For a parametric curve, the important vector is the
tangent vector, v.
For a parametric surface, the important vector is the
normal vector, N.
Since the tangent vectors eu and ev are tangent to a
parametric surface, the normal vector N can be computed as the
cross product of eu and ev.
Normal Vector
The normal vector, N, to a parametric
surface is:
N=eu×ev=∣∣∣∣∣∣∣∣∣∣ı^∂u∂x∂v∂xȷ^∂u∂y∂v∂yk^∂u∂z∂v∂z∣∣∣∣∣∣∣∣∣∣=(∂u∂y∂v∂z−∂u∂z∂v∂y)ı^−(∂u∂x∂v∂z−∂u∂z∂v∂x)ȷ^+(∂u∂x∂v∂y−∂u∂y∂v∂x)k^=⟨∂(u,v)∂(y,z),−∂(u,v)∂(x,z),∂(u,v)∂(x,y)⟩
Notice that the components of the normal vector are 2-dimensional
Jacobian determinants. For the x component the variables are y and
z. For the y component the variables are x and z. For the
z component the variables are x and y. Also notice that in each
component the variables x,y,z occur in
alphabetical order, yz, xz, xy, but
there is a minus on the y component. However, back up a step and
distribute the minus sign in the y component. Then:
In this version, there is no minus sign but the variables are now in
cyclic order, xyzxyz⋯. You can remember
N in either version. However, in computing N we rarely
use the Jacobian determinants, but rather, go back to the cross product formula.
Caution: We previously saw Jacobian determinants
in the computation of the differentials of area and volume. In those cases,
the sign of the determinant did not matter because we were taking the
absolute value to form the Jacobian factor. However, in computing the
components of the normal vector, the sign of each
Jacobian determinant is crucial since if you make a mistake, the
vector will no longer be normal! This is particularly true for the sign of
the y-component.
In dealing with parametric surfaces, you will have to compute the normal
many times and the length of the formulas will keep growing. If you try to
copy the components of eu and ev to take the cross
product, you will invariably make a copying mistake at some time.
Don't re-copy. It is a waste of time and leads
to mistakes. If you carefully calculate eu and ev
and keep things lined up, you just need to put the row ı^,
ȷ^, k^ on top of the values of eu and
ev and take the determinant of the resulting matrix. On paper,
your work would look like this. First, compute the two tangent vectors:
eu=∂u∂R=ev=∂v∂R=∣∣∣∣∣∣∣∣∣∣∣ı^⟨∂u∂x⟨∂v∂xȷ^∂u∂y∂v∂yk^∂u∂z⟩∂v∂z⟩∣∣∣∣∣∣∣∣∣∣∣=N
Now add an ı^, ȷ^ and k^ and some
vertical bars and compute the determinant.
Click Here to delete the determinant.
On the next page, we will see that it is also
useful to know the length of the normal vector. So these examples and
exercises also ask for the length.
Find the normal vector to the sphere of radius ρ=2, parametrized by
R(ϕ,θ)=⟨2sinϕcosθ,2sinϕsinθ,2cosϕ⟩
Then find the magnitude of the normal.
We previously found the tangent vectors:
eϕeθ=∂ϕ∂R=⟨2cosϕcosθ,2cosϕsinθ,−2sinϕ⟩=∂θ∂R=⟨−2sinϕsinθ,2sinϕcosθ,0⟩
Now add the ı^, ȷ^, k^ and some
vertical bars and complete the computation of the cross product:
N=eϕ×eθ=∣∣∣∣∣∣ı^⟨2cosϕcosθ⟨−2sinϕsinθȷ^2cosϕsinθ2sinϕcosθk^−2sinϕ⟩0⟩∣∣∣∣∣∣=ı^(0+4sin2ϕcosθ)−ȷ^(0−4sin2ϕsinθ)+k^(4sinϕcosϕcos2θ+4sinϕcosϕsin2θ)=⟨4sin2ϕcosθ,4sin2ϕsinθ,4sinϕcosϕ⟩
Notice that N points in the same direction as the position vector
since N=2sinϕR(ϕ,θ). This should be expected
since the position vector from the origin to the sphere is perpendicular
to the surface.
