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

In summary, to find the equation for the plane tangent to the given surface at the specified point, you first need to find the values of u' and v' such that f(u', v') = (-1/4, 1/2, 2). Then, the plane is given by the equation (-0.25, 0.5, 2) + {(-t, t+s, 4t) : t, s in R}. Finally, by substituting values for x and y, you can obtain the equation for the tangent plane in the form of z = 1 - 4x.
  • #1
Peach
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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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  • #2
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]
 
  • #3
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.
 
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  • #4
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.
 
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  • #5
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?
 
  • #6
[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}.
 

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

What does it mean to find a plane tangent to a surface at a specific point?

When we talk about finding a plane tangent to a surface at a certain point, we are referring to finding a flat surface that touches the given surface at that point but does not intersect or cross through it. This plane is parallel to the surface at that point and shares the same slope or direction as the surface.

Why is finding the plane tangent to a surface important?

Finding the plane tangent to a surface is important because it allows us to approximate the behavior of the surface at a specific point. This is useful in many fields, such as engineering, physics, and mathematics, where we need to understand the local behavior of a surface.

What information do we need to find the plane tangent to a surface?

In order to find the plane tangent to a surface at a given point, we need to know the coordinates of that point and the equation or function that represents the surface. This allows us to calculate the slope or gradient of the surface at that point, which is necessary for determining the equation of the tangent plane.

Can we find the plane tangent to any type of surface?

Yes, we can find the plane tangent to any type of surface as long as we have the necessary information, such as the coordinates of the point and the equation of the surface. However, the process of finding the tangent plane may differ depending on the complexity of the surface and the available methods for calculating the slope or gradient.

How can we use the plane tangent to a surface in real-world applications?

The plane tangent to a surface has various real-world applications. For example, in engineering, it can be used to approximate the behavior of a curved surface and design structures that can withstand different forces. In physics, it can help us understand the motion of particles or objects on a surface. In mathematics, it can be used to prove theorems and solve problems related to surfaces and their properties.

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