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Find an equation of the tangent plane to the surface \(z=x^{2}-y^{2}\) at the point \(\mathbf{a}=(1,2,-3)\).

Short Answer

Expert verified
The equation of the tangent plane is \( z = 2x - 4y + 3 \).

Step by step solution

01

Understand the Problem

We need to find the equation of the tangent plane to a given surface at a specified point. The surface is represented by the equation \( z = x^2 - y^2 \), and we are interested in the point \( \mathbf{a} = (1, 2, -3) \).
02

Find the Partial Derivatives

To find the tangent plane, we start by determining the partial derivatives of the surface's equation. The partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 2x \), and the partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = -2y \).
03

Evaluate the Partial Derivatives at the Point

Next, we evaluate these partial derivatives at the point \( (1, 2) \). We obtain: \[ \frac{\partial z}{\partial x}(1, 2) = 2 \times 1 = 2 \] \[ \frac{\partial z}{\partial y}(1, 2) = -2 \times 2 = -4 \]
04

Write the Equation of the Tangent Plane

The general equation of the tangent plane at a point \( (x_0, y_0, z_0) \) is given by: \[ z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0) \]Substituting the point \( (1, 2, -3) \) and the evaluated partial derivatives, we get: \[ z + 3 = 2(x - 1) - 4(y - 2) \]
05

Simplify the Tangent Plane Equation

Simplify the equation: \[ z + 3 = 2x - 2 - 4y + 8 \] \[ z + 3 = 2x - 4y + 6 \] Subtract 3 from both sides to obtain: \[ z = 2x - 4y + 3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are an important concept in calculus, especially when dealing with functions of multiple variables. When you have a surface represented by a function like \( z = x^2 - y^2 \), you can think of it as a mountain range or a landscape where the height is determined by \( x \) and \( y \) coordinates.To find out how the surface changes as you move in the \( x \) direction or the \( y \) direction, we use partial derivatives:- The partial derivative with respect to \( x \), denoted \( \frac{\partial z}{\partial x} \), shows the rate of change of \( z \) as \( x \) varies, while keeping \( y \) constant.- Similarly, the partial derivative with respect to \( y \), denoted \( \frac{\partial z}{\partial y} \), shows the rate of change of \( z \) as \( y \) varies, keeping \( x \) constant.In the given exercise, the partial derivatives at the point \((1, 2)\) reveal the slope of the surface at that specific location. Evaluating these partial derivatives gives us the necessary information to describe how steep the surface is in these directions at the point \((1, 2, -3)\).
The partial derivative values help us construct the equation of the tangent plane at this point.
Equation of Surface
The equation of a surface is essentially a mathematical representation that describes all the points making up that surface. In our example, the surface is described by the function \( z = x^2 - y^2 \). This equation tells us that, for any given \( x \) and \( y \), we can determine \( z \), which represents the height at that point.When working with these equations, it is essential to understand:- The surface is defined by all points \((x, y, z)\) that satisfy the equation.- For a given \( x \) and \( y \), \( z \) can be considered as the dependent variable, calculated from \( x^2 - y^2 \).In the context of finding a tangent plane, the equation of the surface provides the base information. We use the function and calculate its partial derivatives to determine how the surface tilts or curves at specific points. This 'tilt' is what allows us to write the equation of the tangent plane at a point. The tangent plane provides a linear approximation of the surface near the point of tangency, enabling us to understand and predict the surface's local behavior.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable. Instead of dealing with lines and curves in a single plane like in single-variable calculus, you deal with surfaces and volumes, which introduces new tools like partial derivatives and gradients to analyze them.Key concepts in multivariable calculus include:- The study of functions of two or more variables, such as \( f(x, y) \).- Understanding how these functions change when their inputs change, which can be represented by partial derivatives.- Using these derivatives to solve real-world problems, such as optimizing surfaces or finding tangent planes, like in our example.In the exercise, multivariable calculus allows us to find the tangent plane to a surface, providing a local linear approximation of the surface. This capability is important in fields such as physics, engineering, and economics, where you often need to predict and understand behavior in a multi-dimensional space.

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Most popular questions from this chapter

Complete the description of the behavior of the second-order approximation near a critical point a by considering the case that \(\operatorname{det} H(\mathbf{a}) \neq 0\) and \(A=0\), where \(H(\mathbf{a})=\left[\begin{array}{ll}A & B \\ B & C\end{array}\right]\). (a) Show that \(\operatorname{det} H(\mathbf{a})<0\). (b) Show that the quadratic term \(A h^{2}+2 B h k+C k^{2}=2 B h k+C k^{2}\) of the second-order approximation can be written as a difference of two perfect squares.

Let \(S\) be the ellipsoid \(x^{2}+\frac{y^{2}}{2}+\frac{z^{2}}{4}=4 .\) Let \(\mathbf{a}=(1,2,2),\) and let \(\mathbf{n}\) be the unit normal vector to \(S\) at a that points outward from \(S\). If \(f(x, y, z)=x y z,\) find the directional derivative \(\left(D_{\mathbf{n}} f\right)(\mathbf{a})\)

The temperature in a certain region of space, in degrees Celsius, is modeled by the function \(T(x, y, z)=20 e^{-x^{2}-2 y^{2}-4 z^{2}},\) where \(x, y, z\) are measured in meters. At the point \(\mathbf{a}=(2,-1,3)\) (a) In what direction is the temperature increasing most rapidly? (b) In what direction is it decreasing most rapidly? (c) If you travel in the direction described in part (a) at a speed of 10 meters/second, how fast is the observed temperature changing at a in degrees Celsius per second?

Find all four second-order partial derivatives of the given function \(f\). $$ f(x, y)=\frac{y}{x+y} $$

Find the directional derivative of \(f\) at the point a in the direction of the vector \(\mathbf{v}\). $$ f(x, y, z)=x^{3}-2 x^{2} y z+x z-3, \mathbf{a}=(1,0,-1), \mathbf{v}=(1,-1,2) $$

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