- #1
wubie
Hello,
First I will post the question.
Now I see that my instructor is trying to progressively guide us through the steps to find the area of the surface S. I have done part a. And I think I know how to do part c and d. But I am confusing myself with part b. Which is frustrating since this should be pretty elementry.
I guess I can show what I have so far:
a)
The vector function is
r(u,v) = (u + v)i + (u - v)j + ((u^2 + v^2)/2)k
and r sub u X r sub v is
(v + u)i - (v - u)j - 2k
For part c) I think that the normal vector at point (4,2,5) is
4i + 2j - 2k where v = 1 and u = 3.
So the equation of the tangent plane at (4,2,5) would be
4(x - 4) + 2(y - 2) - 2(z - 5) = 0.
I haven't done part d) yet but I don't think it will be a problem. However I am not sure what part b is asking.
I know that the normal vector to the tangent plane is
(v + u)i - (v - u)j - 2k.
So what is part b asking then? Is it just a precusor to part c)? I know that the normal vector at point (4,2,5) is
4i - 2j - 2k.
So are the parametric equations
x=4, y=-2, z=-2?
I am assuming that normal vector and the normal line are the same. But then again then normal vector at point (4,2,5) isn't the same as the normal line at point (4,2,5) is it?
Any help is appreciated. Thankyou.
First I will post the question.
Let S be the surface given by the parametric equations
x = u + v, y = u - v, z = (u^2 + v^2)/2, u^2 + v^2 =< 16
a) Represent the suface by a vector equation function r(u,v) and find r sub u X r sub v.
b) Find the parametric equations of the normal line to the surface at (4,2,5).
c) Find an equation of the tangent plane to the suface at (4,2,5).
d) Find the area of S.
Now I see that my instructor is trying to progressively guide us through the steps to find the area of the surface S. I have done part a. And I think I know how to do part c and d. But I am confusing myself with part b. Which is frustrating since this should be pretty elementry.
I guess I can show what I have so far:
a)
The vector function is
r(u,v) = (u + v)i + (u - v)j + ((u^2 + v^2)/2)k
and r sub u X r sub v is
(v + u)i - (v - u)j - 2k
For part c) I think that the normal vector at point (4,2,5) is
4i + 2j - 2k where v = 1 and u = 3.
So the equation of the tangent plane at (4,2,5) would be
4(x - 4) + 2(y - 2) - 2(z - 5) = 0.
I haven't done part d) yet but I don't think it will be a problem. However I am not sure what part b is asking.
I know that the normal vector to the tangent plane is
(v + u)i - (v - u)j - 2k.
So what is part b asking then? Is it just a precusor to part c)? I know that the normal vector at point (4,2,5) is
4i - 2j - 2k.
So are the parametric equations
x=4, y=-2, z=-2?
I am assuming that normal vector and the normal line are the same. But then again then normal vector at point (4,2,5) isn't the same as the normal line at point (4,2,5) is it?
Any help is appreciated. Thankyou.
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