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Chapter 3: Problem 3

For the functions \(f(x, y)\) : (a) sketch the cross-sections of the graph \(z=\) \(f(x, y)\) with the coordinate planes, (b) sketch several level curves of \(f,\) labeling each with the corresponding value of \(c,\) and (c) sketch the graph of \(f\). $$ f(x, y)=y-x^{2} $$

Short Answer

Expert verified
The graph includes cross-section parabolas, level curves of shifted parabolas, and a parabolic cylinder.

Step by step solution

01

Understanding the function

The function given is \( f(x, y) = y - x^2 \). This is a 3D surface where \(z = y - x^2\). Our goal is to sketch its cross-sections, level curves, and 3D graph.
02

Cross-sections with coordinate planes

To find the cross-sections, set either \(x\) or \(y\) constant. - **For the xz-plane** (setting \(y = 0\)): \(z = 0 - x^2 = -x^2\). This is a parabola opening downwards.- **For the yz-plane** (setting \(x = 0\)): \(z = y - 0^2 = y\). This is a linear function, a line with slope 1 through the origin.- **For the xy-plane** (setting \(z = 0\)): \(0 = y - x^2\), which simplifies to \(y = x^2\). This represents a parabola opening upwards.
03

Sketching level curves

To sketch level curves, set the function equal to a constant \(c\), i.e., \(z = c\).- Solve \(c = y - x^2\) for \(y\): \(y = x^2 + c\).These are parabolas opening upwards, shifted upwards by \(c\). For example, if \(c = 1\), you get \(y = x^2 + 1\).
04

Sketching the 3D surface

The 3D graph of \(z = y - x^2\) is a parabolic cylinder. - Recognize that outputs depend on \(x^2\) for a fixed \(y\), shifting the surface vertically by \(y\). Each cross-section parallel to the \(yz\)-plane is a linear function, and parallel to the \(xz\)-plane is a downward-opening parabola.

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Key Concepts

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

Cross-Sections
Cross-sections provide slices of a 3D surface by setting one of the variables to a constant value. When dealing with functions like \( f(x, y) = y - x^2 \), we find the cross-sections by considering each coordinate plane separately.
  • xz-plane: Set \( y = 0 \). The equation becomes \( z = -x^2 \). This results in a parabola that opens downward, resembling an inverted bowl. Imagine looking directly at this parabola from the front; you would notice that it dips below the origin because of the negative sign.
  • yz-plane: Set \( x = 0 \). Here, \( z = y \), forming a straight line with a slope of 1, cutting through the origin. In essence, this is a simple diagonal line, representing the relationship between \( y \) and \( z \) without any transformations.
  • xy-plane: Set \( z = 0 \). Rearranging gives \( y = x^2 \), a standard upward opening parabola. Viewed from above, this parabola sits open on the \( xy \)-plane, looking just like a regular bowl.
Cross-sections help us understand how the surface interacts with the respective planes, providing a foundational view of its shape in parts.
Level Curves
Level curves are crucial in visualizing an entire surface on a single plane. They represent contours where the function \( f(x, y) = y - x^2 \) maintains a constant value, say \( z = c \). By setting \( y - x^2 = c \), we can rearrange it to form \( y = x^2 + c \).
These are parabolas opening upwards, shifted upwards by varying amounts depending on the constant \( c \). Here are some insights:
  • If \( c = 0 \), our level curve is simply \( y = x^2 \).
  • If \( c = 1 \), the parabola shifts upward by 1 unit to \( y = x^2 + 1 \).
  • As \( c \) increases, each level curve represents a higher slice of the 3D surface. Conversely, negative \( c \) shifts the parabolas downward.
Level curves are akin to the lines on a topographic map, giving us a top-down view of the surface's height and shape variations.
3D Surface Sketching
Sketching a 3D surface can be daunting, but understanding the interaction of its parts makes it manageable. For the function \( z = y - x^2 \), the 3D graph resembles a parabolic cylinder. This implies certain characteristics:
  • Each vertical slice parallel to the \( yz \)-plane forms a straight line \( z = y \), evident from the linear relationship when \( x \) is constant. These lines vary vertically, creating planes of different heights aligned with changes in \( y \).
  • Horizontal slices parallel to the \( xz \)-plane give downward-facing parabolas \( z = -x^2 \). As \( y \) changes, these parabolas stack alongside each other, modifying the surface gradient across \( x \).
Visualizing or sketching these features helps in interpreting complex 3D interactions. Think of it as assembling a 3D model from paper cutouts of the cross-sections.
Parabolic Cylinders
A parabolic cylinder is a type of 3D shape where one dimension extends infinitely, while the cross-section is a parabola. In our function \( z = y - x^2 \), the surface forms a parabolic cylinder, stretching infinitely along the \( y \)-axis.
Key features include:
  • Constant cross-sections parallel to the \( yz \)-plane (vertical) depict a line, showing that the cylinder's sides remain uniform and do not curve horizontally.
  • The parabolic cross-sections parallel to the \( xz \)-plane (horizontal) display the curvature typically associated with parabolas, underscoring how the surface bends in the \( xz \) direction.
Understanding parabolic cylinders is a fundamental step in identifying more complex structures in multivariable calculus, providing foundational insights into their geometric properties.

