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Chapter 12: Problem 38

Sketch the curve of intersection of the surfaces, and find a vector equation for the curve in terms of the parameter x = t. $$ y=x, x+y+z=1 $$

Short Answer

Expert verified
The vector equation is \(\mathbf{r}(t) = \langle t, t, 1 - 2t \rangle\).

Step by step solution

01

Identify the Surfaces

We have two surfaces: a plane defined by the equation \(x+y+z=1\) and another surface defined by the equation \(y=x\). Our task is to find and sketch the curve of intersection of these surfaces.
02

Substitute y with x

Since the equation \(y=x\) implies that for every point on the curve \(y\) can be replaced with \(x\), we substitute \(y\) with \(x\) in the plane equation: \(x + x + z = 1\). This simplifies to \(2x + z = 1\).
03

Express z in terms of x

Solve the equation \(2x + z = 1\) for \(z\). This gives us \(z = 1 - 2x\). Now, both \(y\) and \(z\) are expressed in terms of \(x\).
04

Introduce the Parameter t

Let \(x = t\), which serves as our parameter. Thus, \(y = t\) and \(z = 1 - 2t\).
05

Write the Vector Equation

Combine the parameterized coordinates into a vector equation of the curve. The vector equation is \(\mathbf{r}(t) = \langle t, t, 1 - 2t \rangle\), which describes the curve of intersection.

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Key Concepts

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

Vector Equation
A vector equation provides a way to describe a curve or a line in space using a vector format. It's quite useful when dealing with problems in 3D space, such as the one we have here, intersecting a plane and a line. In our exercise, we arrived at the vector equation by expressing all the coordinates (x, y, z) in terms of a single parameter.
  • The parameter here is often denoted by 't'.
  • It gives us a simple and flexible way to represent the entire curve.
This particular vector equation, \[ \mathbf{r}(t) = \langle t, t, 1 - 2t \rangle \], uniquely describes every point along the curve as the parameter 't' varies. The beauty of using a vector equation is how it neatly packages the curve's information, making it easier to plot and analyze in a 3D coordinate system.
Parameterization
Parameterization is a method of expressing a set of quantities as functions of one or more independent variables, known as parameters. In the context of our problem, parameterization involves using a single parameter 't' to express all the variables that define the curve of intersection.
  • Here, we chose the parameter 'x' to be equal to 't', which simplifies our problem.
  • By letting \( y = t \) and \( z = 1 - 2t \), we effectively map the curve.
Using parameterization, we've turned a potentially complex intersection problem into a more manageable vector equation. It allows us to see how each component behaves as the parameter changes, providing insight into both the geometry and direction of the curve.
Plane Equation
The plane equation in this problem is given by \( x + y + z = 1 \). This is a classic representation of a plane in 3-dimensional space. To intersect with another surface, like the line described by \( y = x \), we substitute variables to find a common solution.
  • Each coefficient in the equation has a geometric interpretation, aligning with the plane's orientation in 3D space.
  • The constant term, 1, indicates the plane's position relative to the origin.
By substituting \( y = x \) directly into the plane equation, we transformed our equation into \( 2x + z = 1 \). This makes the task of finding the intersection curve more straightforward, setting us up to parameterize and further analyze it.
Surface Intersection
Surface intersection involves finding a common region, or curve, formed by the overlap of two surfaces. In our case, we're intersecting a plane with a line, which mathematically translates into a specific curve that we can describe using parameterization.
  • The intersection curve contains all points that satisfy both surface equations simultaneously.
  • By utilizing parameterization, we conveniently express this curve to understand its shape and trajectory.
When intersecting, simplifying one of the surface equations often helps. For the line given by \( y = x \) intersecting the plane \( x + y + z = 1 \), substituting and solving gives us a new equation, \( 2x + z = 1 \), showing the intersection explicitly. Understanding surface intersections is crucial in calculus and geometry, revealing how space curves and surfaces relate.

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Most popular questions from this chapter

Find an arc length parametrization of the curve that has the same orientation as the given curve and for which the reference point corresponds to \(t=0 .\) $$ \mathbf{r}(t)=(1+t)^{2} \mathbf{i}+(1+t)^{3} \mathbf{j} ; \quad 0 \leq t \leq 1 $$

These exercises are concerned with the problem of creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves \(C_{1}\) and \(C_{2}\) are joined at a point \(P\) to form a curve \(C,\) then we will say that \(C_{1}\) and \(C_{2}\) make a smooth transition at \(P\) if the curvature of \(C\) is continuous at \(P\). Show that the transition at \(x=0\) from the horizontal line \(y=0\) for \(x \leq 0\) to the parabola \(y=x^{2}\) for \(x>0\) is not smooth, whereas the transition to \(y=x^{3}\) for \(x>0\) is smooth.

Show that in cylindrical coordinates a curve given by the parametric equations \(r=r(t), \theta=\theta(t), z=z(t)\) for \(a \leq t \leq b\) has arc length $$ L=\int_{a}^{b} \sqrt{\left(\frac{d r}{d t}\right)^{2}+r^{2}\left(\frac{d \theta}{d t}\right)^{2}+\left(\frac{d z}{d t}\right)^{2}} d t $$ [Hint: Use the relationships \(x=r \cos \theta, y=r \sin \theta .]\)

Find the velocity, speed, and acceleration at the given time t of a particle moving along the given curve. $$ x=2 \cos t, y=2 \sin t, z=t ; t=\pi / 4 $$

(a) Find the arc length parametrization of the line $$ x=t, \quad y=t $$ that has the same orientation as the given line and has reference point \((0,0) .\) (b) Find the arc length parametrization of the line $$ x=t, \quad y=t, \quad z=t $$ that has the same orientation as the given line and has reference point \((0,0,0) .\)
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