How to Draw Intersection Curve With Surface and Plane

Yura

this is the problem and i've worked for mostof it, merely im not sure i've got the right way to go well-nigh it, or if the ansewr i have institute it right.
here, im using the method for finding the vector equation

given the plane 2x + 6y + z = 6 intersects the paraboloid z = x^2 + y^two, find and proper name this curve of intersection.
__________________________________________

so far, i rearranged the plane equation to a function of z and allow it equal te equation of the paraboloid.

this gave me:
x^2 + y^2 = 6 -2x - 6y
6 = x^2 + 2x + y^two + 6y
vi = [x^2 + 2x + ane^two] + [y^2 + 6y +three^2] -10
16 = [x + 1]^2 + [y + iii]^two
4^2 = [ten + 1]^two + [y + 3]^ii

this and so gave me the projection downwards of the function, a circle. but the function should be an ellipse. (which cant be chosen a part y of x, because it fails the vertical test, non to articulate on this)

to get the part of the curve of intersection, (here is where im non sure about where im going) i wanted x infunction of t and y in role of t (for parametric equations)

x = f(t)
y = g(t)

and then i went:

dominion: (cos(t))^2 + (sin(t))^2 = 1

[(x +1)/4]^ii + [(y + three)/4]^ii = 1

(x +1)/4 = cos(t) => ten + 1 = 4*cos(t) => x = iv*cos(t) - 1
(y + three)/4 = sin(t) => y + 3 = 4*sin(t) => y = iv*sin(t) - three
[(respectively): eqn, starting time eqn, second eqn]
[where o<= t<= 2*pi]

from,
four^two = [x + ane]^2 + [y + iii]^2
=> 4^two = [4*cos(t)]^2 + [iv*sin(t)]^2
iv^2 = xvi*[cos(t)]^2 + 16*[sin(t)]^2
four^2 = 16*([cos(t)]^ii + [sin(t)]^two)
4^2 = sixteen*([cos(t)]^ii + [sin(t)]^2)

substituting 2d eqn 10 and y values into the aeroplane:
z = 6 - 2x - 6y
=> z = half dozen - ii*[iv*cos(t) -1] - vi*[four*sin(t) - 3]
z = 6 - eight*cos(t) +two - 24*sin(t) + xviii
z = -8*cos(t) - 24*sin(t) + 26

this is what i have done then far (end me if im doing something completely wrong please ^^') but i'm a trivial lost in what the text book is telling me

therefore the parametric equations for the equation
here is where everything goes iffy (if not already somewhere above).
when i utilize the parametric equation, because the project is a circumvolve the parametric equations would be:

x = four*cos(t) - one
y = iv*sin(t) - 3
z = -8*cos(t) - 24*sin(t) + 26
[where o<= t<= 2*pi]

and then the corresponding vector equation is:

r(t) = [4*cos(t) - 1]i + [4*sin(t) - 3]j + [-eight*cos(t) - 24*sin(t) + 26]k
[where o<= t<= 2*pi]

lamentable if my explaination is unclear in some parts. im not sure if this method gets me to the answer i desire, but is in that location possibly another mode to solve this trouble? this was the merely style i could think of.
the text book didnt exactly teach the topic merely in that location was an case that was similar to this problem im working on and then i was working alot of this from that example (which was a really elementary instance and only had iv lines for working).

jpr0

I didn't cheque the details of your calculation but the logic seems fine to me.

HallsofIvy

That seems a bit more complicated than what I would do.

You have determined that 10 and y on the curve of intersection satisfy
(ten + 1)ⁱⁱ + (y + 3)^two= 16, a circle with centre at (-one, -3) (in the xy-plane) and radius 4.
Parametric equations for that circumvolve are [itex]x= -1+ 4cos(\theta), y= -three+ 4sin(\theta)[/itex]
Now, we also know that x, y, z, satisfy 2x + 6y + z = 6, the equation of the plane, then z= six- 2x- 6y or [itex]z= half dozen- ii(-1+ 4cos(\theta))- 6(-3+ four sin(\theta))= half dozen+ 2+ 18- 8cos(\theta)- 24sin(\theta)[/itex]
[itex]= 25- viii cos(\theta)- 23 sin(\theta)[/itex].

Yura

@jpr0: thank you for checking

That seems a bit more complicated than what I would do.

yeah, i figured there would have been a much easier manner to go this washed but i couldnt find one. how else could information technology be done?

Parametric equations for that circle are [itex]10= -1+ 4cos(\theta), y= -iii+ 4sin(\theta)[/itex]
At present, we likewise know that x, y, z, satisfy 2x + 6y + z = 6, the equation of the plane, so z= half-dozen- 2x- 6y or [itex]z= 6- 2(-i+ 4cos(\theta))- half dozen(-iii+ 4 sin(\theta))= 6+ 2+ 18- 8cos(\theta)- 24sin(\theta)[/itex]
[itex]= 25- viii cos(\theta)- 23 sin(\theta)[/itex].

was that last line supposed to be 26 and 24 instead of the 25 and 23 (respectively)

in that case should i take stopped at

ten = 4*cos(t) - 1
y = 4*sin(t) - three
z = -8*cos(t) - 24*sin(t) + 26

instead of subbing all the variable values into the vector equation? or is it still correct to do the subbing?

xAxis

Your algebra is skilful but you should always remember of the meaning of equations.
You first solved simultaniously the two equations by equating z-s. Which means you got an equation wrt 10 and y, which is the projection of the intersection, and you recognized by its grade to represent a circle.

