9.2 Motion on a Curve

INTRODUCTION

In the preceding section we saw that the first and second derivatives of the vector function r(t) = 〈f(t), g(t), h(t)〉 = f(t)i + g(t)j + h(t)k can be obtained by differentiating the component functions f, g, and h. In this section we give a physical interpretation to the vectors r′(t) and r″(t).

Velocity and Acceleration

Suppose a particle or body moves along a curve C so that its position at time t is given by the vector function r(t) = f(t)i + g(t)j + h(t)k. If the component functions f, g, and h have second derivatives, then the vectors

are called the velocity and acceleration of the particle, respectively. The scalar function v(t) is the speed of the particle. Since

speed is related to arc length s by s′(t) = v(t). In other words, arc length is given by s = v(t)dt. It also follows from the discussion of Section 9.1 that if P(x1, y1, z1) is the position of the particle on C at time t1, then we may draw v(t1) tangent to C at P. Similar remarks hold for curves traced by the vector function r(t) = f(t)i + g(t)j.

EXAMPLE 1 Velocity and Acceleration Vectors

The position of a moving particle is given by r(t) = t2i + tj + tk. Graph the curve defined by r(t) and the vectors v(2) and a(2).

SOLUTION

Since x = t2, y = t, the path of the particle is above the parabola x = y2. When t = 2 the position vector r(2) = 4i + 2j + 5k indicates that the particle is at the point P(4, 2, 5). Now,

v(t) = r′(t) = 2ti + j + k     and    a(t) = r″(t) = 2i

so that v(2) = 4i + j + k and a(2) = 2i. These vectors are shown in FIGURE 9.2.1. ≡

A graph. 2 curves and 2 vectors are graphed on a three dimensional x y z coordinate system. The first curve is dashed and labeled x = y^2. It is a parabola graphed on the x y plane, passes through the marked point (4, 2, 0) and the origin, and is symmetric with reference to the x axis. The second curve, begins at the origin, and goes up, following the path right above the dashed curve, passes through the labeled point P(4, 2, 5), and exits the top. A dashed vertical line segment begins at the marked point (4, 2, 0), goes vertically up parallel to the z axis, and ends at the labeled point P(4, 2, 5). The first vector, labeled v(2), begins on the second curve at the labeled point P(4, 2, 5), and ends at the point (8, 3, 15 over 2). The second vector, labeled a(2), begins on the second curve at the labeled point P(4, 2, 5), and ends at the point (6, 2, 5).

FIGURE 9.2.1 Velocity and acceleration vectors in Example 1

If a particle moves with a constant speed c, then its acceleration vector is perpendicular to the velocity vector v. To see this, note that v2 = c2 or v · v = c2. We differentiate both sides with respect to t and obtain, with the aid of Theorem 9.1.4(iii):

Thus,

EXAMPLE 2 Velocity and Acceleration Vectors

Suppose the vector function in Example 2 of Section 9.1 represents the position of a particle moving in a circular orbit. Graph the velocity and acceleration vector at t = π/4.

SOLUTION

Recall that r(t) = 2 cos ti + 2 sin tj + 3k is the position vector of a particle moving in a circular orbit of radius 2 in the plane z = 3. When t = π/4 the particle is at the point P(, , 3). Now,

v(t) = r′(t) = –2 sin ti + 2 cos tj

a(t) = r″(t) = –2 cos ti – 2 sin tj.

Since the speed is v(t) = 2 for all time t, it follows from the discussion preceding this example that a(t) is perpendicular to v(t). (Verify this.) As shown in FIGURE 9.2.2, the vectors

are drawn at the point P. The vector v(π/4) is tangent to the circular path and a(π/4) points along a radius toward the center of the circle. ≡

A graph. A circle, a plane and 2 vectors are graphed on a three dimensional x y z coordinate system. The plane, labeled z = 3, is parallel and above the x y plane. The circle is on the plane labeled z = 3 and its center is on the vertical z axis. The circle passes through the labeled point P(sqrt(2), sqrt(2), 3). The first vector, labeled v(pi over 4), begins on the circle at the labeled point P, and is tangent to the circle at the point P as well. The second vector, labeled a(pi over 4), begins on the circle at the point labeled P, and goes along a radius of the circle toward the center of the circle.

