Find the equation of the tangent plane

[Calpalned]

Homework Statement

Find the equation of the plane tangent to $x^2+3y^2+6z^2=67$ at the point $(1, 2,3)$

Homework Equations

$w-w_0 = F_x(x-x_0) + F_y(y-y_0) + F_z(z-z_0)$

The Attempt at a Solution

Using the above formula, I get $w-67 = 2x(x-1) + 6y(y-2) + 12z (z-3)$ = $w-67 = 2(x-1) + 12(y-2) + 36 (z-3)$ Did I use the wrong formula? If so, how can I determine which formula to use?

 

[Brian T]

To start, it looks like your tangent plane equation is in ℝ⁴ when the initial surface is in ℝ³.

You want to use the tangent plane equation for one of the variables in terms of the other two, and rewrite the surface equation accordingly

 

[Calpalned]

I want to quickly confirm: $ℝ^1$ means $y = f(x)$, $ℝ^2$ means $z = f(x,y)$, $ℝ^3$ means $w = f(x,y,z)$.

So I should use the formula $z-z_0 = F_x(x-x_0) + F_y(y-y_0)$?

 

[Calpalned]

Post deleted

 

[Brian T]

ℝ¹ is the set of all numbers (in one dimension, such as a number line) just x
ℝ² is two sets of the previously mentioned (one number line as x, the other as y // whatever you want to call them). y = f(x)
ℝ³ is three sets of the same (so x,y,z) z = f(x,y)

And yes, that would be (one of the) right formula to use

 

[Calpalned]

ℝ¹ is the set of all numbers (in one dimension, such as a number line) just x
ℝ² is two sets of the previously mentioned (one number line as x, the other as y // whatever you want to call them). y = f(x)
ℝ³ is three sets of the same (so x,y,z) z = f(x,y)

And yes, that would be (one of the) right formula to use

I see... I was one dimension too low.

 

[Ray Vickson]

Homework Statement

Find the equation of the plane tangent to $x^2+3y^2+6z^2=67$ at the point $(1, 2,3)$

Homework Equations

$w-w_0 = F_x(x-x_0) + F_y(y-y_0) + F_z(z-z_0)$

The Attempt at a Solution

Using the above formula, I get $w-67 = 2x(x-1) + 6y(y-2) + 12z (z-3)$ = $w-67 = 2(x-1) + 12(y-2) + 36 (z-3)$ Did I use the wrong formula? If so, how can I determine which formula to use?

You are using the wrong $F_x, F_y, F_z$. These are constants because they are evaluated at the point $(x_0,y_0,z_0)$, not at the variable point $(x,y,z)$. In other words, you need to write
$$0 = F_x(\vec{p}_0) \, (x-x_0) + F_y(\vec{p}_0) \, (y-y_0) + F_z(\vec{p}_0) (x-z_0)$$
where $\vec{p}_0 = (x_0,y_0,z_0)$.

 

[Calpalned]

You are using the wrong $F_x, F_y, F_z$. These are constants because they are evaluated at the point $(x_0,y_0,z_0)$, not at the variable point $(x,y,z)$. In other words, you need to write
$$0 = F_x(\vec{p}_0) \, (x-x_0) + F_y(\vec{p}_0) \, (y-y_0) + F_z(\vec{p}_0) (x-z_0)$$
where $\vec{p}_0 = (x_0,y_0,z_0)$.

So $F_x(\vec p_0) =$ the partial derivative of the given equation with respect to $x$, in which I plug in $x=1$ ? If that's the case, the whole answer would be 0 + 0 + 0 = 0

 

[Ray Vickson]

So $F_x(\vec p_0) =$ the partial derivative of the given equation with respect to $x$, in which I plug in $x=1$ ? That will be zero right?

Wrong: $F(x,y,z) = x^2 + 3 y^2 + 6 z^2$, so $F_x = 2x$ is not 0 at $x = 1$!

 

[Calpalned]

Wrong: $F(x,y,z) = x^2 + 3 y^2 + 6 z^2$, so $F_x = 2x$ is not 0 at $x = 1$!

It makes sense how $F_x = 2x$, but where is my blunder?. I thought it was zero because if $F(x,y,z)=x^2+3y^2+6z^2 = 67$, then $F_x = 2x + 0 + 0 = 0$ thus $F_x = 2x = 0$ so $F_x = 0$

 

[Ray Vickson]

It makes sense how $F_x = 2x$, but where is my blunder?. I thought it was zero because if $F(x,y,z)=x^2+3y^2+6z^2 = 67$, then $F_x = 2x + 0 + 0 = 0$ thus $F_x = 2x = 0$ so $F_x = 0$

I have absolutely no idea what you are talking about. We have $F_x = 2x$, so if $x = 1$ we have $F_x = 2$, if $x = 17,000,000$ we have $F_x = 34,000,000$, and If $x = 0$ we have $F_x = 0$.

