Chapter 13: Problem 25
Show that every line that is normal to the sphere $$ x^{2}+y^{2}+z^{2}=1 $$ passes through the origin.
Short Answer
Expert verified
Each normal line to the sphere passes through the origin, as shown by setting \(t = -\frac{1}{2}\) in the line equations.
Step by step solution
01
Understand the Problem
A normal line to a surface at a point is perpendicular to the tangent plane of the surface at that point. Here, we are asked to show that a line which is normal to the sphere, given by \(x^{2} + y^{2} + z^{2} = 1\), passes through the origin.
02
Implicit Differentiation via Gradient
The gradient \(abla f\) of the function \(f(x, y, z) = x^{2} + y^{2} + z^{2} - 1\) is used to find the normal vector to the surface. The gradient is \(abla f = (2x, 2y, 2z)\), giving the direction of the normal line at any point \((x_0, y_0, z_0)\) on the sphere.
03
Forming the Normal Line Equation
The equation of the line that passes through a point \((x_0, y_0, z_0)\) with a direction vector \((a, b, c)\) is: \[x = x_0 + at \y = y_0 + bt \z = z_0 + ct\] Using \((a, b, c) = (2x_0, 2y_0, 2z_0)\), the line becomes: \[x = x_0 + 2x_0 t \y = y_0 + 2y_0 t \z = z_0 + 2z_0 t \] Simplifying gives: \[x = x_0(1 + 2t) \y = y_0(1 + 2t) \z = z_0(1 + 2t) \]
04
Showing the Line Passes Through the Origin
At \(t = -\frac{1}{2}\), the line equations become:\[x = x_0(1 - 1) = 0 \y = y_0(1 - 1) = 0 \z = z_0(1 - 1) = 0 \] Thus, the point at \(t = -\frac{1}{2}\) on the line is \((0, 0, 0)\), meaning the line passes through the origin.
05
Conclusion
Every line normal to the surface of the sphere \(x^{2} + y^{2} + z^{2} = 1\) passes through the origin. This is because when you reduce the line equation parameter \(t\) by \(\frac{1}{2}\), it always results in the origin coordinates \((0, 0, 0)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Vector
The gradient vector is a fundamental concept in calculus, specifically in multivariable calculus. It offers a way to identify the steepest ascent direction on the surface of a function. The gradient vector is composed of the partial derivatives of the function with respect to each variable. For a function described by three variables, such as \( f(x, y, z) \), the gradient is denoted as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). This vector points in the direction where the function increases most rapidly.
In the case of a sphere defined by \( x^2 + y^2 + z^2 = 1 \), which can be a function \( f(x, y, z) = x^2 + y^2 + z^2 - 1 \), the gradient is calculated as \( abla f = (2x, 2y, 2z) \). This vector is crucial because it indicates the direction of the normal line, which is perpendicular to the surface.
In the case of a sphere defined by \( x^2 + y^2 + z^2 = 1 \), which can be a function \( f(x, y, z) = x^2 + y^2 + z^2 - 1 \), the gradient is calculated as \( abla f = (2x, 2y, 2z) \). This vector is crucial because it indicates the direction of the normal line, which is perpendicular to the surface.
- The gradient vector serves as a normal vector at any point on the surface.
- It can be considered a general normal that applies to surfaces defined by level sets of functions.
Implicit Differentiation
Implicit differentiation is a technique used in calculus to differentiate equations where the function is not isolated on one side of the equation. It's particularly useful when dealing with equations like \( x^2 + y^2 + z^2 = 1 \), where variables are intermixed and a direct solution for one variable is not feasible. Instead, we take derivatives with respect to an external variable, often denoted as \( t \), when variables are considered dependent.
In exercises involving a sphere, implicit differentiation helps us find the derivatives of the variables when combined in an equation with a constant, such as a sphere's equation. This comes into play when we need to find the equation of a line or a tangent by differentiating both sides of the equation concerning the chosen parameters.
In exercises involving a sphere, implicit differentiation helps us find the derivatives of the variables when combined in an equation with a constant, such as a sphere's equation. This comes into play when we need to find the equation of a line or a tangent by differentiating both sides of the equation concerning the chosen parameters.
- Helps to differentiate both sides of a non-separable equation.
- Often required when dealing with curves or surfaces where variables are not expressed explicitly.
Equation of a Line
The equation of a line is a basic component of analytical geometry. It can represent a straight line in space extending in two directions from a point. For lines in three dimensions, it involves a direction vector and a point through which the line passes.
The general formula for the equation of a line in space is \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector. This parameter \( t \) represents any real number, allowing the line to extend infinitely in both directions.
In the context of a normal line to a sphere like \( x^2 + y^2 + z^2 = 1 \), the gradient vector provides the direction \((2x_0, 2y_0, 2z_0)\). Hence, the line equation incorporates these directions:
The general formula for the equation of a line in space is \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is the direction vector. This parameter \( t \) represents any real number, allowing the line to extend infinitely in both directions.
In the context of a normal line to a sphere like \( x^2 + y^2 + z^2 = 1 \), the gradient vector provides the direction \((2x_0, 2y_0, 2z_0)\). Hence, the line equation incorporates these directions:
- Defines a path with specific start and direction in three dimensions.
- Used to verify conditions such as passing through a particular point, like the origin.
- Crucial for understanding intersections and tangency-related problems.