Find the points of the paraboloid z=x^{2}+y^{2}-1 at which the normal line to the surface coincides with the line joining the origin to the point. What is the acute angle betwwen the normal and the z-axis at these point
To find the points of the paraboloid where the normal line coincides with the line joining the origin to the point, we need to find the normal vector to the surface and the vector from the origin to the point on the surface, and then check if they are parallel.
Let's start by finding the normal vector to the surface. The equation of the paraboloid is given by z = x^2 + y^2 - 1. In order to find the normal vector, we need to compute the partial derivatives of z with respect to x and y.
∂z/∂x = 2x
∂z/∂y = 2y
The normal vector N is the vector with components (∂z/∂x, ∂z/∂y, -1), which simplifies to (2x, 2y, -1).
Next, we need to find the vector from the origin to the point on the surface. Let's call this vector A, with components (x, y, z).
Since the line joining the origin to the point, and the normal line to the surface coincide at the points of interest, the vectors N and A must be parallel. This means they must be scalar multiples of each other.
Therefore, we have the following equations:
2x = kx
2y = ky
-1 = kz
where k is a scalar. Rearranging the equations, we get:
2x - kx = 0
2y - ky = 0
-1 - kz = 0
From the first two equations, we can factor out x and y respectively:
x(2 - k) = 0
y(2 - k) = 0
For these equations to hold, either x = 0 or y = 0, or both. Let's consider different cases:
Case 1: x = 0
If x = 0, then from the first equation, we have 0(2 - k) = 0, which is satisfied for all values of k. This means that for any value of k, the point on the surface with x = 0 will satisfy the condition.
Case 2: y = 0
If y = 0, then from the second equation, we have 0(2 - k) = 0, which is again satisfied for all values of k. This means that for any value of k, the point on the surface with y = 0 will satisfy the condition.
Case 3: x = 0 and y = 0
If both x and y are zero, then from the first and second equations, we have 0(2 - k) = 0 and 0(2 - k) = 0, which are satisfied for all values of k. This means that for any value of k, the point on the surface with x = 0 and y = 0 will satisfy the condition.
Summarizing the above cases, the points on the paraboloid where the normal line coincides with the line joining the origin to the point are:
1. (0, y, y^2 - 1) for any value of y.
2. (x, 0, x^2 - 1) for any value of x.
3. (0, 0, -1)
To find the acute angle between the normal and the z-axis at these points, we can calculate the dot product between the normal vector N = (2x, 2y, -1) and the unit vector along the z-axis, which is (0, 0, 1).
The dot product of two vectors A and B is given by A · B = |A| |B| cosθ, where θ is the angle between the two vectors.
In this case, since the z-component of the normal vector is -1, the dot product simplifies to -1 * 1 * cosθ = -cosθ.
To find the angle θ, we can take the inverse cosine of the dot product:
θ = cos^(-1)(-cosθ)
Since we are only interested in the acute angle, we can take the absolute value of the result.
Therefore, the acute angle between the normal and the z-axis at these points is given by abs(cos^(-1)(-cosθ)), where θ is the angle between the two vectors.