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Most popular questions from this chapter

Let \(\mathbf{a}=(1,2)\). Show that the function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) given by \(f(\mathbf{x})=\mathbf{a} \cdot \mathbf{x}\) is continuous.

In Exercises 2.1-2.4, sketch the surface in \(\mathbb{R}^{3}\) described by the given equation. $$ x^{2}+y^{2}+z^{2}=4 $$

Let \(U\) be an open set in \(\mathbb{R}^{n}\), and let \(f: U \rightarrow \mathbb{R}\) be a function that is continuous at a point a of \(U\) (a) If \(f(\mathbf{a})>0\), show that there exists an open ball \(B=B(\mathbf{a}, r)\) centered at a such that \(f(\mathbf{x})>\frac{f(\mathbf{a})}{2}\) for all \(\mathbf{x}\) in \(B .\left(\right.\) Hint : Let \(\left.\epsilon=\frac{f(\mathbf{a})}{2} .\right)\) (b) Similarly, if \(f(\mathbf{a})<0\), show that there exists an open ball \(B=B(\mathbf{a}, r)\) such that \(f(\mathbf{x})<\frac{f(\mathbf{a})}{2}\) for all \(\mathbf{x}\) in \(B\). In particular, if \(f(\mathbf{a}) \neq 0\), there is an open ball \(B\) centered at a throughout which \(f(\mathbf{x})\) has the same sign as \(f(\mathbf{a})\).

(Sandwich principles.) Let \(U\) be an open set in \(\mathbb{R}^{n}\), and let a be a point of \(U\). (a) Let \(f, g, h: U \rightarrow \mathbb{R}\) be real-valued functions such that \(f(\mathbf{x}) \leq g(\mathbf{x}) \leq h(\mathbf{x})\) for all \(\mathbf{x}\) in \(U\). If \(f(\mathbf{a})=h(\mathbf{a})\) - let's call the common value \(c\) - and if \(f\) and \(h\) are continuous at a, prove that \(g(\mathbf{a})=c\) and that \(g\) is continuous at \(\mathbf{a}\) as well. (b) Let \(f, g, h\) be real-valued functions defined on \(U,\) except possibly at the point \(\mathbf{a},\) such that \(f(\mathbf{x}) \leq g(\mathbf{x}) \leq h(\mathbf{x})\) for all \(\mathbf{x}\) in \(U,\) except possibly when \(\mathbf{x}=\mathbf{a} .\) If \(\lim _{\mathbf{x} \rightarrow \mathbf{a}} f(\mathbf{x})=\) \(\lim _{\mathbf{x} \rightarrow \mathbf{a}} h(\mathbf{x})=L,\) prove that \(\lim _{\mathbf{x} \rightarrow \mathbf{a}} g(\mathbf{x})=L,\) too.

Find an equation of the plane that contains the points \(\mathbf{p}=(1,0,0), \mathbf{q}=(0,2,0),\) and \(\mathbf{r}=(0,0,3)\).
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