FIGURE 9.2.2 Velocity and acceleration vectors in Example 2

Centripetal Acceleration

For circular motion in the plane, described by r(t) = r0 cos ωti + r0 sin ωtj, r0 and ω constants, it is evident that r″ = –ω2r. This means that the acceleration vector a(t) = r″(t) points in the direction opposite to that of the position vector r(t). We then say a(t) is centripetal acceleration. See FIGURE 9.2.3. If v = v(t) and a = a(t), we leave it as an exercise to show that a = v2/r0.

A circle and 4 vectors of similar magnitudes. The first vector, labeled v(t subscript 1), begins on the circle and is tangent to the circle at the same point. The second vector, labeled a(t subscript 1), is perpendicular to the first vector, begins at the same point, and ends at the center of the circle. The third vector, labeled r(t subscript 2), is perpendicular to the second vector, begins at the center of the circle, and ends at a point on the circle. The fourth vector, labeled v(t subscript 2), begins on the circle and is tangent to the circle at the end point of the third vector, and is perpendicular to the third vector.

FIGURE 9.2.3 Circular motion

Curvilinear Motion in the Plane

Many important applications of vector functions occur in regard to curvilinear motion in a plane. For example, planetary and projectile motion take place in a plane.

In analyzing the motion of short-range ballistic projectiles,* we begin with the acceleration of gravity written in vector form: a(t) = –gj.

If, as shown in FIGURE 9.2.4, a projectile is launched with an initial velocity v0 = v0 cos θi + v0 sin θj from an initial height s0 = s0j, then

where v(0) = v0 implies that c1 = v0. Therefore,

v(t) = (v0 cos θ)i + (–gt + v0 sin θ)j.

Integrating again and using r(0) = s0 yields

(1)

Hence, parametric equations for the trajectory of the projectile are

(2)

We are naturally interested in finding the maximum height H and the range R attained by a projectile. As shown in FIGURE 9.2.5, these quantities are the maximum values of y(t) and x(t), respectively.

A graph. A curve and 4 vectors are graphed on an x y coordinate plane. The curve, representing the trajectory of a projectile, begins at a marked point on the y axis, goes up and to the right, reaches a high point, then goes down to the right, and ends on the x axis. The first vector, labeled s subscript 0 j, begins at the origin, goes up vertically and ends at the starting point of the curve. The second vector is dashed and labeled (v subscript 0 cos theta) i, begins on the y axis at the starting point of the curve, and goes horizontally to the right. The third vector is dashed and labeled (v subscript 0 sin theta) j, begins on the y axis at the starting point of the curve, and goes vertically up. The fourth vector, labeled v subscript 0, begins on the y axis at the starting point of the curve, goes up and to the right, and is tangent to the curve at the starting point. The angle formed by the vectors labeled (v subscript 0 cos theta) i and v subscript 0, is labeled theta.

FIGURE 9.2.4 Trajectory of a projectile

2 graphs. Graph a. Caption. Maximum height H: Find t subscript 1 for which y prime(t subscript 1) = 0; H = y subscript max = y(t subscript 1). Graph. A curve is graphed on an x y coordinate plane. The curve begins at the origin, goes up and to the right, reaches a high point at a marked point, then goes down to the right symmetrically, and ends on the x axis. The vertical distance between the marked high point of the curve and the x axis, is labeled H. Graph b. Caption. Range R: Find t subscript 1 > 0 for which y (t subscript 1) = 0; R = x subscript max = x(t subscript 1). Graph. A curve is graphed on an x y coordinate plane. The curve begins at the origin, goes up and to the right, reaches a high point, then goes down to the right symmetrically, and ends on the x axis. The horizontal distance between the starting and ending points of the curve, is labeled R.

FIGURE 9.2.5 Maximum height and range of a projectile

EXAMPLE 3 Trajectory of a Projectile

A projectile is launched from ground level with an initial speed v0 = 768 ft/s at an angle of elevation θ = 30°. Find (a) the vector function and parametric equations of the projectile’s trajectory, (b) the maximum altitude attained, (c) the range of the projectile, and (d) the impact speed.

