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Breakout Room Session #03 Wk 02 D 02 June 15, 2021
Breakout Room # _________
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)
𝑥2 + 𝑧2 = 4
𝑟(𝑡) = 〈2 cos 𝑡 , 2 sin 𝑡 , 0〉
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).
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, √3, 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.
12:19 AM Thu Jun 17
+ : 0
= Initial velocity
Vo = V(O)
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 = – – mgj, which implies that
a = – gj.
Acceleration of projectile
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 =
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
2gfj + tv, + ro
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
| cos 0)i + (|| vo|| sin e);
= Vo cos di + V, sin Oj.
So, the position vector can be written in the form
r() = – zelj
2914j + tv,
x = || voll cos 0
y = || voll sin 0
+ tv, cos Oi + tv, sin 0j + hj
(v. cos 0)ti +
[ h+ (vo sin @)1 – 220];
Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203
12:19 AM Thu Jun 17
< 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 – żgt]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 45° 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 45° 300 ft 3 ft Not drawn to scale h = 3, Vo = 100, and ө = 45°. 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) – 161]; = (50/2t)i + (3 + 50/2t - 161)j. The velocity vector is v(t) = r'(t) = 50 /2i + (50 V2 – 32t)j. The maximum height occurs when y(t) = 50/2 – 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) – 16(3/2)2 303 – 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 Đ Purchase answer to see full attachment Tags: Spherical coordinates Cylindrical coordinates Rectangular System 3D space curves visualization problem User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
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