This result is the normal vector. Its magnitude will
also be useful in doing integrals. So we compute:
∣N∣=(4sin2ϕcosθ)2+(4sin2ϕsinθ)2+(4sinϕcosϕ)2=16sin4ϕcos2θ+16sin4ϕsin2θ+16sin2ϕcos2ϕ=16sin4ϕ+16sin2ϕcos2ϕ=16sin2ϕ(sin2ϕ+cos2ϕ)=16sin2ϕ=4sinϕ
Notice this result resembles the Jacobian factor for spherical coordinates:
J=ρ2sinϕ. This is coincidence for the spherical geometry.
Find the tangent vectors, normal vector, and length of the normal vector
for the hyperbolic paraboloid z=x2−y2 discussed in a
previous exercise and parametrized by:
R(r,θ)=⟨rcosθ,rsinθ,r2(cos2θ−sin2θ)⟩
Answer
The tangent vectors are:
ereθ=⟨cosθ,sinθ,2r(cos2θ−sin2θ)⟩=⟨−rsinθ,rcosθ,r2(−4sinθcosθ)⟩
The normal vector is:
N=⟨−2r2cosθ,2r2sinθ,r⟩
Its length is:
∣N∣=4r4+r2
To find the tangent vectors, we take the derivative of the parametrization
with respect to the two parameters:
ereθ=∂r∂R=⟨cosθ,sinθ,2r(cos2θ−sin2θ)⟩=∂θ∂R=⟨−rsinθ,rcosθ,r2(−4sinθcosθ)⟩
To find the normal vector we compute the cross product of the two tangent
vectors: (On paper, don't recopy the tangent vectors; just add
ı^, ȷ^ and k^ and vertical bars.)
N=er×eθ=∣∣∣∣∣∣ı^cosθ−rsinθȷ^sinθrcosθk^2r(cos2θ−sin2θ)r2(−4sinθcosθ)∣∣∣∣∣∣=ı^(−4r2sin2θcosθ−2r2cos3θ+2r2cosθsin2θ)−ȷ^(−4r2sinθcos2θ+2r2sinθcos2θ−2r2sin3θ)+k^(rcos2θ+rsin2θ)=ı^(−2r2sin2θcosθ−2r2cos3θ)−ȷ^(−2r2sinθcos2θ−2r2sin3θ)+k^(r)=ı^(−2r2cosθ)−ȷ^(−2r2sinθ)+k^(r)=⟨−2r2cosθ,2r2sinθ,r⟩
Finally, its length is:
∣N∣=(2r2cosθ)2+(2r2sinθ)2+r2=4r4+r2
We can check the normal vector by computing its dot product with each
tangent vector:
N⋅erN⋅eθ=−2r2cos2θ+2r2sin2θ+2r2(cos2θ−sin2θ)=0=2r3cosθsinθ+2r3sinθcosθ+r3(−4sinθcosθ)=0
Find the normal vector, and length of the normal vector for the cylinder
x2+z2=4 discussed in two
previous exercises and parametrized by
R(θ,y)=⟨2cosθ,y,2sinθ⟩.
Answer
The normal vector is:
N=⟨−2cosθ,0,−2sinθ⟩
Its length is:
∣N∣=22
We previously found the tangent vectors:
eθey=∂θ∂R=⟨−2sinθ,0,2cosθ⟩=∂y∂R=⟨0,1,0⟩
Again, to find the normal vector we compute the cross product of the two
tangent vectors:
N=eθ×ey=∣∣∣∣∣∣ı^−2sinθ0ȷ^01k^2cosθ0∣∣∣∣∣∣=ı^(0−2cosθ)−ȷ^(0−0)+k^(−2sinθ)=⟨−2cosθ,0,−2sinθ⟩
Finally, its length is:
∣N∣=(2cosθ)2+0+(−2sinθ)2=8=22
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