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Tangent plane

  • Thread starter quietrain
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  • #1
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Homework Statement


find eqn of tangent plane to surface S at point P(2,1,3)
the curves
r = (2+3t, 1-t2 , 3-4t+t2)
r = (1+u2 ,2u3-1 , 2u+1)
both lie on S.

The Attempt at a Solution



i don't really know how to start. am i suppose to find eqn of S first? then use the formula
n.(r - P) with the partial differientials fx , fy etc?

but how do ifind S? with those 2 curves? thanks!
 

Answers and Replies

  • #2
tiny-tim
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hi quietrain! :smile:

each curve has a tangent line …

and two lines define a plane :wink:
 
  • #3
hunt_mat
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The tangent plane is a 2 dimensional vector space, and you know that the tangent space is defined by the tangent vectors of curves, and to find a tangent vector, just differentiate the curves w.r.t to their parameter and compare the vectors.
 
  • #4
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but how do we know that those 2 curves will intersect at the point P?

are we suppose to let the parameter t be = 0 and u = 1 so that they all become point P?

then i differientiate to get

r = (3, -2t, 2t-4)
r = (2u, 6u2, 2) ?

so these are the tangent vectors at the point P? when i sub t = 0 and u= 1?
so r= (3, 0, -4) , r = (2 , 6 , 2)
so i cross multiply them to get normal vector and then use n.(r-p) = 0 to get the tangent plane equation?

btw, what if there issn't a common value for t or u such that they will become point P?
 
  • #5
tiny-tim
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hi quietrain! :wink:

yes, that's perfect. :smile:
btw, what if there issn't a common value for t or u such that they will become point P?
then the question couldn't start "find eqn of tangent plane to surface S at point P(2,1,3)" :wink:
 
  • #6
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ah i see thanks everyone!
 
  • #7
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The tangent plane is a 2 dimensional vector space, and you know that the tangent space is defined by the tangent vectors of curves, and to find a tangent vector, just differentiate the curves w.r.t to their parameter and compare the vectors.
oh i just wondered, if the equation is not parameterized, then i must do a df/dx df/dy df/dz
partial derivatives and then compare the x with x y with y zwithz values at the point P?
 
  • #8
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anyone?
 
  • #9
HallsofIvy
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If what equation is "not parameterized"? A curve in three dimensions cannot be written as a single equation. That would be a surface. If you are asking about finding the tangent plane to a surface, given by f(x,y,z)= constant, then the normal vector is just the gradient [itex]\nabla f[/itex].

And, by the way, once you have the two tangent vectors to the two curves, the normal vector to the surface they lie in is the cross product of the two vectors.
 
  • #10
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oh , so if i do df/dx ,df/dy,df/dz and sub in the x,y,z of point P , i get the normal vector pointing outwards from that point P that is perpendicular to the tangent plane at that surface point?

but i thought the gradient of f gives me tangent vector and not normal vector?

also, if i sub in the values of point P into df/dx , df/dy, df/dz, i get 1 vector only? how do i get a second vector that also pass through point p and lie in the same plane? so that i can do the cross product to get the normal vector to surface?
 
  • #11
tiny-tim
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hi quietrain! :smile:

please remember that you changed the question …

originally, you asked about a surface defined by curves defined by parametrisation, but then you changed it …
oh i just wondered, if the equation is not parameterized, then i must do a df/dx df/dy df/dz partial derivatives and then compare the x with x y with y zwithz values at the point P?
… you asked about a surface not defined by parametrisation, and HallsofIvy correctly replied …
If what equation is "not parameterized"? A curve in three dimensions cannot be written as a single equation. That would be a surface. If you are asking about finding the tangent plane to a surface, given by f(x,y,z)= constant, then the normal vector is just the gradient [itex]\nabla f[/itex].
oh , so if i do df/dx ,df/dy,df/dz and sub in the x,y,z of point P , i get the normal vector pointing outwards from that point P that is perpendicular to the tangent plane at that surface point?

but i thought the gradient of f gives me tangent vector and not normal vector?
if the surface is defined by parameters that increase along the surface, then the gradient obviously is a tangent

if the surface is defined by a function that is constant over the surface, then the gradient obviously is a normal
 
  • #12
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if the surface is defined by parameters that increase along the surface, then the gradient obviously is a tangent

if the surface is defined by a function that is constant over the surface, then the gradient obviously is a normal
erm i understand the first line, its talking about the question i asked at my 1st post right?

the 2nd line is the one i don't understand :(

lets say the plane x+y+z = 6 , it is a surface that has a constant function over all surface right?(means it is x+y+z = 6 everywhere?) so the gradient is 0 if i do df/dx + df/dy+ df/dz = 0

so what is a normal? i thought normal means the vector pointing perpendicularly outwards of the plane? so it should be 1,1,1 in this case?
 
