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November 8th, 2016, 06:28 AM   #1
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Find normal of surface by using parametric form, help?

I'm not so sure how to put something in parametric form..
Given the equation:$\displaystyle z = x^2 + y^2$

In parametric form that would be:
$\displaystyle r(x,y) = (x^2, y^2, f(x,y)) = (x^2+y^2,x^2+y^2)$

Since the normal is the gradient normalized then the normal is simply:
$\displaystyle grad (r) = (2x, 2y, 2x + 2y) * 1/sqrt((2x)^2+(2y)^2+4*(x+y)^2)$

My book says the answer is:
$\displaystyle (-2x,-2y,1) * 1/sqrt(1+4z)$


If they claim they're parametric function was: f(x,y,z) = (x^2, y^2, -z) = 0
Then why is it a z in the root?

I'm really lost at this, all help appriciated.
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November 8th, 2016, 07:39 AM   #2
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You're given $z=x^2+y^2$. To find a normal vector to a tangent plane, you need to take the gradient of $f(x,y,z)=z-x^2-y^2=0$. Therefore, $\vec{F}=\nabla f=<-2x,-2y,1>$. Your book wants the unit normal vector, so you'll have to calculate $\frac{\vec{F}}{||\vec{F}||}$. Starting with the magnitude, $||\vec{F}||=\sqrt{(-2x)^2+(-2y)^2+1}=\sqrt{4(x^2+y^2)+1}$. Now use the fact that $z=x^2+y^2$ and continue from here.
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Last edited by Compendium; November 8th, 2016 at 07:43 AM.
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November 8th, 2016, 07:52 AM   #3
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Oh ok that's where the z is from!

But can you explain what the parametric form of z = x^2 + y^2 is?
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November 8th, 2016, 08:18 AM   #4
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The way you're describing the problem is confusing. Something enclosed in () is a point. Something enclosed in <> is a vector. You also don't take the gradient of a vector function. You take the gradient of a scalar function, which is called the potential function. In this case, the potential function is $f(x,y,z)=z-x^2-y^2$. The gradient of that function is a vector field.
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November 8th, 2016, 08:23 AM   #5
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If F(x,y,z)=0
dF=$\displaystyle \frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=0
$
$\displaystyle \triangledown F\cdot d\bar{r}=0$
$\displaystyle d\bar{r}$ is constrained to surface so $\displaystyle \triangledown F$ is perpendicular to surface.
As for parameterization:
x=rcos$\displaystyle \theta$
y=rcos$\displaystyle \theta$
z=r^2
But I don't think they want the gradient in polar coordinates.

Or, x and y could be considered parameters:
$\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$
If dz=o, $\displaystyle \triangledown f\cdot d\bar r=0$
where now $\displaystyle \bar r$ varies along curve z= constant and $\displaystyle \triangledown f$ is perpendicular to curve z=constant. This can also be considered gradient, depending on your interpretation. Always confused me. I guess you have to specify whether it's normal to surface F=c or normal to curve z=c.
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November 8th, 2016, 09:32 AM   #6
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I think I get it now.
We move all variables to the left side, then set f = x^2 + y^2 - z
where f is a scalar field (inf many levelfields).
Taking the gradient of a scalarfield gives gradient which is the normal to the level surface at that level.

Thanks for explaining it clearly Compendium.
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