Find Plane Tangent to Surface at (-1/4, 1/2, 2)

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Homework Help Overview

The problem involves finding an equation for the plane tangent to a given surface defined by parametric equations at a specific point in three-dimensional space. The surface is described by the equations x = u^2 - v^2, y = u + v, and z = u^2 + 4v, with the point of interest being (-1/4, 1/2, 2).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to identify the parameters u and v that correspond to the given point on the surface. There is confusion regarding the relationship between the parameters and the tangent plane equation. Some participants explore the derivatives of the function to find the tangent plane.

Discussion Status

Some participants have provided guidance on how to approach finding the tangent plane by suggesting the use of partial derivatives. Others have clarified the notation used for the parameters, indicating that u' and v' are not derivatives but rather specific values corresponding to the point of interest. The discussion reflects a mix of attempts to clarify the problem and explore different interpretations of the equations involved.

Contextual Notes

Participants express frustration over the lack of examples in their textbook that directly address the problem at hand. There is also mention of uncertainty regarding the correct values of u and v needed to proceed with the solution.

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Homework Statement


Find an eqn for the plane tangent to the given surface at the specified point.

x = u^2 - v^2
y= u +v
z = u^2 + 4v

At (-1/4, 1/2, 2)

Homework Equations


A(x-x_0) + B(y-y_0) + C(z-z_0)

The Attempt at a Solution


I thought that I could simply use x = u^2 - v^2 as A, and y = u + v as B, etc. But yeah, it's wrong obviously. I looked through my notes already and my professor didn't go over this so I'm entirely lost. How do I find the plane tangent without knowing what u and v are? Pls help, having a headache over this. The examples in the book didn't help at all either...it didn't have any examples on this.:confused:
 
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You have a function f : RxR -> RxRxR defined by:

f(u,v) = (u^2-v^2, u^v, u^2+4v)

The surface is thus

{f(u,v) : (u,v) is in RxR}

You want to find the tangent plane to this surface at the point (-1/4, 1/2, 2). First, find (u', v') such that f(u', v') = (-1/4, 1/2, 2). Then the plane is given by:

[tex](-0.25,\, 0.5,\, 2)\ +\ \mbox{Span}\left \{\frac{\partial f}{\partial u} (u',\, v'),\ \frac{\partial f}{\partial v}(u',\, v')\right \}[/tex]
 
How do you find f(u',v')? Is that just taking the derivative of all the components?

Edit: Eh nvm. >< I'm not thinking clearly. Thanks for the help, I got it now.
 
Last edited:
u' and v' aren't derivatives. It might have been better if I wrote u0 and v0, it was just easier to do it the way I did. What final answer did you get? You should have got (u',v') = (0,1/2). If you got that you probably got the rest right too.
 
Last edited by a moderator:
Yeah, that's what I thought. So I took u + v = 1/2 and solve for v in terms of u, then plugging in v for the u^2 - v^2 = -1/4.

My final answer is z = 3x + 4y + 3/4?
 
[itex]f_u(u',v') = (2u', 1, 2u') = (0, 1, 0)[/itex]
[itex]f_v(u',v') = (-2v', 1, 4) = (-1, 1, 4)[/itex]

So P = (-0.25, 0.5, 2) + {(-t, t+s, 4t) : t, s in R}.

Can you get for any x and y, a z such that (x, y, z) is in P?

-0.25 - t = x
0.5 + t + s = y

t = -(0.25 + x)
s = y - t - 0.5 = y + x - 0.25

Let z = 2 + 4t = 1 - 4x.

So P = {(x,y,z) : z = 1 - 4x, x,y,z in R} = {(x,y,1-4x) : x,y in R}.
 

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