I have a few questions about the definitions what it is exactly saying. So basically for the cone surface: A cone surface M is determined through a curve R and a point t, the top. The definition is:
#Cylindrical surfaces and cone surfaces
wheat wagon
eternal boltBOT
eternal boltBOT
modern sonnet
Yeah, so like the circles last time, gluing those lines together gives the cone surface (same for cylinder surface)
p is on the surface if and only if there is a point on the curve that crosses the line p and t are on
The diagrams from my textbook shows the more 3D nature of these
wheat wagon
Yeah, so like the circles last time, gluing those lines together gives the cone ...
Ah so here instead of circles, you are “shooting “ lines which ends up looking like a cone ( as in the figure you send) but the important part is that you can find M if there is a point on the curve where the line goes through that point AND the top
modern sonnet
Ah so here instead of circles, you are “shooting “ lines which ends up looking l...
yes
you can image it as vectors pointing to the curve based at t. For each point on the curve you get a vector, hence you get a line
crucially, this recovers what you first learn as a cone when the curve is a circle and t isn't coplanar with the circle
wheat wagon
Yeah makes sense now, thanks! So basically if you actually want to find the equation then you know the direction vector ( if the coordinates of the top are given). And you also know the equation of the curve. Hmm I see
And so for cylindrical surfaces:
the cylindrical surface M is completely determine by a curve R and a line A.
p is on the surface M if and only if there is a point p0 on the curve R where p0p is parallel to the direction vector of A
wheat wagon
The diagrams from my textbook shows the more 3D nature of these
As in fig 4.18
modern sonnet
As in fig 4.18
yes, 4.18 is the cylindrical surface given by q and curve beta
the cone (using symbols from my text) has parameterization $x(u,v):=p+v\delta(u)$, which is just the line connecting $p$ and $\delta(u)$
the cylinder is $x(u,v)=\beta(u)+vq$, again lines in the direction of $q$ passing through $\beta(u)$
The parameterizations mirror the fact it's a bunch of lines
carmine badgerBOT
Omega
wheat wagon
Yeah I see it. So essentially:
- Cylindrical surfaces can be determined if you have given a line and a curve. Basically find a line that is parallel to that one
- Cone surfaces can be also determined if there is a point o on the curve where the line crosses the point p_0 and the top. Essentially you’re creating that surface by shooting those vectors ( lines)
Makes sense now
The definitions probably made it a bit harder for me to understand, but yea
Also I have one other ( probably a simple question) I just wanna ask to be sure
modern sonnet
Yeah I see it. So essentially:
- Cylindrical surfaces can be determined if you h...
Yes, crucially the cylinder lines are parallel (per the parameterization)
modern sonnet
Also I have one other ( probably a simple question) I just wanna ask to be sure
Sure, though might not be able to answer until a few hours/tmr
wheat wagon
Sure, though might not be able to answer until a few hours/tmr
Yeah, no problem. Thank you so much.
I just wanna ask to be sure, sometimes when you’re determining the partial derivatives in what form should it be, for example:
z = sin(x+y^2-4y+4)
If you just take the partial derivatives w.r.t to x and y you get something meaningful, however if you take the partial derivative to z, you end up getting 1= 0. So I am assuming I am doing something wrong or I need to generalize and say F(x,y) = f(x,y)-z and take the partial derivatives of F(x,y)
modern sonnet
Yeah, no problem. Thank you so much.
I just wanna ask to be sure, sometimes wh...
You only take derivatives of the domain variables
wheat wagon
You only take derivatives of the domain variables
What do you mean exactly?
modern sonnet
The domain is the xy plane/subset thereof
Same way you don't differentiate a univariate y=f(x) wrt y
wheat wagon
Same way you don't differentiate a univariate y=f(x) wrt y
Ah yes, makes sense
So you essentially will do
F(x,y) = f(x,y) -z and take from that the partial derivatives
Because then you take the partial derivatives of all domain variables
modern sonnet
So you essentially will do
F(x,y) = f(x,y) -z and take from that the partial der...
z is the output variable.
