Parametrization

[mesty]

I have a question and I dont undersand what its means. The question is as follows:


Consider a surface S, given by the equation

with z>=0.

If nˆ is a normal vector on S with a positive z component, determine a
parameterisation for the boundary curve C for S, such that C is traversed
in a righthand sense relative to nˆ, using t as your parameter.


Could somebody explain what this question is asking me to do?

Thanks

 

[DrPhil]

I have a question and I dont undersand what its means. The question is as follows:


Consider a surface S, given by the equation

with z>=0.

If nˆ is a normal vector on S with a positive z component, determine a
parameterisation for the boundary curve C for S, such that C is traversed
in a righthand sense relative to nˆ, using t as your parameter.


Could somebody explain what this question is asking me to do?

Thanks

S is a surface defined only for \(\displaystyle z \ge 0\), so it has a bounding curve (called "C") in the x-y plane.

\(\displaystyle z = 0 \implies \sqrt{x^2 + y^2} = 2\)

To parametrize using t as the parameter means to find equations for \(\displaystyle x(t)\) and \(\displaystyle y(t)\) that will traverse C in the direction that is righthanded about a vector in the +z diredtion.

Hope that makes it clearer!

 

[mesty]

Thanks DrPhil that makes things much clearer.

Is x=4cos(t),y=4sin(t) and z=0 a correct parametrization?


Thanks.

 

[HallsofIvy]

Thanks DrPhil that makes things much clearer.

Is x=4cos(t),y=4sin(t) and z=0 a correct parametrization?


Thanks.

No, but almost. \(\displaystyle \sqrt{x^2+ y^2}= \sqrt{(4cos(t))^2+ (4sin(t))^2}= \sqrt{16(cos^2(t)+ cos^2(t))}= 4\), not 2.

 

[mesty]

Ok i see now. got it thanks.