  • #13
tiny-tim
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hi quietrain! :smile:
lets say the plane x+y+z = 6 , it is a surface that has a constant function over all surface right?(means it is x+y+z = 6 everywhere?) so the gradient is 0 if i do df/dx + df/dy+ df/dz = 0
yes, if f(x,y,z) = x + y + z, then that plane is the surface f(x,y,z) = 6

but the gradient is a vector, (∂f/∂x,∂f/∂y,∂f/∂z), which in this case is (1,1,1) :wink:
 
  • #14
HallsofIvy
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erm i understand the first line, its talking about the question i asked at my 1st post right?

the 2nd line is the one i don't understand :(

lets say the plane x+y+z = 6 , it is a surface that has a constant function over all surface right?(means it is x+y+z = 6 everywhere?) so the gradient is 0 if i do df/dx + df/dy+ df/dz = 0
The gradient of a function of several variables is a vector, not a number. The gradient of the plane x+ y+ z= 6 is the vector <1 , 1, 1>, not the number 0.
(In Britain, the term "gradient" is also used for the derivative of a function of a single variable. In the United States that is not often done. I know of no case in which a sum of partial derivatives is called a "gradient".)

so what is a normal? i thought normal means the vector pointing perpendicularly outwards of the plane? so it should be 1,1,1 in this case?
Yes, that is the normal. It is your understanding of "gradient" that is wrong.
 
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  • #15
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oh..
i went to read up on surface normal and from wiki, it says

"If a surface S is given implicitly as the set of points (x,y,z) satisfying F(x,y,z) = 0, then, a normal at a point (x,y,z) on the surface is given by the gradient [URL]http://upload.wikimedia.org/math/e/5/1/e5146b7f34a2d4ee4bb6012a376f5f61.png[/URL]
since the gradient at any point is perpendicular to the level set, and F(x,y,z) = 0 (the surface) is a level set of F."

are they talking about a plane here?

and does F(x,y,z) has to be = 0? or a constant will do like what tiny-tim says? or are these 2 things totally different stuff :(

so now if i let f(x,y,z) = x2+y+z
then if i find gradient i would get 2x,1,1? so this is the normal at whatever value x is regardless of y and z ?
 
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  • #16
hunt_mat
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Yes, but it wouldn't be a unit normal.

F(x,y,z) always =0 because I could define a G(x,y,z)=F(x,y,z)-constant and work with that instead.
 
  • #17
tiny-tim
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hi quietrain! :smile:
"If a surface S is given implicitly as the set of points (x,y,z) satisfying F(x,y,z) = 0, then, a normal at a point (x,y,z) on the surface is given by the gradient [URL]http://upload.wikimedia.org/math/e/5/1/e5146b7f34a2d4ee4bb6012a376f5f61.png[/URL]
since the gradient at any point is perpendicular to the level set, and F(x,y,z) = 0 (the surface) is a level set of F."

are they talking about a plane here?
no, F(x,y,z) = 0 can be any surface
and does F(x,y,z) has to be = 0? or a constant will do like what tiny-tim says? or are these 2 things totally different stuff :(
any constant will do …

the different surfaces F(x,y,z) = k are like Russian dolls, they fit snugly inside each other, they never cross each other, and they are crossed by a system of curves which everywhere are normal to each surface, with each normal curve having gradient ∇F, ie (∂f/∂x,∂f/∂y,∂f/∂z)
so now if i let f(x,y,z) = x2+y+z
then if i find gradient i would get 2x,1,1? so this is the normal at whatever value x is regardless of y and z ?
yes :smile:
 
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  • #18
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thanks everyone!
 
  • #19
HallsofIvy
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hi quietrain! :smile:


no, F(x,y,z) = 0 can be any surface


any constant will do …

the different surfaces F(x,y,z) = k are like Russian dolls,
I like the "Russian dolls" analogy!

they fit snugly inside each other, they never cross each other, and they are crossed by a system of curves which everywhere are normal to each surface, with each normal curve having gradient ∇F, ie (∂f/∂x,∂f/∂y,∂f/∂z)


yes :smile:
 

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