Problem 20 - Calculus: Early Transcendental Functions

Question

\(C\) is the intersection of \(z=x^{2}+y^{2}\) and \(z=8-y,\) oriented clockwise from above, \(\mathbf{F}=\left\langle 2 x^{2}, 4 y^{2}, e^{8 z^{2}}\right\rangle\) \(C\) is the intersection of \(x^{2}+y^{2}=1\) and \(z=x-y,\) oriented clockwise from above, \(\mathbf{F}=\left\langle\cos x^{2}, \sin y^{2}, \tan z^{2}\right\rangle\)

Step-by-Step Solution

Answer

The vector fields corresponding to curve \(C\) from the first problem are \(\mathbf{F} = \left\langle 8\cos^{2}\theta, 16\sin^{2}\theta, e^{8(8 - 2\sin \theta)^{2}} \right\rangle\) and for the second problem are \(\mathbf{F} = \left\langle\cos(\cos^{2}\theta), \sin(\sin^{2}\theta), \tan((\cos\theta - \sin\theta)^2) \right\rangle\).

Step 1: Identify the Geometric Intersections

The intersection of the surfaces \(z=x^{2}+y^{2}\) and \(z=8-y\) gives us the equation \(x^{2}+y^{2}=8-y\). This is a circle in the x-y plane with the center at \((0,4)\) and a radius of \(2\). This denotes the curve \(C\) from the first problem.

Step 2: Find Parameterization of Curve C from First Problem

The curve can be parameterized using polar coordinates. Let's take \(x=2\cos \theta\) and \(y=2\sin \theta\), where \(0 \leq \theta \leq 2\pi\). The z-coordinate (denoted as \(z\)) would be \(8-y = 8 - 2\sin \theta\).

Step 3: Calculate the Vector Field for Problem 1

Substitute \(\theta\) into the vector field, you would get \(\mathbf{F} = \left\langle 8\cos^{2}\theta, 16\sin^{2}\theta, e^{8(8 - 2\sin \theta)^{2}} \right\rangle\).

Step 4: Identify Geometric Intersection for Problem 2

For the second problem, the intersection of \(x^{2}+y^{2}=1\) and \(z=x-y\) gives us a unit circle in the x-y plane. This denotes the curve \(C\) for the second problem.

Step 5: Find Parameterization of Curve C from Second Problem

Similar to Step 2, we use polar coordinates to parameterize the curve C for the second problem. Choose \(x=\cos\theta\) and \(y=\sin\theta\), where \(0 \leq \theta \leq 2\pi\). The z-coordinate from the equation \(z=x-y\) would be \(z=\cos\theta - \sin\theta\).

Step 6: Calculate the Vector Field for Problem 2

Substitute \(\theta\) into the vector field for the second problem, we would get \(\mathbf{F} = \left\langle\cos(\cos^{2}\theta), \sin(\sin^{2}\theta), \tan((\cos\theta - \sin\theta) ^2) \right\rangle\).

Key Concepts

Geometric Intersections

Geometric intersections play a crucial role in understanding where different surfaces meet in three dimensions. In this exercise, we explore the intersection of two surfaces—a paraboloid defined by \( z = x^2 + y^2 \) and a plane described by \( z = 8 - y \). These two surfaces intersect to form a curve. By equating the two surface equations, we derive \( x^2 + y^2 = 8 - y \), which simplifies to a circular equation. The curve is a circle centered at \((0,4)\) with a radius of \(2\). It lies in the xy-plane and represents the boundary where these two surfaces overlap.

• This visual intersection helps us visualize complex shapes in simpler terms, often as familiar geometric figures like circles or ellipses.
• Identifying such intersections is essential for visualizing and solving vector calculus problems, particularly when working with integrals and vector fields over such surfaces.

Parameterization

Parameterization is a method to express a curve using a single variable, often called a parameter. By parameterizing, we can easily describe every point on a curve in a mathematical form. For example, in our first example, the curve of intersection, which is a circle, can be expressed using polar coordinates. For the circle defined by \( x^2 + y^2 = 8 - y \), we choose the parameter \( \theta \), where \( x = 2 \cos \theta \) and \( y = 2 \sin \theta \).

• This transformation simplifies the mathematical manipulations and allows us to describe the entire curve as \( \theta \) varies from \( 0 \) to \( 2 \pi \).
• It also helps in integrating over curves since expressions in terms of polar coordinates often simplify algebraic terms, particularly when dealing with symmetric shapes like circles.

In three dimensions, parameterization includes the z-coordinate, here given by substituting back into \( z = 8 - y \). This results in finding \( z = 8 - 2 \sin \theta \), completing the parameterized form \( (x(\theta), y(\theta), z(\theta)) \).

Vector Fields

Vector fields are mathematical constructs where each point in space is associated with a vector. They are essential for modeling various phenomena in physics and engineering, such as fluid flow and electromagnetic fields. In our exercise, we have a vector field \( \mathbf{F} = \langle 2x^2, 4y^2, e^{8z^2} \rangle \) for the first problem. This vector field assigns a vector to each point on the curve, translating how objects might move through space or interact with forces described by the field.

• Vector fields can be visualized as arrows pointing in the direction and magnitude of the vector at each point in space.
• Understanding how the vector field interacts with curves or surfaces, like our parameterized intersection, is critical for problems involving line or surface integrals, where we integrate these fields over paths or areas to find work done or flux.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This system is particularly useful in vector calculus and problems involving circular and rotational symmetries.In our case, using polar coordinates simplifies the equation of the circle in the intersection problem by using the substitution \( x = r \cos \theta \) and \( y = r \sin \theta \). Given \( r = 2 \) for the first problem, we directly translate our circle into this system.

• Polar coordinates reduce complex equations and are effective in analyzing systems with circular or spiral characteristics, making integration and parameterization easier.
• They provide a more intuitive understanding of the geometry when dealing with curves that are naturally circular or spherical.

This form of transformation is beneficial not only for simplifying calculations but also for providing clearer geometrical interpretations of curves and surfaces in calculus.