Hello everyone,
I am stuck on this question, although I suspect it is easy enough to solve, my textbook isn't providing a lot of help, if y'all could explain how to solve this I'd appreciate it!
Find a vector function whose graph is the curve of intersection of the sphere x^2 + y^2 + z^2 = 10 and the plane x = -1.
Thanks!
That does make sense, however, it seems that there must have been a miscalculation somewhere?
These are my answer choices...
-3 sin t, 3 cos t, -1
3 sin t, 1 , -3 sin t
1, 3 cos t, -3 sin t
3 cos t, -3 sin t, 1
-1, 3 sin t, -3 cos t
3 cos t, -1, -3 sin t
plane and sphere don't intersect.
Given those, your answer of -1, 3 cos t, 3 sin t, isn't available...
Oh, also a question I had that is a part of another question, what is the vector function for a circular cylinder of radius 1?
In this case, works as well since which still gives us the equation of the circle we started with.
To answer the second question, I need to know which axis acts as the center of each circular cross section since I can come up with 3 different equations (one for each coordinate axis).
In this case, z can be anything, so lets say . Then we just have the equation for a circle . This is parameterized by and .
Therefore, the parameterization for the cylinder would be .
Does this make sense?
(If this is another multiple choice question, note that the first two elements can be swapped [if needed] and can be postive,negative, or any combination thereof).
Hmm, maybe I should post the full question.
A circular cylinder of radius 1 and the parabolic cylinder z = 2x^2 - 1 intersect as shown in...
Which vector function has the curve of intersection as its graph if x(0) = 1.
The answer choices are...
(cost)i + (sint)j + (cos2t - 1)k
(cost)i + (sint)j + (cos2t)k
(cost)i + (sint)j + (1 - cos2t)k
(sint)i + (cost)j + (cos2t - 1)k
(sint)i + (cost)j + (cos2t)k
(sint)i + (cost)j + (1 - cos2t)k
I'm under the impression that since x(0) has to equal 1, then i must be cost, since cos(0) = 1. If that is correct that eliminates three of the six choices.