hi possible to explain part a more in depth to me? i have no idea whats going on as well as whats a ellipsoidal surface
hi possible to explain part a more in depth to me? i have no idea whats going on as well as whats a ellipsoidal surface
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 25RE
Related questions
Question
hi possible to explain part a more in depth to me? i have no idea whats going on as well as whats a ellipsoidal surface.
![4. Find a parametric representation for each of the following surfaces:
(a) 36(x-1)² + 25 y +9(z+ 2)² = 36 (ellipsoidal surface)
(b) 1+ x + y - z= 0 for 2<z< 9.
(Solution)
(a) Note that we are required to find "a parametric representaion." (There are many
possible answers to this.) One possible way is as follows. Let x-1=t, where t is a
real
follows
that
(5 y)? + (3[ z+ 2])² = 36(1– ²).
parameter.
It
Since
25 y +9(z+ 2) is non-negative, it tells us that 1- t must be non-negative. Thus,
t lies between -1 and 1 (including -1 and 1),0 that is, t E [-1, 1].0 Now,
(5 y)? + (3[ z+ 2])? = 36(1– ?) is satisfied, if we choose 5y=6y1- cos (u) and
3[z+ 2] = 6V1- t sin(u) (for example), where u is any real parameter. [Another
possiblity is 5y= 6V1-2 sin(u) and 3[z+ 2] = 6V1-? cos(2) .] Thus, a possible
2.
parametric representation of the surface 36(x-1)2 + 25 y² + 9(z+ 2)² = 36 is given
by
X=1+t
6.
y=Vi-P cos(u)
for te [-1,1] and ue [0,27].
z=-2+2\1-² sin(u)l]
Note u= 0 and u= 27 gives the same point for the same t.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe594dc0-2b5b-4bcb-9d92-f82cc2a5b645%2Fdec56978-cce2-4875-ace5-37c60bc07e6f%2Ftnxy6sc_processed.png&w=3840&q=75)
Transcribed Image Text:4. Find a parametric representation for each of the following surfaces:
(a) 36(x-1)² + 25 y +9(z+ 2)² = 36 (ellipsoidal surface)
(b) 1+ x + y - z= 0 for 2<z< 9.
(Solution)
(a) Note that we are required to find "a parametric representaion." (There are many
possible answers to this.) One possible way is as follows. Let x-1=t, where t is a
real
follows
that
(5 y)? + (3[ z+ 2])² = 36(1– ²).
parameter.
It
Since
25 y +9(z+ 2) is non-negative, it tells us that 1- t must be non-negative. Thus,
t lies between -1 and 1 (including -1 and 1),0 that is, t E [-1, 1].0 Now,
(5 y)? + (3[ z+ 2])? = 36(1– ?) is satisfied, if we choose 5y=6y1- cos (u) and
3[z+ 2] = 6V1- t sin(u) (for example), where u is any real parameter. [Another
possiblity is 5y= 6V1-2 sin(u) and 3[z+ 2] = 6V1-? cos(2) .] Thus, a possible
2.
parametric representation of the surface 36(x-1)2 + 25 y² + 9(z+ 2)² = 36 is given
by
X=1+t
6.
y=Vi-P cos(u)
for te [-1,1] and ue [0,27].
z=-2+2\1-² sin(u)l]
Note u= 0 and u= 27 gives the same point for the same t.
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