Question

Find the equation of the tangent plane to the given surface at the indicated point. \(x^{2}+y^{2}+z^{2}=16 ;(2,3, \sqrt{3})\)

Short answer

The tangent plane equation is 4x + 6y + 2 root{3}z = 32.

Step-by-step solution

Identify the Surface Function

First, recognize that the given equation represents a surface function. We have the equation of a sphere: x^{2} + y^{2} + z^{2} = 16.

Verify the Point on the Surface

Ensure that the given point (2,3, oot{3}) satisfies the surface equation by substituting it into ymouriginal equation: 2^{2} + 3^{2} + ( oot{3})^{2} = 16. Both sides equal 16, so the point is indeed on the surface.

Determine the Normal Vector

The gradient of the surface function is the normal vector to the surface. Calculate the gradient ∇F(x, y, z) at any point (x, y, z): ∇F = (2x, 2y, 2z).

Evaluate the Gradient at the Point

Substitute the given point into the gradient to find the normal vector at that point: ∇F(2, 3, oot{3}) = (2×2, 2×3, 2× oot{3}) = (4, 6, 2 oot{3}).

Write the Equation of the Tangent Plane

The equation of the tangent plane using point-normal form is: ux - u_{0}x + u_y - u_{0}y + u_z-z_{0} = 0, where (u_{0}, v_{0}, w_{0}) is the given point and (ux_y, ux, ) is the normal vector. Substituting in our values: 4(x - 2) + 6(y - 3) + 2 oot{3}(z - root{3}) = 0.

Expand and Simplify the Tangent Plane Equation

Expand the equation: 4x - 8 + 6y - 18 + 2 oot{3}z - 6 = 0. Simplify it: 4x + 6y + 2 oot{3}z = 32.

Key concepts

Gradient Vector

The gradient vector is a key concept in multivariable calculus that gives us valuable information about a function. It is essentially a vector that points in the direction of the greatest rate of increase of the function. Mathematically, for a function of three variables like our sphere equation, the gradient vector \( abla F = ( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} ) \) involves partial derivatives with respect to each variable.
The sphere equation is \( x^2 + y^2 + z^2 = 16 \). The gradient vector for this function \( F(x, y, z) = x^2 + y^2 + z^2 \) is \( abla F = (2x, 2y, 2z) \). This equation gives us the specificity needed to describe the normal vector at any point on the surface.
For the point (2, 3, \( \sqrt{3} \)), substituting these values into the gradient gives us \( abla F(2, 3, \sqrt{3}) = (4, 6, 2\sqrt{3}) \). This is crucial for determining the orientation of the tangent plane.

Normal Vector

The normal vector is a vector perpendicular to the surface at any given point. In the context of a sphere, and many other surfaces, the gradient vector also acts as the normal vector because it points in the direction perpendicular to the surface.
For the sphere described by \( x^2 + y^2 + z^2 = 16 \), the normal vector at the point (2, 3, \( \sqrt{3} \)) is determined through the gradient \( abla F \). As calculated, the normal vector is \( (4, 6, 2\sqrt{3}) \).
This perpendicularity ensures that any plane passing through this point and oriented according to this normal vector is a tangent plane. Understanding normal vectors is fundamental because they define the plane's orientation in 3D space.

Tangent Planes

A tangent plane is a flat surface that just 'touches' a particular point on a curve or surface. In our example, it touches the sphere at one point and at this point, the plane is neither inside nor outside the sphere.
To form the equation of the tangent plane, we use the normal vector obtained. Recall the normal vector from the previous section: \( (4, 6, 2\sqrt{3}) \). The general equation of a plane in three dimensions can be written as:

• \( 4(x - 2) + 6(y - 3) + 2\sqrt{3}(z - \sqrt{3}) = 0 \)

By expanding and simplifying, our equation becomes \( 4x + 6y + 2\sqrt{3}z = 32 \).
This tangent plane example provides insight into the significant role of normal vectors in determining the plane's position and orientation.

Sphere Equation

A sphere is a perfectly symmetrical geometric object in three dimensions, and its equation defines all points that have the same distance from a central point, known as the center. The standard equation for a sphere with center at the origin is \( x^2 + y^2 + z^2 = r^2 \), where \( r \) is the radius.
In our problem, the equation \( x^2 + y^2 + z^2 = 16 \) describes a sphere centered at the origin with a radius of \( \sqrt{16} = 4 \).
This equation is critical for verifying that our given point, (2, 3, \( \sqrt{3} \)), lies on the sphere. Upon substitution into the sphere equation, the left side equals 16, confirming that the point is indeed on the sphere's surface.
Understanding this concept is vital for solving geometric problems involving tangents and normals to spheres.