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 | 1 | initial version | posted 2014-07-10 16:57:43 -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 an equation of three variables 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 the 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} $.
| | 2 | No.2 Revision |
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 an equation of three variables 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 the 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 \mathbb{R}^3 \rightarrow \mathbb{R} $.
| | 3 | No.3 Revision |
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 $ 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 the 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} $.
| | 4 | No.4 Revision |
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 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 the 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 $ F : \mathbb{R}^3 \rightarrow \mathbb{R} $.
| | 5 | No.5 Revision |
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 the 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} $.
| | 6 | No.6 Revision |
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.
$$ -f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) - f(x_0, y_0) = -L(x, y) $$
$$ \frac{-f_x(x_0, y_0)(x - x_0) - f_y(x_0, y_0)(y - y_0) - f(x_0, y_0)}{1} y_0)}{-1} = \frac{-L(x, y)}{-1} $$
$$ f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + f(x_0, y_0) = L(x, y) $$
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} $.
