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:

[imath]\nabla f(\bold{p}) \cdot \bold{v}[/imath]

My intuition tells me we should do this:

[imath]h(x,y,z) = f(x,y) - z[/imath], then:

[imath]\nabla h(\bold{p}) \cdot \bold{v} = 0[/imath]

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

I assumed we can do this:

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

 

[blamocur]

Professor blamocur, I think he meant this:

[imath]\nabla f(\bold{p}) \cdot \bold{v}[/imath]

My intuition tells me we should do this:

[imath]h(x,y,z) = f(x,y) - z[/imath], then:

[imath]\nabla h(\bold{p}) \cdot \bold{v} = 0[/imath]

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

I assumed we can do this:

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

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

 

[logistic_guy]

Professor blamocur, I think he meant this:

[imath]\nabla f(\bold{p}) \cdot \bold{v}[/imath]

My intuition tells me we should do this:

[imath]h(x,y,z) = f(x,y) - z[/imath], then:

[imath]\nabla h(\bold{p}) \cdot \bold{v} = 0[/imath]

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

I assumed we can do this:

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

thank Mario99

this answer my question

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

thank blamocur