Tangent Planes, level curves, level surfaces

asked 2014-07-10 09:55:49 -0600

Tiffany
656 ●3 ●18 ●34

updated 2014-07-10 12:57:28 -0600

Mark
166 ●1 ●5 ●12 https://sites.google.com/...

I understand how to plug things into the tangent plane lines, and to get the equations, however in class he said that the answers to a and b from part two should be the same, but mine ended up different. We'll start with part a.

$$f(x,y)=x^3y^2$$ $$f_x = 3x^2y^2 \quad f_y = 2yx^3$$ $$p_0 = \langle 1,2\rangle$$ $$ f(x_0,x_0)= 1^3(2^2) = 4$$ $$f_x(x_0,y_0)= 3(1^2)(2^2) = 12$$ $$ f_y(x_0,y_0)= 2(2)(1^3) = 4$$ $$ L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ $$L(x,y) = 4 + 12(x-1) + 4(y-2)$$

for part b:

$$F(x,y,z)=z-x^3y^2$$ $$p_0=\langle1,2,4\rangle$$ $$F_x= -3x^2y^2 \quad F_y= -2yx^3 \quad F_z=1$$ $$ f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$$ $$f_x(x_0,y_0,z_0)= -3(1^2)(2^2)=-12$$ $$f_y(x_0,y_0,z_0)= -2(2)(1^3)= -4$$ So: $$-12(x-1)-4(y-2)+(z-4) =0$$

I know that if you set z=0 and then reverse the side of the equations you can end up with the same equation. Is that what he meant by it should be the same equation? Or did I screw up somewhere along the way.

Also I know he explained the level surface and level curve. From what I got, a level curve is a 2D, and a level surface is a 3D? and basically to find those curves and surfaces you make sure all of the variables are on one side of the equation and set it equal to a constant? If someone could clarify that for me it would be greatly appreciated! :)

Comment: Your answers are the same! To this totally clearly, you've just got to set $L(x,y)=z$ in your first (since that's how function values are interpreted geometrically) and then isolate the $z$ on one side in your second answer.

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answered 2014-07-10 16:57:43 -0600

Justin
1009 ●2 ●20 ●38

updated 2014-07-16 14:25:15 -0600

I am not completely sure of my answer to this, but I will take a stab at part (c) of this problem. Let us first create an outline of what we will do to solve this problem.

Find equation for tangent plane of $F(x, y, z)$.
Derive $L(x, y)$ from this equation.

Part 1

The equation for a tangent plane of a function $ F : \mathbb{R}^3 \rightarrow \mathbb{R} $ can be found by taking the gradient of the function.

$$ F(x, y, z) = z - f(x, y) $$ $$ F_x = - f_x$$ $$ F_y = - f_z$$ $$ F_z = 1$$ $$ \nabla F = \langle F_x, F_y,F_z \rangle = \langle -f_x, -f_y, 1 \rangle $$ $$ \nabla F(x_0, y_0) = \langle -f_x(x_0, y_0), -f_y(x_0, y_0), 1 \rangle $$

We now have $ \nabla F(x_0, y_0) $, which defines the normal vector to the plane we are looking for. Therefore, we can make an equation of this plane by using the components of this vector.

$$ -f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) + (z - z_0) = 0 $$

We shall now derive $L(x, y)$ from this.

Part 2

I will first let $z = L(x, y)$ and $z_0 = f(x_0, y_0)$ in our tangent plane equation that we derived in Part 1.

$$ -f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) + L(x, y) - f(x_0, y_0) = 0 $$

Now I shall solve for $L(x, y)$ by moving subtracting it and dividing by $-1$ on both sides of this equation.

$$ -f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) - f(x_0, y_0) = -L(x, y) $$ $$ \frac{-f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) - f(x_0, y_0)}{-1} = \frac{-L(x, y)}{-1} $$ $$ f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + f(x_0, y_0) = L(x, y) $$

This looks oddly familiar, but I can't place my finger on exactly what it is...! I shall now rearrange a few terms in the equation.

$$ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) $$

Tada! We have now derived $L(x, y)$ from the equation of a plane tangent to the level surface of a function $ F : \mathbb{R}^3 \rightarrow \mathbb{R} $.

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answered 2014-07-10 10:44:19 -0600

Dylan
650 ●2 ●16 ●25

updated 2014-07-10 15:36:03 -0600

I think that what we were doing works like this:

We start with some function of two variables $f(x,y)$, for example the function we did on the sheet, $f(x,y) = x^3y^2$. This function can be graphed in 3-D space by setting $z=f(x,y) = x^3y^2$. From here, the first approach we used in part A was to find the equation for a plane tangent to a point in the function by linearizing the function at that point.

In the second part, we were saying that because $z = f(z,y)$, then $z - f(z,y) = 0$, or in the sheet, that $z - x^3y^2 = 0$. Then we can just call the left hand side of the equation a function of three variables, $x$, $y$, and $z$, and then the equation $z - x^3y^2 = 0$ is just a level surface of that three-variable function. So I think the idea is that we can define a two-variable function as the level surface of a three-variable function, or go the other way and define the level surface of a three-variable function as a two-variable function (assuming we can solve for one of the variables).

In the sheet, the linearization of the function $f(x,y) = x^3y^2$,

$z=L(x,y) = 4 + 12(x- 1) + 4(y-2)$

was really the same thing as the plane tangent to the point $(1,2,4)$ on the function $F(x,y,z) = z -x^3y^2$, and we can algebraically move between them.

I'm pretty sure this is what was going on, but I'm still not sure how to express this as an answer to part C.

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answered 2014-07-10 10:22:46 -0600

SpaceManSpiff
703 ●4 ●22 ●40

updated 2014-07-10 10:29:35 -0600

Yes I'm pretty sure you did this correctly but it's not really a question of setting $z$ equal to zero.

$f(x,y)=x^{3}y^{2}$ is a function whose graph lives in three dimensions. This should make sense because it defines a third variable, $z$, in terms of the other two, $x,y$.

So in other words it's like saying $z=x^{3}y^{2}$. And you bring $x^{3}y^{2}$ over to the other side to create $F(x,y,z)$.

You're just "undoing" this at the end of part b to make the equations from a and b equal to each other by solving for $z$ at the end of part (b). Not setting it equal to zero.

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Tangent Planes, level curves, level surfaces

3 Answers

Part 1

Part 2

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