ques

- anonymous

ques

- Stacey Warren - Expert brainly.com

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- chestercat

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- anonymous

Consider a surface given by
\[\phi(x,y,z)\]
\[x^2+y^2+z^2=9\]
Now the normal to the surface is given by
\[\vec \nabla\phi=2x\hat i+2y\hat j+2z\hat k\]
Now my question is where did the 9 go, I know if u apply the operator to it will become 0 but shouldn't it be something like
\[2x\hat i+2y\hat j+2z\hat k=\vec 0\]

- ganeshie8

We may think of the given surface \(x^2+y^2+z^2=9\) as a specific level surface for the level set of function : \( \phi(x,y,z)=x^2+y^2+z^2\), namely, \(\phi=9\).

- ganeshie8

Next we use the fact the gradient of a function is perpendicular to the level surface to find a normal vector of given surface

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## More answers

- ganeshie8

"gradient" operates on a function
doesn't make much sense to say "gradient of an equation"

- anonymous

Yes, but shouldn't u also apply operator on the right side ??

- ganeshie8

we're simply using the fact that \(\nabla \phi \perp (\text{level surface}) \) to get a normal vector

- anonymous

Let me think about it lol

- ganeshie8

I think your equation is also correct as the gradient of \(g(x,y,z)=0\) is the 0 vector

- ganeshie8

In case you want to see it geometrically, may be consider a function of two variables, sketch the level curves, compute the gradient and verify...

- anonymous

gradient is only perpendicular at any point of the surface if it's a level surface right?

- ganeshie8

Lets say, the function f(x,y) is defined for all real values of ordered pairs (x,y)
then \(\nabla f(a,b)\) is perpendicular to the specific level curve that passes through the point \((a,b)\)

- anonymous

I think I understand it better now,
you can only find gradient of a function,
the function is
\[x^2+y^2+z^2\]
Doesn't make sense to find gradient of an equation
We're really just using the fact
that gradient is perpendicular to the level surface....

- anonymous

as long as the curve is level

- ganeshie8

Nice! It will be more illuminating to see these geometrically...

- anonymous

But I don't understand the point of level surface.
Suppose a point P on a level surface, if the value of the function at the point P is the same as the value of any other point on the surface, it's a level surface, but of course it will be same, because all points on the surface satisfy the equation of the surface

- ganeshie8

Right, as you move on a specific level surface, the value of function remains constant.
The gradient vector changes accordingly to remain perpendicular to the surface.

- anonymous

can u give an example of a non level surface??
even if we take
\[\phi(x,y,z)=x+c\]
where c is a constant
\[\phi(x,y,z)-x=c\]
This in itself defines a new function which makes it a level surface
\[\psi(x,y,z)=c\]
Now the value of the function at all points will be same and equal to c because they all lie on the surface
it seems as if there is nothing like a "non level surface"

- ganeshie8

you get level surface by fixing the value of the function, so every intersection of function and a horizontal plane is a level surface

- ganeshie8

maybe i shouldn't say that, because, that makes sense only for a function of 2 variables..

- ganeshie8

You're right, any equation in "n" variables can be thought of as a "level surface" of a function of "n" variables. The function itself requires "n+1" axes to visualize, but the level surface requires only "n" axes.

- anonymous

What i'm saying is that for a level surface we have the form
\[\phi(x,y,z)=c\]
even if we have an equation such that
both left and right side are entirely made of variables we can always rearrange to have all variables on one side and a constant on the right side
\[x^2+y^2+z^2=xyz\]\[\frac{x^2+y^2+z^2}{xyz}=1\]
of the form
\[\phi(x,y,z)=c\]
So it seems like no matter what we can't have a non level surface ??

- ganeshie8

RIght, you can always define a function such that your surface belongs to its level set

- ganeshie8

Thats how we use gradient to compute a normal vector of ANY surface

- anonymous

oh ok I understand now, so it's a level surface for
\[\psi(x,y,z)=\frac{x^2+y^2+z^2}{xyz}\]
but not for
\[\phi(x,y,z)=x^2+y^2+z^2\]

- ganeshie8

Yep! the connection between gradient and normal vector is with level surfaces

- anonymous

It depends on our function, if it's level or not !!

- ganeshie8

Hmm that sounds okay I guess!
but honestly i never heard that phrase "if its level or not" before haha!

- anonymous

thanks:)

- ganeshie8

np:) I think you will like this video...
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/
when you're free try watching..

- anonymous

Yeh actually my syllabus is really messed up, they are teaching us vector calculus without teaching us about multivariable functions or partial derivatives
For example
\[ d(f(x,y,z))=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\]
I didn't even knew what this meant and we were using it, I had to look it up separately to understand it

- ganeshie8

Ahh then I think those lectures really supplement your course of vector calculus. Except for the french accent, Prof. Denis Auroux is really good !

- anonymous

thx ill check it out

- IrishBoy123

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- IrishBoy123

i spent ages chasing my tail on this so i hope this might be helpful - as i did find some peace of mind, eventually. [at the very least, i hope it makes some sense.]
strictly speaking, you do not apply the del operator to the RHS, only the LHS, ie only to \(\phi\)
because you have defined \(\phi(x,y,z)\) to be constant, you can say that \(d \Phi = 0\) **provided** you remain within the constraint \(\phi(x,y,z) = c\) , that is you limit yourself to x,y,z 's on the so-called level surface. that is why you created the level surface in the first place.
so, for the paraboloid \(z = x^2 + y^2\), you create the level surface \(\phi = z - x^2 - y^2 = 0\) knowing that \(d \Phi = 0\) without needing to do any further analysis provided x,y,z are related by \(z = x^2 + y^2\)
[which, BTW, is completely different from the scalar field \(\phi = z - x^2 - y^2 \) where there is no constraint on x,y,z values you choose, and where \(\vec \nabla \phi\) just gives you the direction/scale of greatest change ]
then, because \(d \Phi = \phi_x dx + \phi_y dy + \phi_z dz\)
it follows that \(\phi_x dx + \phi_y dy + \phi_z dz = 0\))
\(\implies <\phi _x,\phi _y, \phi _z > \bullet = 0\)
and again because of the constraint \(\phi(x,y,z) = c\), you have \(\phi(x+dx,y+dy,z+dz) = c\) so runs along the surface and is the tangent vector, so \(\vec \nabla \phi \) must be the normal vector.
this might sound trivial, but if you moved the paraboloid 5 units up the z axis as \(z = x^2 + y^2 + 5\), you could say
\(\phi = z - x^2 - y^2 = 5\)
or
\(\phi = z - x^2 - y^2 -5 = 0\)
and by definition, because of the constraint, \(d \phi = 0\)
and then apply the operator to the LHS
with the same result
At least that's how i have come to see it...

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