Changing center point of a cylinder

AI Thread Summary
To express a parametric formula for a right circular cylinder centered at (-2, 10, 3) with a radius of 3 and a length of 12, the key is to apply a coordinate translation to the standard parametric equations. The equations can be adjusted by adding the center point coordinates to the standard cylinder equations, resulting in x = -2 + 3cos(θ), y = 10 + 3sin(θ), and z = s, where 0 < θ < 2π and 0 < s < 12. For a tilted cylinder, the same translation applies, allowing for the movement of the cylinder's center without altering its shape. This approach simplifies the process of graphing the cylinder in Maple.
mpittma1
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Homework Statement


How could I express a parametric formula for a right circular cylinder centered at (-2, 10, 3)?
with radius 3 and length 12

Homework Equations


Parametric equations for a right circular cylinder are:

x=rcosΘ
y=rsinθ
z = h

The Attempt at a Solution



Not sure how to start this...
 
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Well, if you have an equation y=f(x), you will find that equation y=f(x-a) is just f(x) shifted to the right by a units.

It is a coordinate translation.
Your problem is the same thing.
 
Simon Bridge said:
Well, if you have an equation y=f(x), you will find that equation y=f(x-a) is just f(x) shifted to the right by a units.

It is a coordinate translation.
Your problem is the same thing.

Thats what i thought, the main problem is that this is an assignment using maple, where i have to graph the right circular cylinder with its center at that point.

when ever i try doing that it doesn't move the center to that point...
 
Please show the adjusted equation followed by the maple code/instructions you are using to draw it.
 
Simon Bridge said:
Please show the adjusted equation followed by the maple code/instructions you are using to draw it.

i posted the same question with the attached maple file here:
 
Last edited:
The link provided does not answer the questions an a way that let's me help you.
Please do not require people to download MB sized files in order to help you.
 
Wasn't requiring anything, was trying to make it easier for you to just download the whole maple file.

unfortunately my colleges server just went down so I can't use maple for the rest of the night...

What i was doing before it crashed though, was I found the null space of the plane 2x-10y-3z=0

and then using that as a basis i made it into an orthonormal basis by using the Gram Schmidt command in maple, then found the third basis that is perpendicular to both w1 and w2 by taking the cross product of w1 and w2.

then using the parametric equation for the cylinder with radius 3 and length 12

{x = rcostheta
C1= {y = rsintheta
{z = s 0<theat<2pi & 0<s<12

then defined a transition matrix P = <w1 w2 w3>

then found the parametric equations for the cylinder by using

C2 = P.C1

then from that i graphed the image and began messing around with placing the center (-2, 10, 3) into the original C1 parametric equation as you suggested in your first response.

It was looking like i was getting close until the stupid program froze up on me..
 
That sounds way too complicated-a method to possibly get the right equation.
Even if it is supposed to work, you are too likely to misplace something.
Why not just apply the transformation directly to the coordinates?

i.e.
for the origin centered cylinder, the x values will vary between x=-r and x=r - which centers the x values at x=0

for the cylinder you want, what will the values of x vary between in order to get the same radius, but centered on x=-2? How does that suggest you should modify the equation?
 
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If you let ##cyl:=[3\cos\theta,3\sin\theta,z]:## and ##v:=[-2,10,3]##, all you have to do is a plot3d of cyl+v.
 
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  • #10
I see what your saying, but its a tilted cylinder that previous to this question i had to use transition matrices to find its parametric equation.

would your suggestion work for such a "titlted" cylinder?
 
  • #11
mpittma1 said:
I see what your saying, but its a tilted cylinder that previous to this question i had to use transition matrices to find its parametric equation.

would your suggestion work for such a "titlted" cylinder?

If you are responding to me you should quote me so I know that. If you can plot your tilted cylinder at the origin, adding v to it will move it wherever you want it.
 
  • #12
LCKurtz said:
If you are responding to me you should quote me so I know that. If you can plot your tilted cylinder at the origin, adding v to it will move it wherever you want it.

Ok, so what you saying is i take system of parametric equations (C2) that make the titled cylinder and simply add the new center point, v, to it?
 
  • #13
mpittma1 said:
Ok, so what you saying is i take system of parametric equations (C2) that make the titled cylinder and simply add the new center point, v, to it?

I haven't checked anything about your tilted cylinder. What I am saying is if you have figured out how to plot it at the origin then, yes, just add v to it to move it. It's just a translation.
 
  • #14
LCKurtz said:
I haven't checked anything about your tilted cylinder. What I am saying is if you have figured out how to plot it at the origin then, yes, just add v to it to move it. It's just a translation.

i understand, don't know why i didnt think of that lol

thanks a lot!
 
  • #15
The cylinder is being rotated as well as translated?!
That information was not in the problem statement.

Thanks to mpittma1 - I was trying not to be so direct :)
 
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