What is the shortest distance from the surface xy+6x+z^2=33
to the origin?
To find the shortest distance from the surface to the origin, we can use the formula for the distance between a point and a plane.
The given equation of the surface can be rewritten as:
z^2 = 33 - xy - 6x
We can treat this equation as the equation of a function f(x,y) = 33 - xy - 6x.
To find the shortest distance, we need to find the point (x, y, z) on the surface that is closest to the origin (0, 0, 0). This can be done by finding the partial derivatives of f(x, y) and setting them equal to zero.
∂f/∂x = -y - 6 = 0
∂f/∂y = -x = 0
From the second equation, we can solve for x and get x = 0. Substituting this result into the first equation, we get -y - 6 = 0, which gives y = -6.
Substituting x = 0 and y = -6 into the equation of the surface, we can solve for z:
z^2 = 33 - (0)(-6) - 6(0)
z^2 = 33
Taking the square root of both sides, we get z = ±√33.
Therefore, there are two points on the surface that are equidistant from the origin: (0, -6, √33) and (0, -6, -√33).
To find the shortest distance, we can calculate the distance between each of these points and the origin using the distance formula:
d = sqrt(x^2 + y^2 + z^2)
For the first point (0, -6, √33), the distance is:
d = sqrt(0^2 + (-6)^2 + (√33)^2)
= sqrt(0 + 36 + 33)
= sqrt(69)
For the second point (0, -6, -√33), the distance is:
d = sqrt(0^2 + (-6)^2 + (-√33)^2)
= sqrt(0 + 36 + 33)
= sqrt(69)
Therefore, the shortest distance from the surface to the origin is sqrt(69).
To find the shortest distance from the surface xy+6x+z^2=33 to the origin, we will use the concept of vector projection and the gradient of the surface equation.
1. Rewrite the equation of the surface in the form:
xy + 6x + z^2 = 33
as
z^2 = 33 - xy - 6x
2. Find the gradient of the surface equation:
Gradient ∇(z^2) = (∂(33 - xy - 6x)/∂x, ∂(33 - xy - 6x)/∂y, ∂(33 - xy - 6x)/∂z)
= (-y - 6, -x, 0)
3. The normal vector N to the surface at any point (x, y, z) is the negative of the gradient:
N = (-(-y - 6), -(-x), 0)
= (y + 6, x, 0)
4. The vector from the origin (0, 0, 0) to any point (x, y, z) on the surface is:
r = (x, y, z)
5. The distance from the origin to the surface is given by the projection of the vector r onto the normal vector N:
Distance = |r| * cos(θ)
= |r| * (r · N) / |r|
6. To find the shortest distance, we need to find the point on the surface where the projection is minimum. This occurs when the vector r is parallel to the normal vector N.
7. Since N = (y + 6, x, 0), we can set x = y + 6 to make the vector r parallel to N.
8. Substituting x = y + 6 into the equation of the surface z^2 = 33 - xy - 6x, we get:
z^2 = 33 - (y + 6)y - 6(y + 6)
= 33 - y^2 - 6y - 6y - 36
= -y^2 - 12y - 3
9. Take the derivative of z^2 with respect to y and set it equal to 0 to find the critical point:
dz^2/dy = -2y - 12 = 0
-2y = 12
y = -6
10. Substituting y = -6 back into x = y + 6, we get:
x = -6 + 6
= 0
11. Similarly, substituting y = -6 into z^2 = -y^2 - 12y - 3, we get:
z^2 = -(-6)^2 - 12(-6) - 3
= -36 + 72 - 3
= 33
12. Therefore, the point (x, y, z) on the surface closest to the origin is (0, -6, √33).
13. The shortest distance from the surface to the origin is the magnitude of the vector OP, where O is the origin and P is the point (0, -6, √33):
Distance = √(0^2 + (-6)^2 + √33^2)
= √(0 + 36 + 33)
= √69
≈ 8.31
Therefore, the shortest distance from the surface xy+6x+z^2=33 to the origin is approximately 8.31 units.