
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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 $f(x,y,z)=zx^2y^2=0$. 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.
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 $f(x,y,z)=zx^2y^2$. The gradient of that function is a vector field.

November 8th, 2016, 08:23 AM  #5 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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. 
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. 

Tags 
find, form, normal, parametric, surface 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Find curve with tangent and normal lines, form triangle  Shamako  Calculus  1  October 10th, 2011 08:04 PM 
How do I find the surface normal  Keysle  Algebra  3  June 3rd, 2011 03:32 AM 
Normal of a nonplanar 3d surface  spazzmonkeys  Linear Algebra  2  May 21st, 2010 06:48 PM 
normal form for real quadratic form  Wojciech_B  Linear Algebra  0  December 7th, 2009 01:52 PM 
Creating a parametric equation for a surface  echo6  Algebra  2  January 23rd, 2008 03:34 AM 