My Math Forum Find normal of surface by using parametric form, help?

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 November 8th, 2016, 06:28 AM #1 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 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.
 November 8th, 2016, 07:39 AM #2 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 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. Thanks from Addez123 Last edited by Compendium; November 8th, 2016 at 07:43 AM.
 November 8th, 2016, 07:52 AM #3 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 Oh ok that's where the z is from! But can you explain what the parametric form of z = x^2 + y^2 is?
 November 8th, 2016, 08:18 AM #4 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 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. Thanks from Addez123
 November 8th, 2016, 08:23 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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. Thanks from Addez123
 November 8th, 2016, 09:32 AM #6 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 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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