F there is just identitcally 0
wheat wagon
F there is just identitcally 0
but the partial derivative of F w.r.t to z would be -1? instead of -1 = 0
modern sonnet
z isnt a domain variable.
its the output
if you had w=f(x,y,z), then you cant differentiate wrt w.
wheat wagon
just do the computation
so I would say F(x,y,z) = f(x,y) -z and then differentiate w.r.t to x, y and z
modern sonnet
if you have a function z=f(x,y)
only dz/dx and dz/dy make sense.
if you have a function w=f(x,y,z)
only dw/dx, dw/dy, dw/dz make sense.
@wheat wagon
wheat wagon
only dz/dx and dz/dy make sense.
Hmm, but I had to find the tangent plane of that function
Oh
Wait nvm
modern sonnet
how did you find the tangent line in the univariate case?
wheat wagon
Yeah it’s z = f(x,y) + …
modern sonnet
yes
wheat wagon
yeah
modern sonnet
$z-z_0=m_x(x-x_0)+m_y(y-y_0)$
carmine badgerBOT
Omega
wheat wagon
Yeah I was thinking about something else my bad
modern sonnet
same structure as univariate
wheat wagon
yep
wheat wagon
So if you have something of the form: f(x,y,z) = 0 and you cannot isolate z, then you can take the partial derivatives of all of them
modern sonnet
So if you have something of the form: f(x,y,z) = 0 and you cannot isolate z, the...
yes, it's an implicit function
wheat wagon
yeah makes sense, it’s the same as calc I
modern sonnet
yes
wheat wagon
Also I had one more question about line of intersection:
K is the line of intersection of M_1 : z= 16-3x^2-y^2 and M2 : z= (x-4)^2 + 3y^2. But my question is that if you set them equal to each other, you will get the line of intersection. But why is that?
modern sonnet
Also I had one more question about line of intersection:
K is the line of inters...
again, same as in 2D
wheat wagon
hm
wheat wagon
again, same as in 2D
Ah I already see what you mean, since it’s an intersection you just wanna look where they are equal to each other and plug the z into the other equation.
wheat wagon
Yeah, makes sense. Sorry for all of these questions just wanna make sure I understood everything.
modern sonnet
nw
wheat wagon
I also find the curve K which is (x-1)^2+y^2 = 1 ( an ellipse)
I can now find the tangent line with finding the derivative by doing implicit differentiation or finding a parametric equation and then find x’(t),…
p(2,0,4)
If you do implicit differentiation you end up with:
2(x-1)+2yy’=0 Which is not true ( 3 = 0)
But I am unsure on how to find the parametric equation, I could say x-1 = cos^2(t) and y = sin^2(t) but idk about z
Also: The solution just tried to find the tangent planes for both surfaces, and the direction vector of the tangent is perpendicular to both, so you can find that by cross product. Is that because the tangent line is in both the tangent planes which means that the normal vector of both tangent planes is perpendicular to it?
modern sonnet
I also find the curve K which is (x-1)^2+y^2 = 1 ( an ellipse)
I can now find t...
an ellipse is a planar curve, so you can just do it for any z value.
at this point, please make a new post if you have new questions
wheat wagon
at this point, please make a new post if you have new questions
Yeah, sorry it was my last one
I mean yeah I know but doing implicit differentiation won’t work
modern sonnet
idk what you mean then
the z values are given by the initial curves, since the intersection is a subset of the surfaces
so $x-1=\cos(t)$, $y=\sin(t)$, $z=16-3(\cos(t)+1)^2-\sin^2(t)$ using $M_1$
carmine badgerBOT
Omega
wheat wagon
I also find the curve K which is (x-1)^2+y^2 = 1 ( an ellipse)
I can now find t...
Yeah, so if u want to know the the slope at that point, you wanna do implicit differentiation but you won’t find anything. that’s what I meant
And also how did you find that z=…?
modern sonnet
Yeah, so if u want to know the the slope at that point, you wanna do implicit di...
you will find something,
projecting the intersection into R^2, you;re just finding dy/dx of the circle (x-1)^2+y^2=1
wheat wagon
yes and you want to find it at the point p (2,0,4) if you do implicit differentation you get:
2(x-1)(1) + 2yy’ = 0
y in this case is 0?
modern sonnet
Make a separate post for this
eternal boltBOT
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