Chapter 16: Problem 1
Determine whether the points \(P\) and \(Q\) lie on the given surface. $$ \begin{array}{l}{\mathbf{r}(u, v)=\langle u+v, u-2 v, 3+u-v\rangle} \\ { P(4,-5,1), Q(0,4,6)}\end{array} $$
Short Answer
Expert verified
Point \( P \) lies on the surface; point \( Q \) does not.
Step by step solution
01
Understand the Exercise
We need to determine if the points \( P(4, -5, 1) \) and \( Q(0, 4, 6) \) are on the given parametric surface represented by the vector function \( \mathbf{r}(u, v) = \langle u+v, u-2v, 3+u-v \rangle \). This involves finding appropriate \( u \) and \( v \) such that \( \mathbf{r}(u, v) = P \) and \( \mathbf{r}(u, v) = Q \).
02
Check Point P(4, -5, 1)
To check if the point \( P(4, -5, 1) \) lies on the surface, we set the equations: \( u+v = 4 \), \( u-2v = -5 \), and \( 3+u-v = 1 \). Solving these equations will help us find \( u \) and \( v \) if the point lies on the surface.
03
Solve Equations for P
Use the equations: \( 3+u-v=1 \Rightarrow u-v=-2 \) and \( u+v=4 \). Solve simultaneously: adding them gives \( 2u = 2 \Rightarrow u = 1 \). Substituting \( u = 1 \) into \( u + v = 4 \) yields \( 1 + v = 4 \Rightarrow v = 3 \). Thus, \( u = 1, v = 3 \) should satisfy \( u-2v = -5 \): \( 1 - 2(3) = 1 - 6 = -5 \). Point \( P \) lies on the surface.
04
Check Point Q(0, 4, 6)
To verify if \( Q(0, 4, 6) \) is on the surface, set \( u+v = 0 \), \( u-2v = 4 \), and \( 3+u-v = 6 \). Solving these will determine if \( u \) and \( v \) exist to meet these conditions.
05
Solve Equations for Q
Start with \( 3 + u - v = 6 \Rightarrow u - v = 3 \) and \( u+v = 0 \). Solving these, adding them gives \( 2u = 3 \Rightarrow u = \frac{3}{2} \). Substituting into \( u + v = 0 \) gives \( \frac{3}{2} + v = 0 \Rightarrow v = -\frac{3}{2} \). Verify: \( u-2v = 4 \Rightarrow \frac{3}{2} - 2(-\frac{3}{2}) = \frac{3}{2} + 3 = \frac{9}{2} \). As this does not equal \( 4 \), \( Q \) is not on the surface.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Functions
In mathematics, a vector function, especially in the context of parametric surfaces, is a brilliant tool to describe surfaces in 3D space. A vector function typically depends on two parameters, often denoted as \( u \) and \( v \), which can vary over a range of values. The given surface is described using the vector function \( \mathbf{r}(u, v) = \langle u+v, u-2v, 3+u-v \rangle \). This representation allows us to parameterize the surface, which is like creating a map that assigns each point on the surface certain parameter values.
By using vector functions, complex surfaces that cannot be expressed using a single equation become more accessible and easier to analyze. In this context, each component of the vector function corresponds to a coordinate in 3D space:
By using vector functions, complex surfaces that cannot be expressed using a single equation become more accessible and easier to analyze. In this context, each component of the vector function corresponds to a coordinate in 3D space:
- \( u+v \) determines the \( x \)-coordinate.
- \( u-2v \) determines the \( y \)-coordinate.
- \( 3+u-v \) determines the \( z \)-coordinate.
Solving Equations
When verifying whether a point, such as \( P(4, -5, 1) \), lies on the surface defined by a vector function, we substitute these coordinates into the vector function and solve for the parameters, \( u \) and \( v \). Solving these equations systematically can confirm the point's presence on the surface.
Taking example point \( P \): the coordinates satisfy the vector function equations \( u+v = 4 \), \( u-2v = -5 \), and \( 3+u-v = 1 \). Solving these equations step-by-step involves simultaneously solving linear equations. Here's how it's done:
Taking example point \( P \): the coordinates satisfy the vector function equations \( u+v = 4 \), \( u-2v = -5 \), and \( 3+u-v = 1 \). Solving these equations step-by-step involves simultaneously solving linear equations. Here's how it's done:
- Combined equations such as \( u-v = -2 \) and \( u+v = 4 \) help simplify and find common solutions.
- Add and subtract the equations to isolate \( u \) or \( v \). For instance, from adding \( (u-v) + (u+v) = -2 + 4 \), we find \( 2u = 2 \), hence \( u = 1 \).
- Substitute back to find \( v \): \( u+v = 4 \) gives \( v = 3 \).
3D Coordinates
Understanding points in three-dimensional space is essential for working with parametric surfaces and vector functions. Each point is represented by a triplet \( (x, y, z) \), which corresponds to its location along the \( x \), \( y \), and \( z \) axes, respectively. This is crucial for understanding whether points like \( P(4, -5, 1) \) and \( Q(0, 4, 6) \) belong to the surface.
With a parametric surface, as structured by our vector function, \( (x, y, z) \) coordinates are found for every choice of parameters \( u \) and \( v \). Practically, this means that:
With a parametric surface, as structured by our vector function, \( (x, y, z) \) coordinates are found for every choice of parameters \( u \) and \( v \). Practically, this means that:
- The \( x \)-coordinate equals \( u + v \).
- The \( y \)-coordinate equals \( u - 2v \).
- The \( z \)-coordinate equals \( 3 + u - v \).