Problem 36
Question
\(36-38\) Find a vector function that represents the curve of intersection of the two surfaces. The cylinder \(x^{2}+y^{2}=4\) and the surface \(z=x y\)
Step-by-Step Solution
Verified Answer
\(\vec{r}(\theta) = \langle 2\cos(\theta), 2\sin(\theta), 2\sin(2\theta) \rangle\)
1Step 1: Understand the Surfaces
The given problem involves two surfaces: a cylinder and a surface defined by an equation. The equation of the cylinder is \(x^2 + y^2 = 4\), which implies it is a cylinder with a radius of 2 centered along the z-axis. The second surface is \(z = xy\), which represents a hyperbolic paraboloid surface.
2Step 2: Parametrize the Cylinder
The cylinder \(x^2 + y^2 = 4\) can be parameterized using polar coordinates. Set \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\), where \(\theta\) is the parameter that varies the angle around the z-axis.
3Step 3: Substitute into the Second Surface
Substitute the parametric values from the cylinder into the equation for the surface. Thus, for the surface \(z = xy\), substitute \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\). This gives \(z = (2\cos(\theta))(2\sin(\theta)) = 4\cos(\theta)\sin(\theta)\).
4Step 4: Simplify the Expression for z
Use the trigonometric identity \(2\sin(\theta)\cos(\theta) = \sin(2\theta)\) to simplify the expression for \(z\). Therefore, \(z = 4\cos(\theta)\sin(\theta) = 2\sin(2\theta)\).
5Step 5: Write the Vector Function
Now that we have expressions for all three coordinates in terms of \(\theta\), write the vector function for the curve of intersection as \(\vec{r}(\theta) = \langle 2\cos(\theta), 2\sin(\theta), 2\sin(2\theta) \rangle\). This represents the vector function of the curve of intersection of the two surfaces.
Key Concepts
Parametric EquationsCylinder SurfaceTrigonometric IdentitiesIntersection of Surfaces
Parametric Equations
Parametric equations allow us to express a set of quantities as explicit functions of one or more independent variables, called parameters. In this problem, the parameterization helps describe the curve created by the intersection of the cylinder and the surface.
To parameterize the cylinder given by the equation \(x^2 + y^2 = 4\), we use polar coordinates, setting \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\). Here, \(\theta\) is a parameter that varies from \(0\) to \(2\pi\), effectively setting the positions around the circular cross-section of the cylinder.
Using parametric equations like these makes it possible to describe curves in three-dimensional space, combining radial, angular, and height components in different ways to suit the surfaces we're examining.
To parameterize the cylinder given by the equation \(x^2 + y^2 = 4\), we use polar coordinates, setting \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\). Here, \(\theta\) is a parameter that varies from \(0\) to \(2\pi\), effectively setting the positions around the circular cross-section of the cylinder.
Using parametric equations like these makes it possible to describe curves in three-dimensional space, combining radial, angular, and height components in different ways to suit the surfaces we're examining.
Cylinder Surface
The cylinder surface in this problem is represented by the equation \(x^2 + y^2 = 4\). This describes a circular cylinder:
This creates a hollow tube-like structure. The beauty of this representation is its rotational symmetry about the z-axis, which is mathematically convenient and leads to neat formulations when combined with parametric equations.
- Centered along the z-axis.
- With a fixed radius of 2.
This creates a hollow tube-like structure. The beauty of this representation is its rotational symmetry about the z-axis, which is mathematically convenient and leads to neat formulations when combined with parametric equations.
Trigonometric Identities
Trigonometric identities are crucial tools for simplifying and transforming expressions, especially when working with curves and surface intersections.
In this problem, the necessary identity used is \(2\sin(\theta)\cos(\theta) = \sin(2\theta)\). This identity allows us to simplify the expression for \(z = 4\cos(\theta)\sin(\theta)\) to \(z = 2\sin(2\theta)\).
In this problem, the necessary identity used is \(2\sin(\theta)\cos(\theta) = \sin(2\theta)\). This identity allows us to simplify the expression for \(z = 4\cos(\theta)\sin(\theta)\) to \(z = 2\sin(2\theta)\).
- The simplification reduces complexity.
- Makes the vector function simpler and more elegant.
Intersection of Surfaces
The intersection of two surfaces is the set of points that satisfy the equations of both surfaces simultaneously. Here, we find the intersection of a cylindrical surface and a hyperbolic paraboloid.
The strategy is to substitute the parametric expressions of the cylinder into the equation of the hyperbolic paraboloid surface \(z = xy\). By plugging in \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\), we determine \(z = 4\cos(\theta)\sin(\theta)\), leading us to the simplified form \(z = 2\sin(2\theta)\).
This substitution translates the problem into a vector function \(\vec{r}(\theta) = \langle 2\cos(\theta), 2\sin(\theta), 2\sin(2\theta) \rangle\), which beautifully encapsulates the curve of intersection, providing a direct line to all points along this curve.
The strategy is to substitute the parametric expressions of the cylinder into the equation of the hyperbolic paraboloid surface \(z = xy\). By plugging in \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\), we determine \(z = 4\cos(\theta)\sin(\theta)\), leading us to the simplified form \(z = 2\sin(2\theta)\).
This substitution translates the problem into a vector function \(\vec{r}(\theta) = \langle 2\cos(\theta), 2\sin(\theta), 2\sin(2\theta) \rangle\), which beautifully encapsulates the curve of intersection, providing a direct line to all points along this curve.
Other exercises in this chapter
Problem 36
Evaluate the integral. $$ \int_{1}^{2}\left(t^{2} \mathbf{i}+t \sqrt{t-1} \mathbf{j}+t \sin \pi t \mathbf{k}\right) d t $$
View solution Problem 36
\(33-38\) Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+3 t \mathbf{k}$$
View solution Problem 37
Evaluate the integral. $$ \int\left(e^{\prime} \mathbf{i}+2 t \mathbf{j}+\ln t \mathbf{k}\right) d t $$
View solution Problem 37
\(33-38\) Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=e^{t} \mathbf{i}+\sqrt{2} t \mathbf{j}+e^{-t} \mathbf{k}$$
View solution