Problem 7 For the functions $f(x, y)$ : ... [FREE SOLUTION]

Chapter 3: Problem 7

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 $f$. $$ f(x, y)=\frac{y}{x^{2}+1} $$

Short Answer

Expert verified

Cross-sections lead to lines along axes; level curves are parabolas. The graph forms a central ridge with fan-like parabolas.

Step by step solution

Sketch Cross-Sections with the Coordinate Planes

To find cross-sections of the graph $ z = f(x, y) $ with coordinate planes, we start with: - **Plane $ y = 0 $:** Substitute $ y = 0 $ in the function: \[ z = \frac{0}{x^2 + 1} = 0. \] Hence, the graph on this plane will be the line $ z = 0 $, a flat line on the $ xz $-plane.- **Plane $ x = 0 $:** Substitute $ x = 0 $ in the function: \[ z = \frac{y}{0^2 + 1} = y. \] This means on the $ yz $-plane, the graph is the line $ z = y $.- **Plane $ z = 0 $:** Solve $ \frac{y}{x^2 + 1} = 0 $: This implies $ y = 0 $. The graph intersects the $ xy $-plane along the $ x $-axis.

Sketch Several Level Curves

Level curves occur at constant $ z = c $, so set $ f(x, y) = c $:\[ \frac{y}{x^2 + 1} = c \quad \Rightarrow \quad y = c (x^2 + 1). \]For different values of $ c $, sketch these curves on the $ xy $-plane:- **For $ c = 0 $:** $ y = 0 $, which is the $ x $-axis.- **For $ c = 1 $:** $ y = x^2 + 1 $, a parabola opening upward.- **For $ c = -1 $:** $ y = -x^2 - 1 $, a parabola opening downward.Sketch these parabola-like shapes for multiple $ c $ showing the pattern of the level sets.

Sketch the Graph of $ f $

Using the information from previous steps:- The cross-sections on $ y = 0 $ and $ z = 0 $ both lie flat on respective planes.- The level curves in Step 2 influence how the function stretches across the plane; they form a continuous surface.- Note the behavior as $ x \to \pm \infty $, where $ \frac{y}{x^2+1} \to 0 $ for any finite $ y $.A three-dimensional sketch will have a central ridge along the $ yz $ plane from Step 1 and parabolic fans radiating out as described in Step 2.

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Key Concepts

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

Cross-sections

Understanding cross-sections is like taking slices out of a 3D object to see its interior structure. In the context of multivariable calculus, a cross-section involves slicing through a surface with a plane, which allows you to see how the surface behaves on this specific plane.
For the function given, the cross-sections with the coordinate planes are calculated by setting one variable to a constant value and analyzing the resulting 2D line or curve.
Here, for:

**Plane $y = 0$:** We substitute $y = 0$ into $f(x, y)$, resulting in $z = \frac{0}{x^2 + 1} = 0$. This means the cross-section in this plane is a flat line on the $xz$-plane at $z = 0$.
**Plane $x = 0$:** Substituting $x = 0$ into the function gives $z = y$. This forms a straight line on the $yz$-plane matching the equation $z = y$.
**Plane $z = 0$:** Solving $\frac{y}{x^2 + 1} = 0$ for $y$ leads to $y = 0$, showing the cross-section in the $xy$-plane is also a straight line along the x-axis.

These intersections help visualize how the surface interacts with each coordinate plane, allowing better understanding of its 3-dimensional shape.

Level curves

Level curves help us visualize a 3D surface within a 2D plane by slicing through the surface at constant height, or constant $z$-value, which we denote as $c$.
To identify these curves for the function $f(x, y)$, we set equal to $c$: \[ \frac{y}{x^2 + 1} = c \quad \Rightarrow \quad y = c (x^2 + 1). \]
Here's how the level curves sketch out:

**For $c = 0$:** The equation simplifies to $y = 0$, translating to the $x$-axis in the $xy$-plane.
**For $c = 1$:** The level curve becomes $y = x^2 + 1$, which is a parabola opening upwards, starting at $y = 1$ when $x = 0$.
**For $c = -1$:** Here, the curve flips to $y = -x^2 - 1$, creating a downward-opening parabola, starting at $y = -1$ when $x = 0$.

Drawing multiple of these curves across different values of $c$ creates a family of shapes that indicate how the surface extends across the $xy$-plane. These representations are crucial for understanding how the surface changes and moves as the $z$-value varies.

Graph sketching

Graph sketching in multivariable calculus is about translating formulaic information into a visual format, offering a clearer understanding of the surface behavior of a function. By combining insights from cross-sections and level curves, we can form a comprehensive 3D graph of the function.
The critical steps involve:

**Using Cross-sections**: Observing that along specific planes, sections of the graph like $z = 0$ for $y = 0$ and the line $z = y$ for $x = 0$ form defining shapes on their respective planes.
**Level Curves Influence**: Parabolic shapes outlined by the level curves illustrate how parts of the surface bow upward or downward, indicating rise or fall in the topography of the 3D graph.
**Tendency as $x$ approaches $\pm \infty$**: The value $\frac{y}{x^2+1}$ trends towards zero, signaling that surface flattens beyond the central peaks and valleys.

With these details, we can sketch a 3D graph that includes a ridge along the $yz$-plane, with parabolic fans showing the surface's behavior as described by the level curves. This sketch gives a visual representation of how the function behaves in space, marrying algebraic calculations with geometric visualization.

91影视

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 \(f\). $$ f(x, y)=\frac{y}{x^{2}+1} $$

Short Answer

Step by step solution

Sketch Cross-Sections with the Coordinate Planes

Sketch Several Level Curves

Sketch the Graph of \( f \)

Key Concepts

Cross-sections

Level curves

Graph sketching

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