SOLUTION

(a) The initial height and velocity are, respectively, s0 = 0 and

(3)

Integrating a(t) = –32j and using (3) give

(4)

Integrating (4) and using r(0) = s0 = 0 then give

The components of this vector function,

, (5)

are the parametric equations of the projectile’s trajectory.

(b) From (5) we see that y′(t) = 0 when –32t + 384 = 0 or t = 12 s. The maximum height attained by the projectile is

(c) From (5) we see that y(t) = 0 when The time that projectile hits the ground is t = 24 s and the corresponding range is

(d) Finally, from (4) we see that and so the impact speed is

≡

In Example 3, note that the impact speed is the same as the launch speed v0 = 768 ft/s. Verify that this is still the case if we change the angle of elevation to, say, 50°. See Problem 16 in Exercises 9.2.

REMARKS

We have seen that the rate of change of arc length ds/dt is the same as the speed v(t) = r′(t). However, as we shall see in the next section, it does not follow that the scalar acceleration d2s/dt2 is the same as a(t) = r″(t). See Problem 24 in Exercises 9.2.

9.2 Exercises Answers to selected odd-numbered problems begin on page ANS-23.

In Problems 1–8, r(t) is the position vector of a moving particle. Graph the curve and the velocity and acceleration vectors at the indicated time. Find the speed at that time.

  9. Suppose r(t) = t2i + (t3 – 2t)j + (t2 – 5t)k is the position vector of a moving particle. At what points does the particle pass through the xy-plane? What are its velocity and acceleration at these points?
  10. Suppose a particle moves in space so that a(t) = 0 for all time t. Describe its path.
  11. A shell is fired from ground level with an initial speed of 480 ft/s at an angle of elevation of 30°. Find:
    1. a vector function and parametric equations of the shell’s trajectory,
    2. the maximum altitude attained,
    3. the range of the shell, and
    4. the speed at impact.
  12. Rework Problem 11 if the shell is fired with the same initial speed and the same angle of elevation but from a cliff 1600 ft high.
  13. A used car is pushed off an 81-ft-high sheer seaside cliff with a speed of 4 ft/s. Find the speed at which the car hits the water.
  14. A small projectile is launched from ground level with an initial speed of 98 m/s. Find the possible angles of elevation so that its range is 490 m.
  15. A football quarterback throws a 100-yd “bomb” at an angle of 45° from the horizontal. What is the initial speed of the football at the point of release?
  16. If a projectile is launched from level ground, then the initial conditions are r(0) = 0, and the vector function (1) becomes

    (6)

    This function is equivalent to the solution x(t), y(t) of the system of linear differential equations (10) in Problem 21 of Exercises 4.6 subject to the same initial conditions. Underlying that earlier discussion, as well as in the derivation of (1), is the assumption that the projectile is not subject to air resistance. Use (6) to show that the impact speed of a projectile is the same as the initial speed for any angle of elevation θ, 0° < θ < 90°.

  17. When air resistance is ignored, we saw in Problem 21 of Exercises 4.6 that a projectile launched at an angle θ, 0° < θ < 90° from level ground (r(0) = 0), then its horizontal range and maximum height are, respectively,

    From the first formula, it follows that when the projectile is launched at distinct complementary angles the horizontal range is the same and that the maximum range is attained when the angle of elevation is θ = 45°. But if the projectile is launched from an initial height s0 > 0 the foregoing formulas and statements are not valid.

    1. With v0 = 480 ft/s, s0 = 1600 ft rework Problem 12, but this time use the complementary angle of elevation θ = 60°. Compare the maximum height, range, and impact speed of the projectile with the answers to parts (b), (c) and (d) of Problem 12.
    2. Use a graphing utility or CAS to plot the trajectory of the projectile defined by the parametric equations x(t) and y(t) in part (a) of Problem 12. Repeat for the parametric equations in part (a) of this problem. Superimpose both of these curves on the same coordinate system.
  18. As mentioned in Problem 17, if a projectile is launched from an initial height s0 > 0 the maximum range of the projectile is not attained using θ = 45° as the angle of elevation.
    1. A projectile is launched from an initial height of v0 = 480 ft/s, s0 = 1600 ft, at an angle of elevation of θ = 45°. Use a calculator or CAS to find the time the projectile hits the ground and the corresponding range.
    2. If the angle of elevation in part (a) is changed to θ = 39.76°, show that the range is greater than that in part (a).
    3. Use a graphing utility or CAS to plot the trajectories of the projectiles in parts (a) and (b). Superimpose both of these curves on the same coordinate system.
  19. If a projectile is launched from level ground, the initial conditions are r(0) = 0, and if linear air resistance is taken into consideration, the vector function analogue of (1) is

