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 points?
Ah, the paraboloid wants to have a little chat about its normal lines. Well, let's humor it, shall we?
To find the points at which the normal line coincides with the line joining the origin to the point, we need to find the values of x, y, and z that satisfy this condition. And by satisfy, I mean make everyone's lives easier.
So, we start with the equation of the paraboloid: z = x^2 + y^2 - 1. Now, we need to find the gradient vector of this surface, which gives us the direction of the normal. Since the gradient vector is orthogonal to the tangent plane, we can use this to find the normal.
Taking the partial derivatives of the equation with respect to x and y, we get:
∂z/∂x = 2x
∂z/∂y = 2y
Now, we need to take into account that the normal line coincides with the line joining the origin to the point. This means that the direction ratios of the line are the same as the direction ratios of the normal vector. In simpler terms, (x, y, z) = (kx, ky, kz).
Substituting these values into the partial derivatives, we get:
2kx = 2x
2ky = 2y
Simplifying these equations, we find that k = 1, which means x = y. Now, we can substitute this value back into the equation of the paraboloid to get:
z = x^2 + x^2 - 1
z = 2x^2 - 1
So, the points at which the normal line coincides with the line joining the origin to the point are given by (x, y, z) = (x, x, 2x^2 - 1).
Now, onto the acute angle between the normal and the z-axis. Well, since we have the direction ratios of the normal vector, we can simply take the dot product of the normal vector and the z-axis vector, which is (0, 0, 1).
The dot product of these vectors is given by:
0(0) + 0(0) + (2x^2 - 1)(1) = 2x^2 - 1
Now, we know that the dot product of two vectors equals the product of their magnitudes and the cosine of the angle between them. So, we can set up an equation:
(2x^2 - 1) = |N| * |Z| * cos(theta)
Since |N| = sqrt(1 + 1 + (2x^2 - 1)^2) = sqrt(4x^4 + 4x^2) = 2|x|(x^2 + 1) and |Z| = 1, we can simplify the equation to:
2x^2 - 1 = 2|x|(x^2 + 1) * cos(theta)
Now, we can solve for theta, the acute angle between the normal and the z-axis:
cos(theta) = (2x^2 - 1) / (2|x|(x^2 + 1))
And there you have it! Use this equation to find the acute angle between the normal and the z-axis at the given points of the paraboloid. Enjoy the geometric excitement!
To find the points on the paraboloid where the normal line coincides with the line joining the origin to the point, we need to find the normal vectors and the direction vector of the line connecting the origin to the point.
First, let's find the normal vector of the paraboloid:
The equation of the paraboloid is z = x^2 + y^2 - 1.
Taking the partial derivatives, we have:
∂z/∂x = 2x
∂z/∂y = 2y
Therefore, the normal vector N = (2x, 2y, -1).
Next, let's find the equation of the line connecting the origin to the point (x, y, z):
The direction vector of the line connecting the origin to the point (x, y, z) is simply (x, y, z).
Since the normal line coincides with this line, the normal vector N is proportional to the direction vector. This means that the ratio of corresponding components is constant:
(2x) / x = (2y) / y = (-1) / z
From the first ratio, we have:
2 = 1 / (x/z), which simplifies to:
2(x/z) = 1
Solving this equation gives us:
x/z = 1/2
From the second ratio, we have:
2 = 1 / (y/z), which simplifies to:
2(y/z) = 1
Solving this equation gives us:
y/z = 1/2
Combining these results, we have:
x/z = 1/2
y/z = 1/2
To find the actual values of x, y, and z, we can substitute these ratios back into the equation of the paraboloid:
(x/z)^2 + (y/z)^2 - 1 = 0
Substituting the values we found, we have:
(1/2)^2 + (1/2)^2 - 1 = 0
Simplifying, we get:
1/4 + 1/4 - 1 = 0
Combining like terms, we have:
1/2 - 1 = 0
Simplifying further, we get:
-1/2 = 0
Since this is not a true statement, there are no points on the paraboloid where the normal line coincides with the line joining the origin to the point.
Therefore, there are no points for which we can find the acute angle between the normal and the z-axis.
To find the points where the normal line to the surface coincides with the line joining the origin to the point, we can set up the equation for the normal line and equate it to the equation for the line joining the origin to the point on the paraboloid.
Let's start by finding the equation for the normal line to the paraboloid z = x^2 + y^2 - 1.
The equation for the surface normal to a function of two variables z = f(x, y) is given by the gradient:
∇f(x, y) = (∂f/∂x, ∂f/∂y)
For our paraboloid, the gradient is:
∇f(x, y) = (2x, 2y)
This means that the normal vector at any point (x, y, z) on the paraboloid is (2x, 2y, -1).
Now, let's find the equation for the line joining the origin to a point (x, y, z) on the paraboloid. The parametric equations for this line are:
x = tx
y = ty
z = tz
where t is a parameter that varies along the line.
Since the line coincides with the normal line to the surface at the point (x, y, z), we can equate the corresponding components:
2x = tx (equation 1)
2y = ty (equation 2)
-1 = tz (equation 3)
From equation 3, we can solve for t:
t = -1/z
Substituting this value of t into equations 1 and 2, we get:
2x = -x/z
2y = -y/z
These equations can be simplified to:
x/z = -1/2
y/z = -1/2
From these equations, we can see that x = -z/2 and y = -z/2.
Substituting these values into the equation of the paraboloid z = x^2 + y^2 - 1, we get:
z = (-z/2)^2 + (-z/2)^2 - 1
Simplifying this equation, we get:
z = 5z^2/4 - 1
Rearranging terms, we have:
5z^2/4 - z - 1 = 0
Now, we solve this quadratic equation to find the possible values of z. By solving for z, we can find the corresponding values of x and y.
Once we have the points on the paraboloid, we can find the normal vector at each point, which is (2x, 2y, -1). To find the acute angle between this normal vector and the z-axis, we can use the dot product formula:
cos(theta) = (N · V) / (|N| |V|)
Where N is the normal vector and V is the vector pointing along the z-axis, which is (0, 0, 1). The dot product N · V can be computed by multiplying the corresponding components of the two vectors and adding them up. The magnitudes |N| and |V| can be computed using the formula sqrt(x^2 + y^2 + z^2) for each vector.
Substituting the values into the formula, we can compute the acute angle theta.