## 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 *t*_{1} to *t*_{2}, 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 *v*_{0}. 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 *c*_{1} = *c*_{2} = 0. So Also and since when *t* = 0, we see that *c*_{3} = *c*_{4} = 0. Finally, then,

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

*x* = cos 2*t*, *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**