    This function is equivalent to the solution x(t), y(t) of the system of linear differential equations (13) in Problem 22 of Exercises 4.6 subject to the same initial conditions. Here β > 0 is a constant of proportionality related to the air resistance or drag. If and θ = 38°, use a calculator or CAS to find the impact speed of the projectile. See part (b) of Problem 22 in Exercises 4.6.

  20. Suppose the angle of elevation of the projectile in Problem 19 is changed to θ = 52°. Do you think that the impact speed of the projectile is the same, greater than, or less than the value found in that problem? Prove your assertion.
  21. A projectile is fired from a cannon directly at a target that is dropped from rest simultaneously as the cannon is fired. Show that the projectile will strike the target in midair. See FIGURE 9.2.6. [Hint: Assume that the origin is at the muzzle of the cannon and that the angle of elevation is θ. If rp and rt are position vectors of the projectile and target, respectively, is there a time at which rp = rt?]
    A visual representation. A projectile is fired from a cannon pointing up and to the right, at a target that is dropped vertically from rest. A dashed vertical line begins at the muzzle of the cannon, goes up and to the right, and ends at the falling target. A curve, representing the trajectory of the projectile, begins at the muzzle of the cannon, goes up and to the right, reaches a high point, then goes down to the right, and ends at a point on a dashed vertical line representing the path of the falling target.

    FIGURE 9.2.6 Cannon in Problem 21

  22. In army field maneuvers sturdy equipment and supply packs are simply dropped from planes that fly horizontally at a slow speed and a low altitude. A supply plane flies horizontally over a target at an altitude of 1024 ft at a constant speed of 180 mi/h. Use (1) to determine the horizontal distance a supply pack travels relative to the point from which it was dropped. At what line-of-sight angle α should the supply pack be released in order to hit the target indicated in FIGURE 9.2.7?
    A visual representation. A supply pack is being dropped from a plane flying horizontally at a low speed and a low altitude of 1024 feet. A downward sloping line connects the plane and the target on the ground. The angle formed by the dashed horizontal flight path of the plane and the downward sloping line, is labeled alpha. A dashed curve, representing the trajectory of the supply pack, begins on the plane, goes down and to the right with increasing steepness, above the downward sloping line, and ends on the target on the ground.

    FIGURE 9.2.7 Supply plane in Problem 22

  23. Suppose that r(t) = r0 cos ωti + r0 sin ωtj is the position vector of an object that is moving in a circle of radius r0 in the xy-plane. If v(t) = v, show that the magnitude of the centripetal acceleration is a = a(t) = v2/r0.
  24. The motion of a particle in space is described by

    r(t) = b cos ti + b sin tj + ctk, t ≥ 0.

    1. Compute v(t).
    2. Compute dt and verify that ds/dt is the same as the result of part (a).
    3. Verify that d2s/dt2 ≠ a(t).
  25. The effective weight we of mass m at the equator of the Earth is defined by we = mg – ma, where a is the magnitude of the centripetal acceleration given in Problem 23. Determine the effective weight of a 192-lb person if the radius of the Earth is 4000 mi, g = 32 ft/s2, and v = 1530 ft/s.
  26. Consider a bicyclist riding on a flat circular track of radius r0. If m is the combined mass of the rider and bicycle, fill in the blanks in FIGURE 9.2.8. [Hint: Use Problem 23 and force = mass × acceleration. Assume that the directions are upward and to the left.] The resultant vector U gives the direction the bicyclist must be tipped to avoid falling. Find the angle ϕ from the vertical at which the bicyclist must be tipped if her speed is 44 ft/s and the radius of the track is 60 ft.
    A visual representation. A cyclist is shown from the front. The cyclist is tipped to the left from the vertical to avoid falling. A horizontal vector, labeled (dash, 0), begins on the cyclist and goes to the left. A vertical vector, labeled (0, dash), begins at the same point and goes up. An upward sloping left arrow, labeled U = (dash, dash), follows the direction the bicyclist is tipped. The vector U is also labeled resultant. The angle formed by the vertical vector and the vector U is labeled phi. The following information is given about the vertical vector: force exerted by track = opposite of the combined weight of bike and person. A curved arrow under the cyclist, going counter clockwise, is labeled centripetal force.