You have a surface $F(x,y,z) = \text{const.}$ If a point $(x,y,z) = (a,b,c)$ is on the surface, a neighboring point $(x,y,z) = (a+\Delta a, b + \Delta b, c + \Delta c)$ is on the same surface if the function $G(\Delta a , \Delta b, \Delta c) = 0$, where
$$G(\Delta a, \Delta b, \Delta c) \equiv F(a + \Delta a, b + \Delta b, c + \Delta c) - F(a,b,c) .$$
In other words, in order to remain on the surface the value of $F$ should not change.

For small ("infinitesimal") $\Delta a, \Delta b, \Delta c$ the condition of staying on the surface becomes
$$0 = F_x \Delta a + F_y \Delta b + F_z \Delta c ,$$
where $F_x = F_x(a,b,c)$, etc. Whether or not $\Delta a, \Delta b, \Delta c$ are small, that equation gives a plane in the "variables" $\Delta a, \Delta b, \Delta c$. When the $\Delta$-variables are small, that plane essentially coincides with the surface, but when they are large the plane and the surface diverge (i.e., grow increasingly farther apart). That is why we call it the tangent plane: it agrees with the surface over a small patch.

 

[Mark44]

It might help to distinguish the function F and its partials from their values at a particular point.
Function: F(x, y, z) = x² + 3y² + 4z²
A partial: F_x(x, y, z) = 2x

At the point (1, 2, 3)
F(1, 2, 3) = 1² + 3 * 2² + 4 * 3² = 49
F_x(1, 2, 3) = 2 * 1 = 2

 

[Calpalned]

I have absolutely no idea what you are talking about. .

$F_x = 0$ if I take the partial derivative with respect to f of $f(x,y,z) = x^2+3y^2+6z^2 = 67$ The derivative of $3y^2, 6z^2$ and $67$ are all equal to zero, so $F_x = 2x + 0 + 0 = 0$ and therefore $F_x = 0$. I want to know what the error is in this method.

The correct method that you indicated makes sense too. $F(x,y,z) = x^2+3y^2+6z^2$ therefore $F_x = 2x$ However, how did you ignore the constant value of 67?

 

[Mark44]

$F_x = 0$ if I take the partial derivative with respect to f of $f(x,y,z) = x^2+3y^2+6z^2 = 67$

With respect to f? The independent variables are x, y, and z. Also, F is the function, so you shouldn't be talking about f(x, y,z).

The derivative of $3y^2, 6z^2$ and $67$ are all equal to zero, so $F_x = 2x + 0 + 0 = 0$ and therefore $F_x = 0$. I want to know what the error is in this method.

I think you might be confused about what exactly the function is. I believe that F is defined this way: F(x, y, z) = x² + 3y² + 6z². This function maps R³ to R¹, which was alluded to by Brian T earlier in this thread. That means that a graph of the function would require four dimensions.

In this problem, we are dealing with a level surface for which F(x, y, z) = 67. This does not mean that the function is x² + 3y² + 6z² = 67.

The correct method that you indicated makes sense too. $F(x,y,z) = x^2+3y^2+6z^2$ therefore $F_x = 2x$ However, how did you ignore the constant value of 67?

Because it is not part of the function's definition.

 

[Ray Vickson]

$F_x = 0$ if I take the partial derivative with respect to f of $f(x,y,z) = x^2+3y^2+6z^2 = 67$ The derivative of $3y^2, 6z^2$ and $67$ are all equal to zero, so $F_x = 2x + 0 + 0 = 0$ and therefore $F_x = 0$. I want to know what the error is in this method.

The correct method that you indicated makes sense too. $F(x,y,z) = x^2+3y^2+6z^2$ therefore $F_x = 2x$ However, how did you ignore the constant value of 67?

I am going to try to clear this up once and for all, so please read the following carefully and slowly. Make sure you read only what I actually write, and not what you imagine I write or think I should write.

Let's start with a function $f(x,y,z) = x^2 + 3 y^2 + 6z^2$, and note that I write $f$, not $F$, because I want so start fresh, like it were a new problem. Here is a list of five points in $(x,y,z)$-space:
$$\begin{array}{ccc} (x,y,z) & f(x,y,z) & f_x(x,y,z)\\ (0,0,0) & 0 & 0 \\ (1,1,1) & 10 & 2 \\ (-2,2,1) & 22 & -4 \\ (1,2,3) & 67 & 2 \\ (4,1,3) & 73 & 8 \end{array}$$
Here, of course, $f_x (x,y,z) \equiv \partial f(x,y,z) / \partial x$.

The point $(1,2,3)$ is on the surface $f(x,y,z) = 67$, while the other four points are not on the surface. Did we need to know anything about the surface in order to compute $f$ and $f_x$? Did the value '67' affect the computations of $f$ and $f_x$ in any way at all? Do we always have $f_x = 2x$, whether or not the point $(x,y,z)$ is on the surface $f = 67$?