Tangent planes and normal vectors

  • Thread starter Mdhiggenz
  • Start date
  • #1
327
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Homework Statement



Find all points on the surface at which the tangent plane is horizontal

z=x3y2

Things I know:
Tangent plane is horizontal then therefore the normal must be vertical in order to be perpendicular.

Dot product of the tangent plane with normal is = 0

Normal plane is given by the partial with respect to x,y,z evaluated at some points P(x0,y0,z0)

Putting that together I get
f(x,y,z)=x3y2-z
fx=3x2y2
fy=2x3y
fz=-1

Here is where I am completely stuck I know that I still don't have a normal vector because I do not have any points to plug in to the partials.



Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



Find all points on the surface at which the tangent plane is horizontal

z=x3y2

Things I know:
Tangent plane is horizontal then therefore the normal must be vertical in order to be perpendicular.

Dot product of the tangent plane with normal is = 0

Normal plane is given by the partial with respect to x,y,z evaluated at some points P(x0,y0,z0)

Putting that together I get
f(x,y,z)=x3y2-z
fx=3x2y2
fy=2x3y
fz=-1

Here is where I am completely stuck I know that I still don't have a normal vector because I do not have any points to plug in to the partials.

Your original equation is of the form ##z = f(x,y)##. The tangent plane will be horizontal at any point ##(x,y,z)## where ##f_x(x,y)=0## and ##f_y(x,y)=0##.
 
  • #3
HallsofIvy
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Or, same thing, The vector you got, [itex]<3x^2y^2, 2x^3y, -1>[/itex] must be of the form <0, 0, a> to be vertical. Obviously we can take a= -1 but we need [itex]3x^2y^2= 0[/itex] and [itex]2x^3y= 0[/tex]. That will be true if x= 0 or if y= 0. In other words above the x and y axes.
 
  • #4
327
1
hmm, so I can equal them together, and get

x=(3/2)y and y=2/3x

Don't really understand what the values represent?
 

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