    FIGURE 9.2.8 Bicyclist in Problem 26

  27. The velocity of a particle moving in a fluid is described by means of a velocity field v = v1i + v2j + v3k, where the components v1, v2, and v3 are functions of x, y, z, and time t. If the velocity of the particle is v(t) = 6t2xi – 4ty2j + 2t(z + 1)k, find r(t). [Hint: Use separation of variables.]
  28. Suppose m is the mass of a moving particle. Newton’s second law of motion can be written in vector form as

    where p = mv is called linear momentum. The angular momentum of the particle with respect to the origin is defined to be L = r × p, where r is its position vector. If the torque of the particle about the origin is τ = r × F = r × dp/dt, show that τ is the time rate of change of angular momentum.

  29. Suppose the Sun is located at the origin. The gravitational force F exerted on a planet of mass m by the Sun of mass M is

    F is a central force—that is, a force directed along the position vector r of the planet. Here k is the gravitational constant, r = r, u =r/r is a unit vector in the direction of r, and the minus sign indicates that F is an attractive force—that is, a force directed toward the Sun. See FIGURE 9.2.9.

    1. Use Problem 28 to show that the torque acting on the planet due to this central force is 0.
    2. Explain why the angular momentum L of a planet is constant.
      A visual representation. A planet of mass lowercase m, orbits along an elliptic path around the Sun of mass uppercase M. A line segment connecting the planet and the sun is labeled r. A vector beginning on the planet and going toward the Sun along the line r, is labeled F.

      FIGURE 9.2.9 Force F in Problem 29

Discussion Problems

  1. In this problem the student will use the properties in Sections 7.4 and 9.1 to prove Kepler’s first law of planetary motion: The orbit of a planet is an ellipse with the Sun at one focus. We assume that the Sun has mass M and is located at the origin, r is the position vector of a body of mass m moving under the gravitational attraction of the Sun, and u = r/r is a unit vector in the direction of r.
    1. Use Problem 29 and Newton’s second law of motion F = ma to show that

    2. Use part (a) to show that r × r″ = 0.
    3. Use part (b) to show that (r × v) = 0.
    4. It follows from part (c) that r × v = c, where c is a constant vector. Show that c = r2(u × u′).
    5. Show that (u · u) = 0 and consequently u · u′ = 0.
    6. Use parts (a), (e), and (d) to show that

      (v × c) = kM .

    7. By integrating the result in part (f) with respect to t, we get v × c = kMu + d, where d is another constant vector. Dot both sides of this last expression by the vector r = ru and use Problem 61 in Exercises 7.4 to show that

      where c = c, d = d, and θ is the angle between d and r.

    8. Explain why the result in part (g) proves Kepler’s first law.
    9. At perihelion, the point in the orbit where the body is closest to the Sun, the vectors r and v are perpendicular and have magnitudes r0 and v0, respectively. Use this information and parts (d) and (g) to show that c = r0v0 and .
  2. Suppose a projectile is launched from an initial height s0 at an angle of elevation θ. If air resistance is ignored, show that the horizontal range is given by

    Note that when s0 = 0 this formula reduces to the range R given in Problem 17.

  3. Consider the formula in Problem 31 as a function of the single variable θ. Show that the range of a projectile launched from an initial height s0 is a maximum for the angle of inclination

    Note that when s0 = 0 this formula reduces to the value given in Problem 17, that is, θ = π/4 or 45°. Use this formula to verify the angle of inclination used in part (b) of Problem 18.

 

*A projectile is shot or hurled rather than self-propelled. In the analysis of long-range ballistic motion, the curvature of the Earth must be taken into consideration.