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) = ,
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];
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12:19 AM Thu Jun 17
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