:

Let r (t) = f (t) i + g (t) j + h (t) k be a vector function and L a vector. We say that r has limit L as t approaches t ₀ and write

if, for every number e > 0, there exists a corresponding number d > 0 such that for all t 0 < | t - t ₀| < d Þ |r (t) - L| < e .

EXAMPLE 1:

If r (t) = t ² i + (t - 1) j + (t ³ + 3t) k, then find the following limit.

SOLUTION:

DEFINITION:

A vector function r (t) is continuous at a point t = t ₀ in its domain if

The function is continuous if it is continuous at every point in its domain.

FACT:

The vector r (t) = f (t) i + g (t) j + h (t) k is continuous at t = t ₀ if and only if f, g, and h are continuous at t ₀.

Suppose r (t) = f (t) i + g (t) j + h (t) k is the position vector of a particle moving along a curve in space and that f, g, and h are differentiable functions of t. Then the difference between the particle's position at time t and time t + D t is D r = r (t + D t) - r(t). (See figure 2)

figure 2

In terms of the components D r = r (t + D t) - r (t) = [f (t + D t) i + g (t + D t) j + h (t + D t) k] - [f (t) i + g (t) j + h (t) k] = [f (t + D t) - f (t)] i + [g (t + D t) - g (t)] j + [h (t + D t) - h(t)] k.

As D t ® 0, three things seem to happen simultaneously.

1.

Q ® P along the curve.

2.

The secant line PQ seems to be approaching a limiting position tangent to the curve at P.

3.

D r/D t approaches the limit.

FACT:

FACT:

The vector dr/dt, when different from the 0, is also the vector tangent to the curve.

FACT:

The tangent line to the curve at a point (f (t ₀), g (t ₀), h (t ₀)) is defined to be the line through the point parallel to dr/dt at t = t ₀.

FACT:

dr/dt ¹ 0 is a must for a smooth curve to make sure the curve has a continuously turning tangent at each point.

EXAMPLE 2:

Find the parametric equations for the tangent line to the curve

r (t) = (cos t) i + (sin t) j + (sin 2t) k at t = p /2.

SOLUTION:

First find the point on the curve. Plug t = p /2 into each component to get the point (x, y, z). The point is (0, 1, 0).

Now find the derivative.

Now evaluate the derivative at t = p /2.

Here are the parametric equations for the tangent line.

x = -t

y = 1

z = -2t

FACT:

If r is the position vector of a particle moving along a smooth curve in space, then v(t) = dr/dt is the particle's velocity vector, tangent to the curve. At any time t, the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative, a = dv/dt, when it exists, is the particle's acceleration vector.

1.

Velocity:

2.

Speed = | v |

3.

Acceleration:

4.

Directions of motion at time t is v/| v |.

EXAMPLE 3:

Given r (t) = (e ^{- t} )i + (2cos 3t)j + (2sin 3t)k, t = 0, find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of t. Write the particle's velocity at that time as the product of its speed and direction.

SOLUTION:

v = (-e ^{- t})i + (-6sin 3t)j + (6cos 3t)k

v (0) = -i + 6k

a = (e ^{- t})i + (-18cos 3t)j + (-18sin 3t)k

a (0) = i - 18j

EXAMPLE 4:

Given

find the angle between the velocity and acceleration vectors at time t = 0.

SOLUTION:

EXAMPLE 5:

Given r (t) = (sin t)i + tj + (cos t)k, t ³ 0. Find the time or times in the given time interval when the velocity and acceleration vectors are orthogonal.

SOLUTION:

v = (cos t)i + j - (sin t)k

a = (-sin t)i - (cos t)k

v · a = -cos t sin t + sin t cos t = 0

v and a are orthogonal for all values of t.

EXAMPLE 6:

Given r (t) = (t ² + 1)i + tj + (t ³ - 1)k, t ³ 0. Find the time or times in the given time interval when the velocity and acceleration are orthogonal.

SOLUTION:

v = 2ti + j + 3t ²k

a = 2i + 6tk

v · a = 4t + 18t ³ = 0 ® 2t (2 + 9t ²) = 0 ® 2t = 0 ® t = 0

v and a are orthogonal when t = 0.

CONSTANT FUNCTION RULE:

(any constant vector c)

If u and v are differentiable vector functions of t, then the following can be used.

SCALAR MULTIPLE RULES:

(any number c)

(any differentiable scalar function f (t)).

SUM RULE:

DIFFERENCE RULE:

DOT PRODUCT RULE:

CROSS PRODUCT RULE:

CHAIN RULE:

If r is a differentiable function of t and t is a differentiable function of s, then

When we track a particle moving on a sphere centered at the origin, the position vector has constant length equal to the radius of the sphere. In fact, the velocity vector dr/dt, tangent to the path of motion, is tangent to the sphere and perpendicular to r. This is how we will decide if a differentiable vector function is of constant length.

FACT:

If u is a differentiable vector function of t of constant length, then

EXAMPLE 7:

Show that u = (sin t)i + (cos t)j + 3k has constant length and is orthogonal to its derivative.

SOLUTION:

Therefore, u has constant length and it is orthogonal to its derivative.

FACT:

A differentiable vector function R (t) is an antiderivative of a vector function r (t) on an interval I if

at each point of I.

FACT:

The indefinite integral of r with respect to t is the set of all antiderivatives of r, denoted by

If R is any antiderivative of r, then

FACT:

If the components of r (t) = f (t)i + g (t)j + h (t)k are integrable over [a, b], then so is r, and the definite integral of r from a to b is

EXAMPLE 8:

SOLUTION:

EXAMPLE 9:

Solve the initial value problem.

SOLUTION:

When t = 0, r = 100j, so 100j = 0i + 0j + C or c = 100j.

EXAMPLE 10:

Solve the following initial value problem.

SOLUTION:

8i + 8j = -32(0)k + C ® C = 8i + 8j.

When t = 0, r (0) = 100k.

100k = 0i + 0j - 0k + C ® C = 100k

r (t) = (8t)i + (8t)j + (100 - 16t ²)k