MHB Cartesian equation of the tangent plane

[mathmari]

Hey!

A differentiable function $f(x,y,z)$ has $\nabla f (x_0, y_0, z_0) \neq (0,0,0)$ and zero instant rate of change from $(x_0, y_0, z_0)$ in the direction $\left( \frac{2}{3},-\frac{1}{3},-\frac{2}{3}\right)$. Which could be the cartesian equation of the tangent plane of the level surface of the function at the point $(x_0, y_0, z_0)$ ?
I have done the following :

Since the function has zero instant rate of change from $(x_0, y_0, z_0)$ in the direction $\left( \frac{2}{3},-\frac{1}{3},-\frac{2}{3}\right)$ we have that:
$$\nabla f(x_0, y_0, z_0) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=0 \\ \Rightarrow \frac{2}{3} f_x(x_0, y_0, z_0)-\frac{1}{3} f_y(x_0,y_0,z_0)-\frac{2}{3} f_z(x_0,y_0,z_0)=0$$

The level surfaces of $f(x,y,z)$ are of the form $f(x,y,z)=c$.

How could we find the equation of the tangent plane of the level surface at $(x_0,y_0,z_0)$ ? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

I'm a bit confused about the wording 'level curve'.
Doesn't an equation of the form $f(x,y,z)=c$ identify a surface?
Take for instance $f(x,y,z)=x^2+y^2+z^2$. Then $f(x,y,z)=c$ is a sphere, isn't it?

And if it's really a curve, doesn't it have more than a single tangent plane?
Is perhaps the rectifying plane intended then (the plane that contains the tangent and the binormal vector of the curve)?

Can you clarify? (Wondering)

 

[mathmari]

Hey mathmari! (Smile)

I'm a bit confused about the wording 'level curve'.
Doesn't an equation of the form $f(x,y,z)=c$ identify a surface?
Take for instance $f(x,y,z)=x^2+y^2+z^2$. Then $f(x,y,z)=c$ is a sphere, isn't it?

And if it's really a curve, doesn't it have more than a single tangent plane?
Is perhaps the rectifying plane intended then (the plane that contains the tangent and the binormal vector of the curve)?

Can you clarify? (Wondering)

Ohh.. (Blush) Level curves should be level surafces. I changed it in post #1.

 

[I like Serena]

Ohh.. (Blush) Level curves should be level surafces. I changed it in post #1.

Okay... then we have one vector with zero instant rate of change... but don't we need two of them? Independent from each other? (Wondering)
If so, we could define the second vector to a set of unknowns, after which we can set up an equation.

Or can it be that the non-zero instant rate of change is intended?
In that case we would have sufficient information to set up the equation for the tangent plane. (Thinking)

For the record, the equation of the tangent plane would be in either case:
$$
\nabla f_0\cdot \mathbf x = \nabla f_0\cdot \mathbf x_0
$$
since $\nabla f_0$ is normal to the tangent plane.

 

[mathmari]

Okay... then we have one vector with zero instant rate of change... but don't we need two of them? Independent from each other? (Wondering)
If so, we could define the second vector to a set of unknowns, after which we can set up an equation.

Or can it be that the non-zero instant rate of change is intended?
In that case we would have sufficient information to set up the equation for the tangent plane. (Thinking)

For the record, the equation of the tangent plane would be in either case:
$$
\nabla f_0\cdot \mathbf x = \nabla f_0\cdot \mathbf x_0
$$
since $\nabla f_0$ is normal to the tangent plane.

It is zero instant rate of change. So, we consider a vector $(a,b,c)$ ?

Do we take $\nabla f_0\cdot (a,b,c)$ ? (Wondering)

 

[I like Serena]

It is zero instant rate of change. So, we consider a vector $(a,b,c)$ ?

Do we take $\nabla f_0\cdot (a,b,c)$ ? (Wondering)

Hmm... perhaps we can define:
$$(a,b,c) = \nabla f(x_0, y_0, z_0)$$
That means that the equation of the tangent plane is (from the formula in my previous post):
$$(a,b,c) \cdot (x,y,z) = (a,b,c) \cdot (x_0, y_0, z_0)$$
And additionally we have the restriction:
$$\nabla f(x_0, y_0, z_0) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=(a, b, c) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=0$$
(Thinking)

 

[mathmari]

Hmm... perhaps we can define:
$$(a,b,c) = \nabla f(x_0, y_0, z_0)$$
That means that the equation of the tangent plane is (from the formula in my previous post):
$$(a,b,c) \cdot (x,y,z) = (a,b,c) \cdot (x_0, y_0, z_0)$$
And additionally we have the restriction:
$$\nabla f(x_0, y_0, z_0) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=(a, b, c) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=0$$
(Thinking)

So, we have to find a relation for the vector $(a,b,c)$ ? Or do we not have to get rid of thes unknowns? (Wondering)

 

[I like Serena]

So, we have to find a relation for the vector $(a,b,c)$ ? Or do we not have to get rid of thes unknowns? (Wondering)

As it is, we have 6 unknown constants in the tangent equation: $a,b,c,x_0,y_0,z_0$.
And we have 1 equation that puts a restriction on $a,b,c$.
To be honest, the problem seems to be a bit 'incomplete'.
Either way, it seems to me that we might as well let it stand like that. (Thinking)

 

[mathmari]

So, we have that $(a, b, c) \cdot (\frac{2}{3},-\frac{1}{3},-\frac{2}{3})=0 \Rightarrow 2a-b-2c=0 \Rightarrow b=2a-2c$. Therefore, we get
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \Rightarrow a(x-x_0)+(2a-2c)(y-y_0)+c(z-z_0)=0 $$
right? (Wondering)

 

[I like Serena]

I believe that's the best that we can do (together with the definition $(a,b,c)=\nabla f_0$). (Nod)

 

[mathmari]

I believe that's the best that we can do (together with the definition $(a,b,c)=\nabla f_0$). (Nod)

Great! Thanks a lot! (Smile)