Respuesta :
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
Answer:
The unit vector is :
[tex]u=0.873i+0.436j-0.218k[/tex]
Step-by-step explanation:
Given information :
Surface equation : [tex]z^2-2x^4-y^4=16[/tex]
Point [tex](P)=(2,2,8)[/tex]
Now, the normal is given by the directional derivative ([tex]\Delta[/tex])
Hence, [tex]\Delta = \frac{d}{dx}i+ \frac{d}{dy}j+\frac{d}{dz}k[/tex]
So, for the given function we can write :
[tex]\Delta=\frac{d}{dx}(-2x^4)+ \frac{d}{dy}(-y^4)+\frac{d}{dx}(z^2)[/tex]
[tex]\Delta=-8x^3 i-4y^3j+2z[/tex]
Now, putting the values of point p in above derivative
We get the normal vector;
[tex]n=-64i-32j+16k[/tex]
by multiplying (-1) the normal vector in xy plane will be
[tex]n=64i+32j-16k[/tex]
Now for the unit vector we need:
[tex]|n| = \sqrt (64^2+32^2+16^2)[/tex]
[tex]|n| = 73.3212[/tex]
Now , the unit vector :
[tex]u=n/|n|\\[/tex]
[tex]u=\frac{64i+32j-16k}{73.3212}[/tex]
Hence, the unit vector is :
[tex]u=0.873i+0.436j-0.218k[/tex]
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