Linear approximation for two variables: Difference between revisions

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==The normal line==
==The normal line==


In analytical geometry we can prove that <math>ax + by + cx = 0</math>, the vector <math>(a,b,c)</math> is perpendicular to the plane. We can do the same for the tangent plane and obtain the same vector from the tangent plane's equation.
In analytical geometry we can prove that <math>ax + by + cx = 0</math>, the vector <math>(a,b,c)</math> is perpendicular to the plane. We can do the same for the tangent plane and obtain the same vector from the tangent plane's equation. If we look at the previous equation, we obtain:


<math>
<div style="text-align:center;">
<math>z = \frac{\partial f}{\partial x}(x_0,y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y - y_0) - f(x_0,y_0)</math>
</div>
 
'''Normal vector:''' <math>(\frac{\partial f}{\partial x}(x_0,y_0), \frac{\partial f}{\partial y}(x_0,y_0), -1)</math>

Revision as of 02:10, 8 May 2022

To approximate a function of two variables with a tangent plane is the natural extension of approximating a function of one variable with a tangent line. In the same way that zooming in a function of one variable makes it render closer to a straight line, with a tangent plane we see that the level curves become closer to straight parallel lines if we zoom in enough.

(not to scale)

As with single variable functions, a multivariable function has to be continuous and differentiable for us to use the tangent plane approximation. In case it's continuous but not differentiable a plane exists, but it's not the same as the tangent plane because if the function is not differentiable there can't be a tangent plane. For example: [math]\displaystyle{ f(x,y) = \sqrt{x^2 + y^2} }[/math]. A plane at [math]\displaystyle{ (0,0) }[/math] is not going to be horizontal. It's going to be angled in one direction or another.

The tangent plane

In analytical geometry a plane is defined with [math]\displaystyle{ Z = O + t_1\overrightarrow{v_1} + t_2\overrightarrow{v_2} }[/math]. In the vector form each point of it is given by a point of origin, two parameters and two linearly independent vectors. In the general form we have an equation that should have been seen in school at some point [math]\displaystyle{ Ax + By + Cz + d = 0 }[/math].

Assuming the function to be differentiable at a point, we have:

[math]\displaystyle{ f(x_0,y_0) }[/math] the point of origin.

[math]\displaystyle{ (x - x_0) }[/math] and [math]\displaystyle{ (y - y_0) }[/math] two pairs of points, belonging to the function's domain, that give the direction in [math]\displaystyle{ x }[/math] and in [math]\displaystyle{ y }[/math].

[math]\displaystyle{ \frac{\partial f}{\partial x}(x_0,y_0) }[/math] and [math]\displaystyle{ \frac{\partial f}{\partial y}(x_0,y_0) }[/math] the variation in each direction, which corresponds to [math]\displaystyle{ t_1 }[/math] and [math]\displaystyle{ t_2 }[/math] in the vector form.

Therefore, the equation of the tangent plane is:

[math]\displaystyle{ z - f(x_0,y_0) = \frac{\partial f}{\partial x}(x_0,y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y - y_0) }[/math]

With this equation we find all points of a tangent plane. We disregard the [math]\displaystyle{ d }[/math] because this plane is not any plane, it's tied to a function of two variables.

The normal line

In analytical geometry we can prove that [math]\displaystyle{ ax + by + cx = 0 }[/math], the vector [math]\displaystyle{ (a,b,c) }[/math] is perpendicular to the plane. We can do the same for the tangent plane and obtain the same vector from the tangent plane's equation. If we look at the previous equation, we obtain:

[math]\displaystyle{ z = \frac{\partial f}{\partial x}(x_0,y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y - y_0) - f(x_0,y_0) }[/math]

Normal vector: [math]\displaystyle{ (\frac{\partial f}{\partial x}(x_0,y_0), \frac{\partial f}{\partial y}(x_0,y_0), -1) }[/math]