Parametrization

mesty

New member
Joined
Sep 6, 2013
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14
I have a question and I dont undersand what its means. The question is as follows:


Consider a surface S, given by the equation
gif.latex
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
 
I have a question and I dont undersand what its means. The question is as follows:


Consider a surface S, given by the equation
gif.latex
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 z0\displaystyle z \ge 0, so it has a bounding curve (called "C") in the x-y plane.

z=0    x2+y2=2\displaystyle z = 0 \implies \sqrt{x^2 + y^2} = 2

To parametrize using t as the parameter means to find equations for x(t)\displaystyle x(t) and y(t)\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!
 
Thanks DrPhil that makes things much clearer.

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


Thanks.
 
Thanks DrPhil that makes things much clearer.

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


Thanks.
No, but almost. x2+y2=(4cos(t))2+(4sin(t))2=16(cos2(t)+cos2(t))=4\displaystyle \sqrt{x^2+ y^2}= \sqrt{(4cos(t))^2+ (4sin(t))^2}= \sqrt{16(cos^2(t)+ cos^2(t))}= 4, not 2.
 
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