Exploring the Paraboloid: A Parametric Surface Investigation

Thanks for the advice.In summary, the parametric surface r(u,v)=<vsinu, vcosu, v^2> can be identified as a paraboloid. The point (1,1,2) lies on the surface, and the grid curves with constant h and u that contain this point are <√2sinu, √2cosu, 2> and <v√2/2, v√2/2, v^2>, respectively. The tangent vector to both of these grid curves at the point (1,1,2) is <-1, 1, 0> and <√2/2, √2/2, 2
  • #1
BoundByAxioms
99
0

Homework Statement


Consider the parametric surface r(u,v)=<vsinu, vcosu, v^2>
a) Identify the shape of the surface

b) The point (1,1,2) is on the surface. Find:

i) A grid curve wit hv constant that contains this point

ii) A grid curve with u constant that contains this point

c) Find the tangent vector to both grid curves you just found at the point (1,1,2)

d) Find the angle between the grid curves at the point (1,1,2).

e) Find a parametric equation for the plane containing the wo tangent vectors from part c, and containing the point (1,1,2).

f) Describe the relationship between the plane and the surface.




Homework Equations



Dot product. And n[tex]\cdot[/tex](r-r[tex]_{0}[/tex])=0.


The Attempt at a Solution



For a-c I am pretty confident that my answers are right, but d-f is where I need some help.

a. Paraboloid

b.
i. <[tex]\sqrt{2}[/tex]sinu, [tex]\sqrt{2}[/tex]cosu, 2>

ii. <v[tex]\frac{\sqrt{2}}{2}[/tex], v[tex]\frac{\sqrt{2}}{2}[/tex], v[tex]^{2}[/tex]>

c. <[tex]\sqrt{2}[/tex],0,2> and <[tex]\frac{\sqrt{2}}{2}[/tex], [tex]\frac{\sqrt{2}}{2}[/tex], 2>

d. I used the definition of the dot product, and dotted i and ii, using [tex]\frac{\pi}{4}[/tex] as u, and [tex]\sqrt{2}[/tex] as v. I got that [tex]\theta[/tex]=0, which doesn't seem right to me.

e. I'm not sure on this one either. I could perhaps the second equation that I posted, but I don't know what I'd use for n.

f. Since I'm not sure on e, I'm not sure of f. As soon as I know e though, I'm sure I can do f.
 
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  • #2
BoundByAxioms said:
c. <[tex]\sqrt{2}[/tex],0,2> and <[tex]\frac{\sqrt{2}}{2}[/tex], [tex]\frac{\sqrt{2}}{2}[/tex], 2>

Hi BoundByAxioms! :smile:

(have a square-root: √ and a squared: ² and a pi: π :smile:)

Both your tangents are wrong.

The second one, I think you've just made a error of arithmetic.

But I don't see how you got the first one at all. :confused:

Try again! :smile:
 
  • #3
tiny-tim said:
Hi BoundByAxioms! :smile:

(have a square-root: √ and a squared: ² and a pi: π :smile:)

Both your tangents are wrong.

The second one, I think you've just made a error of arithmetic.

But I don't see how you got the first one at all. :confused:

Try again! :smile:

Hmm, then I fear that I am lost. Could you give me a hint? Thanks.
 
  • #4
To find the tangent vector in the "u" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to u. To find the tangent vector in the "v" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to v. To find them at the given point, substitute [itex]u= \pi/4[/itex] and [itex]v= \sqrt{2}[/itex].
 
  • #5
HallsofIvy said:
To find the tangent vector in the "u" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to u. To find the tangent vector in the "v" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to v. To find them at the given point, substitute [itex]u= \pi/4[/itex] and [itex]v= \sqrt{2}[/itex].

So, when r(u,v)=<vcos(u), vsin(u), v2>, the derivative with respect to u is <-vsin(u), vcos(u), 0>. And the derivative with respect to v is <cos(u), sin(u), 2v>. Now, I plug in [tex]\sqrt{2}[/tex] for v and [tex]\pi/4[/tex] for u. So I get, <-1, 1, 0> and <[tex]\sqrt{2}/2[/tex], [tex]\sqrt{2}/2[/tex], 2[tex]\sqrt{2}[/tex]>. Is this correct so far?
 
  • #6
BoundByAxioms said:
So, when r(u,v)=<vcos(u), vsin(u), v2>, the derivative with respect to u is <-vsin(u), vcos(u), 0>. And the derivative with respect to v is <cos(u), sin(u), 2v>. Now, I plug in [tex]\sqrt{2}[/tex] for v and [tex]\pi/4[/tex] for u. So I get, <-1, 1, 0> and <[tex]\sqrt{2}/2[/tex], [tex]\sqrt{2}/2[/tex], 2[tex]\sqrt{2}[/tex]>. Is this correct so far?

So far … so good! :smile:

(except … I don't know whether they expect you to give the unit vector … I note that the question asks for "the" rather than "a" tangent vector. :confused:)
 
  • #7
How would you find d? Do you uese the tangent lines or do you find the angle of the grid curve with the points plugged in or the tangent line?
 
  • #8
Welcome to PF!

LeonJHardman said:
How would you find d? Do you uese the tangent lines or do you find the angle of the grid curve with the points plugged in or the tangent line?

Hi Leon! Welcome to PF! :smile:

As BoundByAxioms said, you use the definition of the dot product (so you can find cosθ).

And yes, you use the tangent lines … the angle between two curves is the angle between their tangent lines. :smile:
 
  • #9
Oh thx. So the answer should be 90 unless i did something wrong.
 
  • #10
LeonJHardman said:
Oh thx. So the answer should be 90 unless i did something wrong.

I think so:

<-1,1,0> [tex]\bullet[/tex] <[tex]\frac{\sqrt{2}}{2}[/tex], [tex]\frac{\sqrt{2}}{2}[/tex], 0> = 0. And since the dot product is zero, the angle between them is [tex]\frac{\pi}{2}[/tex]. Remember to always answer questions in radians.
 
  • #11
Thank you very much. I'll remember that.
 

1. What is a paraboloid?

A paraboloid is a three-dimensional surface that is created by rotating a parabola around its axis.

2. What is a parametric surface?

A parametric surface is a surface that is defined by a set of equations that use parameters to describe its points.

3. What are the applications of exploring paraboloids?

Exploring paraboloids can have various applications in fields such as mathematics, engineering, and architecture. They can be used to design and analyze structures, optimize shapes for efficiency, and create visually appealing designs.

4. How do you investigate a paraboloid using parametric equations?

To investigate a paraboloid using parametric equations, one must first define the parameters and equations that describe the surface. Then, different values can be plugged in for the parameters to generate points on the surface and explore its properties.

5. What are some interesting properties of paraboloids?

Paraboloids have many interesting properties, such as being a doubly-ruled surface, meaning it can be created by straight lines that intersect at two points. It also has a constant curvature along any given direction, making it a useful shape for many applications.

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