Calculus AB and Calculus BC

CHAPTER 8 Further Applications of Integration

B. MOTION ALONG A PLANE CURVE

BC ONLY

In Chapter 4, §K, it was pointed out that, if the motion of a particle P along a curve is given parametrically by the equations x = x(t) and y = y(t), then at time t the position vector R, the velocity vector v, and the acceleration vector a are:

Image

The components in the horizontal and vertical directions of R, v, and a are given respectively by the coefficients of i and j in the corresponding vector. The slope of v is Image its magnitude,

Image

is the speed of the particle, and the velocity vector is tangent to the path. The slope of a is Image The distance the particle travels from time t1 to t2, is given by

Image

How integration may be used to solve problems of curvilinear motion is illustrated in the following examples.

BC ONLY

EXAMPLE 4

Suppose a projectile is launched from the origin at an angle of elevation α and initial velocity v0. Find the parametric equations for its flight path.

SOLUTION: We have the following initial conditions:

Position: x(0) = 0; y(0) = 0.

Velocity: Image

We start with equations representing acceleration due to gravity and integrate each twice, determining the constants as shown:

Image

If desired, t can be eliminated from this pair of equations to yield a parabola in rectangular coordinates.

EXAMPLE 5

A particle P(x, y) moves along a curve so that

Image at any time t Image 0.

At t = 0, x = 1 and y = 0. Find the parametric equations of motion.

SOLUTION: Since Image we integrate to get Image and use

x(0) = 1 to find that C = 2. Therefore, Image and

Image

Since y(0) = 0, this yields C = 1, and so (2) becomes

Image

Thus the parametric equations are

Image

BC ONLY

EXAMPLE 6

The particle in Example 5 is in motion for 1 second, 0 ≤ t ≤ 1. Find its position, velocity, speed, and acceleration at t = 1 and the distance it traveled.

SOLUTION: In Example 5 we derived the result Image the parametric representation of the particle’s position. Hence its position at t = 1 is Image

From P(t) we write the velocity vector:

Image

Hence, at t = 1 the particle’s velocity is Image

Speed is the magnitude of the velocity vector, so after 1 second the particle’s speed is

Image

The particle’s acceleration vector at t = 1 is

Image

On the interval 0 ≤ t ≤ 1 the distance traveled by the particle is

Image

BC ONLY

EXAMPLE 7

A particle P(x, y) moves along a curve so that its acceleration is given by

Image

when t = 0, the particle is at (1, 0) with Image

(a) Find the position vector R at any time t.

(b) Find a Cartesian equation for the path of the particle, and identify the conic on which P moves.

SOLUTIONS:

(a) Image and since Image when t = 0, it follows that c1 = c2 = 0. So Image Also Image and since Image when t = 0, we see that c3 = c4 = 0. Finally, then,

Image

(b) From (a) the parametric equations of motion are

x = cos 2t, y = 2 sin t.

By a trigonometric identity,

Image

P travels in a counterclockwise direction along part of a parabola that has its vertex at (1, 0) and opens to the left. The path of the particle is sketched in Figure N8–1; note that −1 ≤ x ≤ 1, −2 ≤ y ≤ 2.

Image

FIGURE N8–1