Y'all have and then expressed it parametrically, and substituted in either plane or paraboloid equation to get the equation of the intersection. You then just name that curve and that'due south the solution.
Y'all also shouldn't worry about whether something is function or non in questions like these. You are looking for equations of the lines\surfaces.

Locst

Does this mean that the equation for the intersection of the plane and surface is z= 26-8*cos(t)-24*sin(t)?

I don't understand this, equally this equation does non stand for a circle. Couldn't y'all just leave it as 4^2 = [10 + 1]^ii + [y + iii]^two. Could somebody explicate why we demand to detect the parametric equations?

Also, if I wanted to use Lagrange multipliers to find points on this curve which are nearest to and furthest from the origin, how would I go nigh that? I know that if you want to find the distance D from a bespeak (in this case the origin) to a plane, y'all need to minimise D subject to the constraint of the equation of the plane, even so for the intersection above we do non take a aeroplane but a curve where the plane and surface intersect. We besides need to find a maximum.

Sorry if that makes no sense!

Evil Tomato

Shouldn't this be in Homework Aid, folks?

Not quite ( and someone please correct me if I'g incorrect - I'm doing the aforementioned class as Locst and Yura so I'm not a master! )

The equation for the z-component of the ellipse is z= 26-8*cos(t)-24*sin(t), and yous need the similar equations for ten and y earlier you can merits to have an equation (or gear up of parametric equations) for the intersection.

4^two = [x + 1]^ii + [y + 3]^2 is the projection of this ellipse into the x-y aeroplane. Information technology has no z-component, and is most definitely not the equation for the intersection.

To notice the maximum and minimum distances from the origin you need to find maximum and minimum values for d^2=x^2+y^two+z^ii bailiwick to the constraints 2x + 6y + z = 6 and z = x^two + y^2. (Commonly this would be d=(10^2+y^2+z^ii)^-1, but you can square both sides to remove the nasty foursquare root operator.)

Just set up your simultaneous equations with two LaGrange multipliers and solve them to become values for x, y and z. You end up with five equations in v unknowns. In that location is a good example of this on page 1006 of Stewart 5th ed.

Finally, since the assignment is due in in a little under ii hours, best of luck! :redface:

Russel

Yura

oh im done. i was merely getting some reassurance for my method i used. (thus why i didnt mail this into the homework forum)

HallsofIvy

Does this hateful that the equation for the intersection of the airplane and surface is z= 26-8*cos(t)-24*sin(t)?
I don't understand this, as this equation does not represent a circle. Couldn't you lot just leave it as 4^2 = [ten + 1]^ii + [y + 3]^ii. Could somebody explain why we need to detect the parametric equations?

A ane-dimensional object such as a curve (or circumvolve in item) in 3 dimensions, cannot exist written in terms of a single equation. Each equation reduces the "degrees of freedom" by one so anything like z= f(ten,y) must represent a iii-1= ii dimensional object- a surface. To represent a curve in iii dimensions you need either two equations (like representing a line as the intersection of two planes) so that you have 2 equations in three unknows- 3- ii= 1 "caste of freedom"- or, using parametric equations, to give you lot 3 equations if four unknowns (ten, y, z, and the parameter), and then again you take 4- 3= ane "degree of freedom". You could not "just leave it as 4^two = [x + i]^two + [y + 3]^2" because that tells you nothing about the z-coordinate. In three dimensions that is the equation of a cylinder, just as, in two dimensions, y= 1 is the equation of a line.

[/quote]Also, if I wanted to use Lagrange multipliers to discover points on this curve which are nearest to and furthest from the origin, how would I go about that? I know that if y'all want to find the distance D from a point (in this case the origin) to a plane, you lot need to minimise D subject to the constraint of the equation of the airplane, withal for the intersection above we do not take a plane only a curve where the plane and surface intersect. Nosotros as well demand to detect a maximum.

Sorry if that makes no sense! [/QUOTE]
Then you would minimize D subject to the bespeak existence on that bend.

HallsofIvy

Does this hateful that the equation for the intersection of the plane and surface is z= 26-8*cos(t)-24*sin(t)?
I don't empathize this, equally this equation does not represent a circle. Couldn't you only exit information technology as 4^ii = [x + 1]^2 + [y + 3]^2. Could somebody explain why we need to detect the parametric equations?

A 1-dimensional object such as a curve (or circumvolve in detail) in 3 dimensions, cannot be written in terms of a single equation. Each equation reduces the "degrees of liberty" by ane then anything similar z= f(10,y) must stand for a 3-i= ii dimensional object- a surface. To represent a curve in 3 dimensions you demand either 2 equations (like representing a line as the intersection of two planes) so that you have 2 equations in iii unknows- three- 2= 1 "degree of freedom"- or, using parametric equations, to give yous 3 equations if four unknowns (x, y, z, and the parameter), so again you have 4- 3= one "degree of freedom". Y'all could not "merely leave information technology every bit 4^2 = [ten + 1]^2 + [y + three]^2" because that tells you nothing nearly the z-coordinate. In iii dimensions that is the equation of a cylinder, but as, in 2 dimensions, y= i is the equation of a line.

[/quote]Too, if I wanted to apply Lagrange multipliers to find points on this curve which are nearest to and furthest from the origin, how would I get about that? I know that if you lot want to find the distance D from a point (in this instance the origin) to a plane, you demand to minimise D bailiwick to the constraint of the equation of the plane, however for the intersection to a higher place we do not have a plane only a curve where the plane and surface intersect. Nosotros also demand to find a maximum.

Pitiful if that makes no sense! [/QUOTE]
Then you would minimize D subject to the indicate being on that curve.

How to Draw Intersection Curve With Surface and Plane

Bend of intersection of plane and surface

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