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Hi, I'm working through Schaums outlines linear algebra, and one of the excercises from the first chapter asks to find the normal vector
and tnagent plane H to the surface at the point P(1,3,2).Heres my working:
parameterizing:
at P(1,3,2):
at P(1,3,2):
at P(1,3,2):
The answer given is
, although no method is given.I'm fairly sure vector calculus isn't assumed as a pre-requisite so I have two questions:
-Whats wrong with my method? (why has it given the wrong answer?)
-Whats the easier (or should I say correct?) way to do this? Since the first chapter only deals with basic vector properties and identities I assume it can be done using only this.
Thanks alot.
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Dear Onyx
Here is the solution
the point P(1,3,2) lies on the plane
x^2* y +3yz=20 rewriting the equation as follows
x^2* y +3yz-20=0 so let
f=x^2* y +3yz-20
df/dx=2xy so df/dx (p) = 2(1)(3)=6
df/dy=x^2+3z so df/dy (p) = 7
df/dz=3y so df/dz (p) =9
so the normal n = 6i+7j+9k
and the tnagent plane H to the surface is 6(x-1)+7(y-3)+9(z-2)=0 or
6x+7y+9z=45
Best Regards
Riad Zaidan
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Thanks, but I don't understand the method. Doesn't a function of three variables such as:
describe a surface in
, and not ?Then the gradient of f,
would be a vector in - I know this is what we are looking for but don't understand how it relates to the function of three variables, f, or even how you can simply rewrite the equation as you have and declare it as a function. I would have though parameterizing would have been the only way, but obviously this is wrong (I'm still not sure why though).Offline
Dear Onyx
I think I found the problem:
The point (1,3,2) does not lie on the surface x^2* y +3yz=20 so you may change 20 into 21
in order to have a point on that surface then you have the following:
z=21/3y - x^2/3 or f(x,y)=7/y - x^2 / 3 so
df/dx=-2x/3 so df/dx (p) = -2/3
df/dy=-7/(y^2) so df/dy (p) =- 7/9 therefore
the tnagent plane H to the surface is -2/3(x-1)- 7/9(y-3)-(z-2)=0 or
6x+7y+9z=45 and the normal to the plane is n=6i+7j+9k as wanted.
But why the value of 20 does not infuence my way because when diff. 20 or 21 gives 0.
Note: afunction of four variables is very difficult to be drawn in the plane , so we use the level syrfaces on which f is constant.
Best Regards
Riad Zaidan
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Thanks, is there any theorem that says whenever we have a function,
, generating a surface in , then when we express this function implicitly as , the normal vector to the surface can be given by:This is the tecnique you have used here, though I've never came across it.
Also can someone tell me what was wrong with my method of parameterization?
Thanks
Last edited by Onyx (2009-08-26 00:15:34)
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Hi dear Onyx:
You asked the following:
Also can someone tell me what was wrong with my method of parameterization?
The answer is :
the problem was of the point P(1,3,2) does not lie on the given surface , so when you change the constant 20 into 21 in order to have the point on the surface and repeat the procedures you have done , every thing will be right. Try this by yourself.
Best Regards
Riad Zaidan
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Thanks, do you have any idea about this?..
Thanks, is there any theorem that says whenever we have a function,
, generating a surface in , then when we express this function implicitly as , the normal vector to the surface can be given by:This is the tecnique you have used here, though I've never came across it.
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Anyone?
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