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Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$h(x, y)=2 x^{2}+3 y^{2}$$

Short Answer

Expert verified
Answer: The function h(x, y) = 2x^2 + 3y^2 represents an elliptic paraboloid. Its domain is all real values of x and y, denoted as {(x, y) | x, y 鈭 鈩潁, and its range comprises all non-negative values of h(x, y), or {h(x, y) | h(x, y) 鈮 0}.

Step by step solution

01

Identify the type of surface

The given function is in the form of a quadratic function in two variables: $$h(x, y) = ax^2 + by^2$$ where a=2 and b=3. This is the equation of an elliptic paraboloid.
02

Determine the domain of the function

The domain of the function is the set of all possible values of x and y for which the function is defined. Since the equation has no restrictions on the values of x and y, they can take any real values. Therefore, the domain of the function is: $$\text{Domain} = \{(x, y) | x, y \in \mathbb{R}\}$$
03

Determine the range of the function

The range of the function is the set of all possible values of h(x, y) for which the function is defined. Since this is an elliptic paraboloid and it opens upwards, the minimum value of h(x, y) occurs when x = 0 and y = 0, that is h(0, 0) = 0. Since there is no maximum value for h(x, y), the range of the function is: $$\text{Range} = \{h(x, y) | h(x, y) \ge 0\}$$
04

Sketch the graph of the function

To sketch the graph of the function, follow these steps: 1. Since it is an elliptic paraboloid, start by drawing the z-axis vertically, with the x and y-axes on the horizontal plane. 2. The vertex of the elliptic paraboloid is at the origin (0, 0, 0). 3. As x and y take positive and negative values, the paraboloid curves upwards in each quadrant of the xy-plane. 4. The paraboloid is elongated along the x-axis since the coefficient of the x^2 term (a = 2) is smaller than the y^2 term (b = 3), making the paraboloid narrower along the y-axis. 5. Using the x and y values, sketch the elliptic paraboloid opening upwards, with a narrower shape along the y-axis and a wider shape along the x-axis. Using this information, you should now have a visual representation of the surface and its properties, including the domain and range.

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Key Concepts

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

Calculus
Calculus provides powerful tools for analyzing and understanding the behavior of various functions, particularly those involving rates of change and accumulation. While in single-variable calculus, we delve into the slopes of curves, areas under curves, and their applications, in multivariable calculus, these concepts are extended to functions of several variables, like in our exercise.

In the context of graphing surfaces such as the elliptic paraboloid given by the function \(h(x, y) = 2x^2 + 3y^2\), calculus aids in determining critical points, optimizing functions, and understanding how changes in the variables influence the shape and position of the surface. When sketching graphs in calculus, partial derivatives can also be used to determine the curvature at various points on the surface, aiding in the accurate rendering of its shape.
Quadratic functions in two variables
Quadratic functions in two variables form a fundamental class of functions studied in algebra and are represented generally by \( ax^2 + bxy + cy^2 + dx + ey + f \), where \(a, b, c\) are not all zero. If \(b = 0\) and the signs of \( a\) and \( c\) are the same, as in our exercise \(h(x, y) = 2x^2 + 3y^2\), the graph is an elliptic paraboloid.

Structure of an Elliptic Paraboloid

Unlike a simple parabola, an elliptic paraboloid extends in two dimensions and resembles a satellite dish. It's essentially a 3D shape formed by the revolution of a parabola around its central axis. Understanding the coefficients \(a\) and \(b\) in the equation helps explain the 'stretch' of the paraboloid along the \(x\)- and \(y\)-axes.
Domain and Range of a Function
The domain of a function consists of all the input values for which the function is defined, while the range is the set of all possible output values. In several variables, the domain is typically a subset of \( \mathbb{R}^2\) or \( \mathbb{R}^3\), as appropriate. For our function \(h(x, y) = 2x^2 + 3y^2\), the domain is all of \( \mathbb{R}^2\), because \(x\) and \(y\) can take any real values.

Considering the range, since \(h(x, y)\) produces a paraboloid opening upward, its smallest value is \(0\) at the origin. It increases without bound as \(x\) and \(y\) move away from the origin, leading to a range of \( [0, \infty) \). Understanding domain and range is crucial when analyzing functions, as it informs us about the limitations and possibilities of function values.
Graphing Surfaces
Graphing surfaces in multivariable calculus often requires a fusion of algebraic understanding and geometrical intuition. It's the process of visually representing the set of points in three-dimensional space that satisfy an equation in two variables. This visualization aids in conceptualizing complex functions and allows for a clearer understanding of their properties.

For the elliptic paraboloid \(h(x, y) = 2x^2 + 3y^2\), sketching involves plotting the vertex, the axis of symmetry, and several points on the surface by choosing values for \(x\) and \(y\) and calculating the corresponding \(z\)-values (here represented by \(h(x, y)\)). Then, one connects these points to reveal the general 'bowl' shape of the paraboloid. Proper sketching is important for grasping the nature of quadratic surfaces, such as understanding how they stretch in different directions based on their coefficients.

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Most popular questions from this chapter

Suppose you make monthly deposits of \(P\) dollars into an account that earns interest at a monthly rate of \(p \% .\) The balance in the account after \(t\) years is \(B(P, r, t)=P\left(\frac{(1+r)^{12 t}-1}{r}\right),\) where \(r=\frac{p}{100}\) (for example, if the annual interest rate is \(9 \%,\) then \(p=\frac{9}{12}=0.75\) and \(r=0.0075) .\) Let the time of investment be fixed at \(t=20\) years. a. With a target balance of \(\$ 20,000,\) find the set of all points \((P, r)\) that satisfy \(B=20,000 .\) This curve gives all deposits \(P\) and monthly interest rates \(r\) that result in a balance of \(\$ 20,000\) after 20 years. b. Repeat part (a) with \(B=\$ 5000, \$ 10,000, \$ 15,000,\) and \(\$ 25,000,\) and draw the resulting level curves of the balance function.

Use Lagrange multipliers to find these values. \(f(x, y, z)=(x y z)^{1 / 2}\) subject to \(x+y+z=1\) with \(x \geq 0\) \(y \geq 0, z \geq 0\)

Determine whether the following statements are true and give an explanation or counterexample. a. The domain of the function \(f(x, y)=1-|x-y|\) is \(\\{(x, y): x \geq y\\}\). b. The domain of the function \(Q=g(w, x, y, z)\) is a region in \(\mathbb{R}^{3}\). c. All level curves of the plane \(z=2 x-3 y\) are lines.

Looking ahead- tangent planes Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) .\) A point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=8-x y z=0 ; P(2,2,2)$$

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) In two dimensions, the motion of an ideal fluid (an incompressible and irrotational fluid) is governed by a velocity potential \(\varphi .\) The velocity components of the fluid, \(u\) in the \(x\) -direction and \(v\) in the \(y\) -direction, are given by \(\langle u, v\rangle=\nabla \varphi .\) Find the velocity components associated with the velocity potential \(\varphi(x, y)=\sin \pi x \sin 2 \pi y\)

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