# Multivariable calculus cylinder equation

• August 27th 2009, 12:58 PM
rida426
Multivariable calculus cylinder equation
It is as I know the defination of cylinder..

Q:Find the equation of cylinder if the curve is f(x , y) =0 and
n=<a,b,c> and c not equal to zero.

Thanks !!
• August 28th 2009, 07:41 AM
Opalg
Quote:

Originally Posted by rida426
It is as I know the defination of cylinder..

Q:Find the equation of cylinder if the curve is f(x , y) =0 and
n=<a,b,c> and c not equal to zero.

If I understand it right, the equation f(x,y)=0 represents the cross-section of the cylinder by the plane z=0. Think of that as being some sort of curve (maybe a circle) centred at the origin. Then the cross-section of the cylinder for other values of z ought to look like the same curve, but centred at a point on the line in the direction of the vector (a,b,c).

The point on this line having its third coordinate equal to z is $\left(\tfrac acz,\tfrac bcz,z\right)$. That makes me think that the equation of the cylinder should be $f\!\left(x-\tfrac acz, y-\tfrac bcz\right)=0$. But I don't know how to give a convincing explanation of that.
• September 3rd 2009, 03:21 PM
rida426
how the point on the line that is parallel to z-axis is (a/cz,b/cz,z) ??

check me if at any point I am wrong:

n=<a,b,c>

line equation: r=r+tv;

r=<x0+y0+z0> + t<a,b,c>

z=z0, (not a fixed value i.e whole z-axis upward)

line: x=x0+at , y=y0+bt, z=z0+ct.

=> x0=x-at, y0=y-bt, z0=z-ct ;

t=(z-z0)/c put it in x0 and y0 equations

x0=x- a(z-z0)/c, y0=y- b(z-z0)/c

???

I am not going to the right direction I think .......