Problem 12
Question
Find the vector function that describes the curve \(C\) of intersection between the given surfaces. Sketch the curve \(C\). Use the indicated parameter. $$ x^{2}+y^{2}-z^{2}=1, y=2 x ; x=t $$
Step-by-Step Solution
Verified Answer
The vector function is \( \mathbf{r}(t) = \langle t, 2t, \pm \sqrt{5t^2 - 1} \rangle \).
1Step 1: Analyze the given surfaces
We have two surfaces: a hyperboloid given by the equation \( x^2 + y^2 - z^2 = 1 \) and a plane given by the equation \( y = 2x \). Our goal is to find the curve of intersection between these two surfaces.
2Step 2: Substitute for the plane equation
Since the plane equation is \( y = 2x \), we substitute \( y = 2x \) into the hyperboloid equation. This gives us \( x^2 + (2x)^2 - z^2 = 1 \).
3Step 3: Simplify the hyperboloid equation
Substituting \( y = 2x \) into the hyperboloid equation results in: \[ x^2 + 4x^2 - z^2 = 1 \]. Simplifying this, we have: \[ 5x^2 - z^2 = 1 \].
4Step 4: Express variables in terms of the parameter
Given that \( x = t \), substitute \( x = t \) into the equation \( 5x^2 - z^2 = 1 \). We have:\[ 5t^2 - z^2 = 1 \]. This simplifies to \( z^2 = 5t^2 - 1 \), and therefore \( z = \pm \sqrt{5t^2 - 1} \).
5Step 5: Write the vector function
The position vector function describing the curve \( C \) is \( \mathbf{r}(t) = \langle t, 2t, \pm \sqrt{5t^2 - 1} \rangle \). This expression encompasses both branches of the intersection curve in the given parameter \( t \).
6Step 6: Sketch the curve
For a sketch, plot the vector function \( \mathbf{r}(t) \). It represents a curve having its projection as a straight line in the \( xy \)-plane (since \( y = 2x \)) and a hyperbolic component in the \( z \)-axis.
Key Concepts
Understanding HyperboloidsExploring the Curve of IntersectionUnderstanding Vector FunctionsDecoding Parametric Equations
Understanding Hyperboloids
A hyperboloid is an interesting and significant surface in the realm of three-dimensional geometry. The equation associated with a hyperboloid is generally of the form \(x^2 + y^2 - z^2 = 1\) or similar variants. It consists of two types: a hyperboloid of one sheet and a hyperboloid of two sheets.
In our context, the hyperboloid given by \(x^2 + y^2 - z^2 = 1\) is known as a hyperboloid of one sheet.
In our context, the hyperboloid given by \(x^2 + y^2 - z^2 = 1\) is known as a hyperboloid of one sheet.
- Its key feature is that for any fixed value of \(z\), the cross-sections are circles.
- It resembles a cooling tower or a smooth hourglass shape.
- Hyperboloids have a unique characteristic of having straight lines that lie entirely on the surface.
Exploring the Curve of Intersection
The curve of intersection is where two surfaces meet in three-dimensional space. In this exercise, the surfaces are a hyperboloid and a plane.
This curve often reflects both shapes' characteristics and can be an intriguing geometrical entity.
This curve often reflects both shapes' characteristics and can be an intriguing geometrical entity.
- For the surfaces \(x^2 + y^2 - z^2 = 1\) and \(y = 2x\), the intersection is a curve where both equations hold true simultaneously.
- Substituting \(y = 2x\) into the hyperboloid simplifies it significantly, allowing us to describe the curve more easily.
- The resulting curve appears as a twisted path extending through space, maintaining the planar relationship \(y = 2x\).
Understanding Vector Functions
Vector functions are vital in representing curves in space. They utilize vectors to map from a parameter (usually \(t\)) to points on the curve.
- A vector function, like \(\mathbf{r}(t) = \langle t, 2t, \pm \sqrt{5t^2 - 1} \rangle \), provides each point's coordinate on the curve.
- It allows us to express the curve parametrically using \(t\) as a simple input, yielding a range of coordinates that describe the path.
- This transformation helps identify specific points or segments of the curve by varying \(t\).
Decoding Parametric Equations
Parametric equations define a curve by expressing each coordinate as a separate function of an independent parameter, usually \(t\).
- In the vector function \(\mathbf{r}(t)\), each component (\(x\), \(y\), and \(z\)) is given as a function of \(t\).
- This approach allows the curve to be described in terms of a single parameter, which can simplify calculations and visualizations.
- By using parametric equations, we have the freedom to consider multiple values for \(z\) simultaneously, thereby encompassing variations within the solution.
Other exercises in this chapter
Problem 12
Find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=4 x+x y^{2}-5 y ;(3,-1), \quad \theta=\pi / 4 $$
View solution Problem 12
\(\mathbf{r}(t)\) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any \(t\). $$ \mathbf{r}(t)=\tan
View solution Problem 13
Evaluate the given integral by means of the indicated change of variables. $$ \begin{aligned} &\iint_{R}(x+y) d A \text { , where } R \text { is the region boun
View solution Problem 13
In Problems 13-16, use Stokes' theorem to evaluate \(\iint_{S}\) (curl \(\left.\mathbf{F}\right) \cdot \mathbf{n} d S\). Assume that the surface \(S\) is orient
View solution