﻿ ﻿MOTION ALONG A PLANE CURVE - Further Applications of Integration - 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: 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 its magnitude, is the speed of the particle, and the velocity vector is tangent to the path. The slope of a is The distance the particle travels from time t1 to t2, is given by 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: We start with equations representing acceleration due to gravity and integrate each twice, determining the constants as shown: 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 at any time t 0.

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

SOLUTION: Since we integrate to get and use

x(0) = 1 to find that C = 2. Therefore, and Since y(0) = 0, this yields C = 1, and so (2) becomes Thus the parametric equations are 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 the parametric representation of the particle’s position. Hence its position at t = 1 is From P(t) we write the velocity vector: Hence, at t = 1 the particle’s velocity is Speed is the magnitude of the velocity vector, so after 1 second the particle’s speed is The particle’s acceleration vector at t = 1 is On the interval 0 ≤ t ≤ 1 the distance traveled by the particle is BC ONLY

EXAMPLE 7

A particle P(x, y) moves along a curve so that its acceleration is given by when t = 0, the particle is at (1, 0) with (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) and since when t = 0, it follows that c1 = c2 = 0. So Also and since when t = 0, we see that c3 = c4 = 0. Finally, then, (b) From (a) the parametric equations of motion are

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

By a trigonometric identity, 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. FIGURE N8–1

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