Solved Examples on Tangent Plane to a Surface
Example 1: Find the equation of the tangent plane to the surface z = x2 + y2 at the point (1, 2, 5).
Solution:
To find the equation of the tangent plane, we need to find a point on the plane and a normal vector to the plane.
Point on the Plane: The point (1, 2, 5) lies on the surface z = x2 + y2, so it also lies on the tangent plane.
Normal Vector: The normal vector to the surface z = x2 + y2 is given by the gradient of the surface function (∇f = fx, fy, -1), where fx and fy are the partial derivatives of f with respect to x and y, respectively.
For the given surface z = x2 + y2, fx = 2x and fy = 2y. So, ∇f = (2x, 2y, -1).
Evaluating ∇f at the point (1, 2), we get ∇f(1, 2) = (2(1), 2(2), -1) = (2, 4, -1)
Now, we have a point on the plane (1, 2, 5) and a normal vector (2, 4, -1). Using the point-normal form of the equation of a plane, the equation of the tangent plane is:
2(x-1) + 4(y-2) – (z-5) = 0
Simplify to obtain the final equation of the tangent plane:
2x + 4y – z – 1 = 0
Example 2: Find the equation of the tangent plane to the surface z = 3x22 – y2 at the point (2, -1, 11).
Solution:
To find the equation of the tangent plane, we need to find a point on the plane and a normal vector to the plane.
Point on the Plane: The given point is (2, -1, 11), which lies on the surface z = 3x2 – y2, so it also lies on the tangent plane.
Normal Vector: The normal vector to the surface z = 3x2 – y2 is given by the gradient of the surface function (∇f = fx, fy, -1), where fx and fy are the partial derivatives of f with respect to x and y, respectively.
For the given surface z = 3x2 – y2, fx = 6x and fy = -2y. So, ∇f = (6x, -2y, -1).
Evaluating ∇f at the point (2, -1), we get ∇f(2, -1) = (6(2), -2(-1), -1) = (12, 2, -1)
Now, we have a point on the plane (2, -1, 11) and a normal vector (12, 2, -1).
Using the point-normal form of the equation of a plane, the equation of the tangent plane is:
12(x-2) + 2(y+1) – (z-11) = 0
Simplify to obtain the final equation of the tangent plane:
12x + 2y – z + 13 = 0
Example 3: Find the equation of the tangent plane to the surface z = xy + ex at the point (1, 0, 1).
Solution:
To find the equation of the tangent plane, we need to find a point on the plane and a normal vector to the plane.
Point on the Plane: The given point is (1, 0, 1), which lies on the surface z = xy + ex, so it also lies on the tangent plane.
Normal Vector: The normal vector to the surface z = xy + ex is given by the gradient of the surface function (∇f = fx, fy, -1), where fx and fy are the partial derivatives of f with respect to x and y, respectively.
For the given surface z = xy + ex, fx = y + ex and fy = x. So, ∇f = (y + ex, x, -1).
Evaluating ∇f at the point (1, 0), we get ∇f(1, 0) = (0 + e1, 1, -1) = (e, 1, -1)
Now, we have a point on the plane (1, 0, 1) and a normal vector (e, 1, -1). Using the point-normal form of the equation of a plane, the equation of the tangent plane is:
e(x-1) + (y-0) – (z-1) = 0
Simplify to obtain the final equation of the tangent plane:
ex + y – z – e + 1 = 0
Tangent Plane to a Surface
A tangent plane is a flat surface that touches a curve or surface at a single point, sharing the same slope or direction at that point, facilitating local approximation in calculus. This article discusses tangent planes, which are flat surfaces that touch curves or surfaces at specific points. It explains their definition, how to calculate them, and their geometric interpretation. It also explores their applications in various fields like engineering, physics, and computer graphics.
Table of Content
- Definition of Tangent Plane
- How to Find the Tangent Plane to a Surface
- Tangent Plane Equation
- Geometric Interpretation of the Tangent Plane
- Applications of Tangent Plane to a Surface
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