prove that v is tangent to surface M

[logistic_guy]

here is the question

Prove that: $\displaystyle \bold{v} = (v_1, v_2, v_3)$ is tangent to $\displaystyle M: z = f(x,y)$ at a point $\displaystyle \bold{p}$ of $\displaystyle M$ if and only if $\displaystyle v_3 = \frac{\partial f}{\partial x}(p_1,p_2)v_1 + \frac{\partial f}{\partial y}(p_1,p_2)v_2$.

i know the idea but i don't know how to start it because i need three points to take the gradient on that surface $\displaystyle M$. it give only $\displaystyle x$ and $\displaystyle y$. this mean the surface $\displaystyle M$ is in $\displaystyle \bold{R}^2$. typo maybe? because i'm expect to take the gradient of $\displaystyle f(x,y,z)$ not $\displaystyle f(x,y)$

 

[blamocur]

i don't know how to start it because i need three points to take the gradient on that surface M.

Why? Do you mean 3 coordinates? -- you have those as (x,y,f(x,y)), don't you?

 

[logistic_guy]

Why? Do you mean 3 coordinates? -- you have those as (x,y,f(x,y)), don't you?

this give me $\displaystyle \frac{\partial f}{\partial z} = 0$. at some point i'll use the dot product. $\displaystyle \frac{\partial f}{\partial z}v_3 = 0v_3 = 0$. so $\displaystyle v_3$ will be deleted and i can't solve for $\displaystyle v_3$ as the question asking me.

 

[blamocur]

this give me $\displaystyle \frac{\partial f}{\partial z} = 0$. at some point i'll use the dot product. $\displaystyle \frac{\partial f}{\partial z}v_3 = 0v_3 = 0$. so $\displaystyle v_3$ will be deleted and i can't solve for $\displaystyle v_3$ as the question asking me.

Where does this come from ?

 

[mario99]

Where does this come from ?

Professor blamocur, I think he meant this:

$\nabla f(\bold{p}) \cdot \bold{v}$

My intuition tells me we should do this:

$h(x,y,z) = f(x,y) - z$, then:

$\nabla h(\bold{p}) \cdot \bold{v} = 0$

This allows us to solve for $v_3$ and also shows $\bold{v}$ is orthogonal to $\nabla h(\bold{p})$.

I assumed we can do this:

$\displaystyle \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x}$ to get his final form. $\displaystyle \left(\text{the same for} \frac{\partial h}{\partial y}\right)$

 

[blamocur]

Professor blamocur, I think he meant this:

$\nabla f(\bold{p}) \cdot \bold{v}$

My intuition tells me we should do this:

$h(x,y,z) = f(x,y) - z$, then:

$\nabla h(\bold{p}) \cdot \bold{v} = 0$

This allows us to solve for $v_3$ and also shows $\bold{v}$ is orthogonal to $\nabla h(\bold{p})$.

I assumed we can do this:

$\displaystyle \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x}$ to get his final form. $\displaystyle \left(\text{the same for} \frac{\partial h}{\partial y}\right)$

Not sure what the OP meant, but I agree with your post.

 

[logistic_guy]

Professor blamocur, I think he meant this:

$\nabla f(\bold{p}) \cdot \bold{v}$

My intuition tells me we should do this:

$h(x,y,z) = f(x,y) - z$, then:

$\nabla h(\bold{p}) \cdot \bold{v} = 0$

This allows us to solve for $v_3$ and also shows $\bold{v}$ is orthogonal to $\nabla h(\bold{p})$.

I assumed we can do this:

$\displaystyle \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x}$ to get his final form. $\displaystyle \left(\text{the same for} \frac{\partial h}{\partial y}\right)$

thank Mario99

this answer my question

Not sure what the OP meant, but I agree with your post.

thank blamocur