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Breakout Room Session #03 Wk 02 D 02 June 15, 2021

Breakout Room # _________

Members:

________________________________________________________________________

1. Welcome to the world of 3-dimensional Calculus.

(2/2) = 4 points

2. Welcome to the world of 3-dimensional Calculus.

You may draw you own illustrations or use CalcPlot3D to show the graph that satisfies each of the

corresponding conditions; that is graph the solution set for each of these three problems (6 points total)

๐ฅ=2

๐ฆ=3

๐ง=1

๐

2

1โค ๐โค2

1โค ๐งโค2

0โค ๐โค

๐ฅ2 + ๐ง2 = 4

๐(๐ก) = โฉ2 cos ๐ก , 2 sin ๐ก , 0โช

No need to explain. I will just look at your solutions. It is best to share what you think with your network to

get some feedback prior to entering the Breakout Room on Thursday.

3. Projectile Motion.

This problem is from your text on page 842 #30. Illustrate the situation based on the description

described in the text and determine the solution. (3/2) + 1 = 6 points (visualization, explanation, and

final calculations leading to what is requested).

Illustration (1)

Explain and Deliver (3/2)

Before I could arrive at my solution, I had to . . .

Here are the calculations that are outlined in the

description of my process.

4. Use CalcPlot3D to Illustrate the space curve defined by

r(t) = < 2 sin t , 2 cos t, 4 sin2t >,

determine the equation of the tangent line at the point P(1, โ3, 1), and plot the space

curve and tangent line using the space curve option.

This problem was #14 (page 852).

If all your calculations are right, I should be able to see the space curve and a tangent line.

As with the last visualization problem, no explanation is needed. You and I can see whether you are correct.

Again, work with your network if you have questions with CalcPlot3D. You also have youtube.com videos.

Just search for CalcPlot3D space curves if you or your network has questions.

(4 pts.)

12:19 AM Thu Jun 17

86%

TO

o

+ : 0

840

Chapter 12

Vector-Valued Functions

y vo

= Initial velocity

V(41)

Vo = V(O)

a

Projectile Motion

To derive the parametric equations for the path of a projectile, assume that gravity is

the only force acting on the projectile after it is launched. So, the motion occurs in a

vertical plane, which can be represented by the xy-coordinate system with the origin

as a point on Earth’s surface, as shown in Figure 12.17. For a projectile of mass m, the

force due to gravity is

F = – mgj

Force due to gravity

where the acceleration due to gravity is g = 32 feet per second per second, or

9.8 meters per second per second. By Newton’s Second Law of Motion, this same force

produces an acceleration a = a(t) and satisfies the equation F = ma. Consequently, the

acceleration of the projectile is given by ma = โ – mgj, which implies that

a = โ gj.

Acceleration of projectile

a

v(t)

a

Initial height

X

Figure 12.17

EXAMPLE 5 Derivation of the Position Vector for a Projectile

A projectile of mass m is launched from an initial position ro with an initial velocity Vo.

Find its position vector as a function of time.

Solution Begin with the acceleration a(t) – gj and integrate twice.

– gj dt = -gtj + C

v(o) = falo) dt =

– fvwo) de = f(-sej + C,) d =

r(t)

2gfj + Cyt+CZ

You can use the initial conditions v(0) = V, and r(0) = r, to solve for the constant

vectors C and C2. Doing this produces

C, = vo and C2 = ro.

Therefore, the position vector is

1

r(t)

= =

2gfj + tv, + ro

Position vector

Il voll = vo = initial speed

|| 1. || = h = initial height

ั

In many projectile problems, the constant vectors r, and v, are not given explicitly.

Often you are given the initial height h, the initial speed yo, and the angle o at which

the projectile is launched, as shown in Figure 12.18. From the given height, you can

deduce that r. hj. Because the speed gives the magnitude of the initial velocity, it

follows that vo || v || and you can write

vo xi + yj

(Il voll

| cos 0)i + (|| vo|| sin e);

= Vo cos di + V, sin Oj.

So, the position vector can be written in the form

VO

A

xi

h

ro

r() = – zelj

2914j + tv,

+ ro

Position vector

x = || voll cos 0

y = || voll sin 0

Figure 12.18

1

^

281;

+ tv, cos Oi + tv, sin 0j + hj

858

=

(v. cos 0)ti +

[ h+ (vo sin @)1 โ 220];

,

1288

Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

1

QA

12:19 AM Thu Jun 17

86%

<
T
+
Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203
12.3
Velocity and Acceleration
841
THEOREM 12.3 Position Vector for a Projectile
Neglecting air resistance, the path of a projectile launche om an initial
height h with initial speed V, and angle of elevation 0 is described by the
vector function
r(t) = (v, cos 0)ti + [h + (vo sin 6)t โ ลผgt]j
where g is the acceleration due to gravity.
EXAMPLE 6
Describing the Path of a Baseball
$10 ft
A baseball is hit 3 feet above ground level at
100 feet per second and at an angle of 45ยฐ with
respect to the ground, as shown in Figure 12.19.
Find the maximum height reached by the
baseball. Will it clear a 10-foot-high fence
located 300 feet from home plate?
Solution You are given
45ยฐ
300 ft
3 ft
Not drawn to scale
h = 3,
Vo = 100, and
ำฉ =
45ยฐ.
32 feet per
Figure 12.19
So, using Theorem 12.3 with g
second per second produces
r(t) =
= (100 cos %). + [3 + (100 sin 4) โ 161];
=
(50/2t)i + (3 + 50/2t - 161)j.
The velocity vector is
v(t) = r'(t) = 50 /2i + (50 V2 โ 32t)j.
The maximum height occurs when
y(t) = 50/2 โ 32t
is equal to 0, which implies that
25/2
= 2.21 seconds.
16
t=
So, the maximum height reached by the ball is
(252
y
161
16
16
649
8
z 81 feet.
Maximum height when t = 2.21 seconds
The ball is 300 feet from where it was hit when
3/2
~ 4.24 seconds. At this time, the height
X(t) = 300 50 2t = 300.
Solving this equation for t produces t =
of the ball is
y = 3 + 50/2(3/2) โ 16(3/2)2
303 โ 288
859
/ /
1288
= 15 feet.
Height when tz 4.24 seconds
Therefore, the ball clears the 10-foot fence for a home run.
CHA
x Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203
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Spherical coordinates
Cylindrical coordinates
Rectangular System
3D space curves
visualization problem
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