Chapter 14: Problem 28
Check that the point (-1,1,2) lies on the given surface. Then, viewing the surface as a level surface for a function \(f(x, y, z),\) find a vector normal to the surface and an equation for the tangent plane to the surface at (-1,1,2). $$x^{2}-x y z=3$$
Short Answer
Expert verified
Point (-1,1,2) lies on the surface; the tangent plane is -4x + 2y + z = 8.
Step by step solution
01
Verify if Point Lies on Surface
We need to check if the point \((-1,1,2)\) satisfies the equation of the surface \(x^2 - xyz = 3\). Substitute \(x = -1\), \(y = 1\), and \(z = 2\) into the equation:\[-1^2 - (-1)(1)(2) = 1 + 2 = 3\].Since this equation holds true, the point \((-1,1,2)\) lies on the surface.
02
Define the Function for Level Surface
The surface given by \(x^2 - xyz = 3\) can be viewed as a level surface of the function \(f(x, y, z) = x^2 - xyz\) at the level \(c = 3\).
03
Compute Gradient (Normal) Vector
To find the normal vector, compute the gradient \(abla f(x, y, z)\). The gradient is given by:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\].Calculate each partial derivative:\[ \frac{\partial f}{\partial x} = 2x - yz, \quad \frac{\partial f}{\partial y} = -xz, \quad \frac{\partial f}{\partial z} = -xy \].Thus, the gradient is \(abla f(x, y, z) = (2x - yz, -xz, -xy)\).
04
Evaluate Gradient at Given Point
Substitute \((-1,1,2)\) into the gradient vector:\[abla f(-1, 1, 2) = (2(-1) - (1)(2), -(-1)(2), -(-1)(1)) = (-2 - 2, 2, 1) = (-4, 2, 1)\].So, \((-4, 2, 1)\) is the normal vector at the point \((-1,1,2)\).
05
Equation of Tangent Plane
The equation of the tangent plane at \((x_0, y_0, z_0)\) with normal vector \((a, b, c)\) is given by:\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\].Substituting \((-4, 2, 1)\) as the normal vector and \((-1, 1, 2)\) as the point:\[-4(x + 1) + 2(y - 1) + 1(z - 2) = 0\].Simplify the equation:\[-4x - 4 + 2y - 2 + z - 2 = 0\].Combine like terms:\[-4x + 2y + z = 8\].This is the equation of the tangent plane.
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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 vector calculus that plays a crucial role in a variety of applications, including optimizing functions and determining tangent planes. In the context of a function of several variables, the gradient is a vector that consists of all the partial derivatives with respect to each variable.
The gradient vector, often denoted as \( abla f \), provides the direction of steepest ascent. Here's how it works:
The gradient vector, often denoted as \( abla f \), provides the direction of steepest ascent. Here's how it works:
- If you have a function \( f(x, y, z) \), the gradient is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
- By computing these partial derivatives, you create a vector field that points in the direction where \( f(x, y, z) \) increases most rapidly.
- \( \frac{\partial f}{\partial x} = 2x - yz \)
- \( \frac{\partial f}{\partial y} = -xz \)
- \( \frac{\partial f}{\partial z} = -xy \)
Normal Vector
The normal vector is an incredibly important tool in understanding the geometry of surfaces. It is a vector that is perpendicular to a surface at a particular point. In this exercise, the normal vector is derived from the gradient of the function representing the surface.
For a level surface defined by a function \( f(x, y, z) \), the normal vector at any point \((x_0, y_0, z_0)\) on the surface can be found by evaluating the gradient vector \( abla f(x_0, y_0, z_0) \).
For a level surface defined by a function \( f(x, y, z) \), the normal vector at any point \((x_0, y_0, z_0)\) on the surface can be found by evaluating the gradient vector \( abla f(x_0, y_0, z_0) \).
- In the problem, we calculated \( abla f(-1, 1, 2) = (-4, 2, 1) \).
- This vector \((-4, 2, 1)\) is perpendicular to the surface at the point \((-1, 1, 2)\).
Level Surface
A level surface is a simple yet vital concept in multivariable calculus, widely used to describe sets of points that satisfy certain conditions. Consider a function \( f(x, y, z) \). A level surface is composed of all the points where the function takes on a constant value, like \( f(x, y, z) = c \).
For the exercise, the surface equation \( x^2 - xyz = 3 \) can be treated as a level surface where the function \( f(x, y, z) = x^2 - xyz \) equals the constant \( 3 \).
For the exercise, the surface equation \( x^2 - xyz = 3 \) can be treated as a level surface where the function \( f(x, y, z) = x^2 - xyz \) equals the constant \( 3 \).
- This means every point on this surface satisfies \( f(x, y, z) = 3 \).
Partial Derivative
Partial derivatives are the building blocks of calculus when dealing with functions of multiple variables. Each partial derivative measures how a function changes as one variable changes while holding the others constant.
Taking the partial derivative with respect to a variable gives insight into how the function behaves along one of its axes.
Taking the partial derivative with respect to a variable gives insight into how the function behaves along one of its axes.
- For a function \( f(x, y, z) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), examines how the function changes as \( x \) changes while \( y \) and \( z \) stay fixed.
- In the exercise, we computed the partial derivatives of \( f(x, y, z) = x^2 - xyz \) to find the gradient.
- The partial derivatives were \( 2x - yz \), \( -xz \), and \( -xy \) for \( x \), \( y \), and \( z \), respectively.