Exploring the Paraboloid: A Parametric Surface Investigation

[BoundByAxioms]

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$$\cdot$$(r-r$$_{0}$$)=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. <$$\sqrt{2}$$sinu, $$\sqrt{2}$$cosu, 2>

ii. <v$$\frac{\sqrt{2}}{2}$$, v$$\frac{\sqrt{2}}{2}$$, v$$^{2}$$>

c. <$$\sqrt{2}$$,0,2> and <$$\frac{\sqrt{2}}{2}$$, $$\frac{\sqrt{2}}{2}$$, 2>

d. I used the definition of the dot product, and dotted i and ii, using $$\frac{\pi}{4}$$ as u, and $$\sqrt{2}$$ as v. I got that $$\theta$$=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.

 

[tiny-tim]

c. <$$\sqrt{2}$$,0,2> and <$$\frac{\sqrt{2}}{2}$$, $$\frac{\sqrt{2}}{2}$$, 2>

Hi BoundByAxioms!

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

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.

Try again!

 

[BoundByAxioms]

Hi BoundByAxioms!

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

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.

Try again!

Hmm, then I fear that I am lost. Could you give me a hint? Thanks.

 

[HallsofIvy]

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 $u= \pi/4$ and $v= \sqrt{2}$.

 

[BoundByAxioms]

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 $u= \pi/4$ and $v= \sqrt{2}$.

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 $$\sqrt{2}$$ for v and $$\pi/4$$ for u. So I get, <-1, 1, 0> and <$$\sqrt{2}/2$$, $$\sqrt{2}/2$$, 2$$\sqrt{2}$$>. Is this correct so far?

 

[tiny-tim]

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 $$\sqrt{2}$$ for v and $$\pi/4$$ for u. So I get, <-1, 1, 0> and <$$\sqrt{2}/2$$, $$\sqrt{2}/2$$, 2$$\sqrt{2}$$>. Is this correct so far?

So far … so good!

(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. )

 

[LeonJHardman]

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?

 

[tiny-tim]

Welcome to PF!

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!

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.

 

[LeonJHardman]

Oh thx. So the answer should be 90 unless i did something wrong.

 

[BoundByAxioms]

Oh thx. So the answer should be 90 unless i did something wrong.

I think so:

<-1,1,0> $$\bullet$$ <$$\frac{\sqrt{2}}{2}$$, $$\frac{\sqrt{2}}{2}$$, 0> = 0. And since the dot product is zero, the angle between them is $$\frac{\pi}{2}$$. Remember to always answer questions in radians.

 

[LeonJHardman]

Thank you very much. I'll remember that.