Answer:
u = 0.873 i + 0.436 j - 0.218 k
Step-by-step explanation:
Given:
- The given surface as follows:
z^2 - 2*x^4 - y^4 = 16
- The point P = ( 2 , 2 , 8 )
Find:
Find a unit vector n that is normal to the surface at point P such that it points in the direction of x-y plane.
Solution:
- Any vector that is normal to a surface in any coordinate system is given by its directional derivative ∀:
- Where ∀ is the partial derivatives of the function with respect to x, y , and z directions as follows:
∀ = d/dx i + d/dy j + d/dz k
Where, f is the given function as follows:
f ( x , y , z ) = z^2 - 2*x^4 - y^4 - 16
- The directional derivative is given as:
D_f = ∀.f = -8*x^3 i - 4*y^3 j + 2*z
- The above result is the normal vector that at any general point on the surface. So at point P the normal vector n would be:
n = -8*2^3 i - 4*2^3 j + 2*8 k
n = -64 i - 32 j + 16 k
- Now we have to check the direction of the normal vector whether it passes the x-y plane. For any vector to pass the x-y plane or z=0. Its kth component must be "negative" because as we along the vector the position of each point must be closer to x-y plane or approach z = 0. Hence, the normal vector in the direction of x-y plane is:
n = -1*( -64 i - 32 j + 16 k )
n = 64 i + 32 j -16 k
- The corresponding unit vector (u) of the normal vector (n) is given as :
u = n / | n |
Where, | n | is the magnitude of the normal vector:
| n | = sqrt ( 64^2 + 32^2 + 16^2)
| n | = sqrt (5376)
| n | = 73.3212
- The unit vector is given as:
u = 64 i + 32 j -16 k / 73.3212
u = 0.873 i + 0.436 j - 0.218 k
