/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 sketch the curve \(f(x, y)=c\) t... [FREE SOLUTION] | 91Ó°ÊÓ

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sketch the curve \(f(x, y)=c\) together with \(\nabla f\) and the tangent line at the given point. Then write an equation for the tangent line. \begin{equation}x y=-4, \quad(2,-2)\end{equation}

Short Answer

Expert verified
Tangent line: \( y = x - 2 \).

Step by step solution

01

Identify the Function and Level Curve

The given equation is \( x y = -4 \). This can be seen as a level curve of the function \( f(x, y) = x y \) with a constant \( c = -4 \). At the point \( (2, -2) \), we have \( x \cdot y = 2 \cdot (-2) = -4 \), satisfying the equation.
02

Find the Gradient \( \nabla f \)

The gradient of the function \( f(x, y) = xy \) is \( abla f = (y, x) \). This implies \( abla f(x, y) = (-2, 2) \) at the point \( (2, -2) \). The gradient vector is perpendicular to the level curve at the given point.
03

Equation of the Tangent Line

The tangent line at a point can be calculated using the formula \( (x_1, y_1) \cdot abla f = 0 \), considering \( abla f \) as the normal vector. So, \( (x-2)(-2) + (y+2)(2) = 0 \). Simplifying, we get \(-2x + 2y + 4 = 0\). Dividing by 2, the equation of the tangent line is \(-x + y + 2 = 0\) or \( y = x - 2 \).
04

Sketch the Curve and Components

For visualizing: plot the hyperbola \( xy = -4 \). At \( (2, -2) \), draw the tangent line \( y = x - 2 \) and show the gradient vector \( (-2, 2) \). The gradient points perpendicular to the tangent line and indicates the direction of steepest ascent.

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Key Concepts

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

Level Curve
In multivariable calculus, the concept of a level curve is an essential tool that helps us understand functions of two variables. A level curve is essentially a set of points in the plane where the function has a constant value. This means if you have a function like \( f(x, y) \), the level curve of that function is the set of points \( (x, y) \) that satisfy the equation \( f(x, y) = c \) for some constant \( c \).

In the provided exercise, we dealt with the level curve \( xy = -4 \). This implies that at every point on this curve, the product of the coordinates \( x \) and \( y \) is always \(-4\). Level curves are helpful because they create a map that can be visualized like contour lines on a geographic topographic map. Each line represents a constant height. Here, instead of height, it's a constant value of the function.

Understanding level curves allows us to envisage the behavior of functions across the plane and how they relate to each other.
Gradient Vector
The gradient vector plays a pivotal role in understanding multivariable functions. It is essentially a vector that captures both the direction and the rate of the steepest increase of a function. For a function \( f(x, y) \), the gradient, denoted as \( abla f \), is a vector composed of the function's partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

In the exercise, we found the gradient of the function \( f(x, y) = xy \). The gradient is \( abla f = (y, x) \). At the point \( (2, -2) \), this evaluates to \( (-2, 2) \). The gradient is perpendicular to the level curve at any given point, providing crucial information about the function's local behavior.
  • The gradient points in the direction of the greatest increase.
  • The magnitude of the gradient vector shows how steep the increase is.
  • It's perpendicular to the level curves, forming a kind of cross-section of the surface.
Understanding the gradient helps in analyzing how functions change, and is instrumental in optimization problems.
Tangent Line
The tangent line is a fundamental concept often used to approximate a curve at a given point. It touches the curve precisely at one point and has the same slope as the curve does at that point. In multivariable calculus, the tangent line can be derived using the gradient vector.

For a curve defined by a level curve, the gradient vector at any point on the curve acts as a normal vector since it is perpendicular. In our exercise, we calculated the tangent line by using the condition \( (x_1 - x_0) abla f_x + (y_1 - y_0) abla f_y = 0 \). This condition basically ensures that the line is orthogonal to the gradient vector.

The equation for the tangent line came out to be \( y = x - 2 \) at the point \( (2, -2) \). This line gives us an approximation of the hyperbola near the point where it's calculated. Remember, the tangent line provides linear guidance on where the curve is heading immediately around the point of tangency.
Hyperbola
A hyperbola is a type of conic section that appears as an open curve with two distinct branches. It occurs naturally when slicing a double cone with a plane in a way that results in these disconnected curves. In mathematics and its applications, hyperbolas have unique properties that make them useful in various scenarios like astronomy and navigation.

In the context of our exercise, the hyperbola is visible in the level curve \( xy = -4 \). This specific hyperbolic shape is a result of the multiplication operation constrained to equal a constant, providing a typical example of a rectangular hyperbola. The general equation for a rectangular hyperbola is \( xy = c \), where \( c \) is some constant.
  • The asymptotes of a rectangular hyperbola are the axes \( x \) and \( y \), meaning its branches grow ever closer to these lines but never intersect them.
  • Its symmetry showcases an interesting balance within the graph, splitting it neatly into two halves.
  • Hyperbolas provide clear examples of how nonlinear relationships manifest visually.
Understanding hyperbolas helps bring insight into phenomena that exhibit exponential or inverse properties, broadening comprehension of real-world patterns.

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Most popular questions from this chapter

In Exercises \(1-3,\) begin by drawing a diagram that shows the relations among the variables. If \(w=x^{2}+y-z+\sin t\) and \(x+y=t,\) find $$ \begin{array}{lll}{\text { a. }\left(\frac{\partial w}{\partial y}\right)_{x, z}} & {\text { b. }\left(\frac{\partial w}{\partial y}\right)_{z, t}} & {\text { c. }\left(\frac{\partial w}{\partial z}\right)_{x, y}} \\ {\text { d. }\left(\frac{\partial w}{\partial z}\right)_{y, t}} & {\text { e. }\left(\frac{\partial w}{\partial t}\right)_{x, z}} &{\text { f. }\left(\frac{\partial w}{\partial t}\right)_{y, z}}\end{array} $$

In Exercises \(71-74,\) find a function \(z=f(x, y)\) whose partial derivatives are as given, or explain why this is impossible. $$\frac{\partial f}{\partial x}=3 x^{2} y^{2}-2 x, \quad \frac{\partial f}{\partial y}=2 x^{3} y+6 y$$

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$ \begin{array}{l}{m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}} \\\ {b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right)}\end{array} $$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically.

Show that each function in Exercises \(83-90\) satisfies a Laplace equation. $$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}$$

In Exercises \(71-76\) , you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$ \begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3,-3 / 2 \leq x \leq 3 / 2} , {-3 / 2 \leq y \leq 3 / 2}\end{array} $$

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