Problem 2
Question
Consider a transformation \(T\) defined by \(x=\sqrt{v-u}\) \(y=v+u\). Find the images of the points \((1,1),(1,3)\), and \((\sqrt{2}, 2)\) in the \(x y\) -plane under \(T^{-1}\).
Step-by-Step Solution
Verified Answer
The images are (0, 1), (1, 2), and (0, 2).
1Step 1: Understand the Transformation
The transformation maps coordinates \(u, v\) to \(x, y\) using the equations \(x = \sqrt{v-u}\) and \(y = v+u\). We need to find the inverse of this transformation to map \(x, y\) back to \(u, v\).
2Step 2: Set Up the Inverse Transformation Equations
To find \(T^{-1}\), solve for \(u\) and \(v\) in terms of \(x\) and \(y\). From \(x = \sqrt{v-u}\), we have \(v-u = x^2\). And from \(y = v+u\), we can write two equations: \(v-u = x^2\) and \(v+u = y\).
3Step 3: Solve for \(u\)
Add the equations: \(v-u + v+u = x^2 + y\), which simplifies to \(2v = x^2 + y\). Solving for \(v\) gives \(v = \frac{x^2 + y}{2}\).
4Step 4: Solve for \(v\)
Subtract the equations: \(v+u - (v-u) = y - x^2\), which simplifies to \(2u = y - x^2\). Solving for \(u\) gives \(u = \frac{y - x^2}{2}\).
5Step 5: Calculate the Image of (1,1)
Plug \((x, y) = (1, 1)\) into the inverse equations: \(v = \frac{1^2 + 1}{2} = 1\) and \(u = \frac{1 - 1^2}{2} = 0\). The image is \(T^{-1}(1, 1) = (0, 1)\).
6Step 6: Calculate the Image of (1,3)
Plug \((x, y) = (1, 3)\) into the inverse equations: \(v = \frac{1^2 + 3}{2} = 2\) and \(u = \frac{3 - 1^2}{2} = 1\). The image is \(T^{-1}(1, 3) = (1, 2)\).
7Step 7: Calculate the Image of (\(\sqrt{2}\), 2)
Plug \((x, y) = (\sqrt{2}, 2)\) into the inverse equations: \(v = \frac{(\sqrt{2})^2 + 2}{2} = 2\) and \(u = \frac{2 - (\sqrt{2})^2}{2} = 0\). The image is \(T^{-1}(\sqrt{2}, 2) = (0, 2)\).
Key Concepts
Coordinate TransformationInverse FunctionsMapping in the Coordinate Plane
Coordinate Transformation
A coordinate transformation is an operation that changes the coordinate system used to represent geometric objects. In this context, transformations like the function \( T \) map given coordinates \((u, v)\) to new coordinates \((x, y)\). Such transformations are useful in scenarios where the geometry or position of objects needs to be analyzed in different ways.
Coordinate transformation involves a set of equations that relate the coordinates of one system to another. In the given exercise, the transformation \( T \) is defined by the equations \( x = \sqrt{v-u} \) and \( y = v+u \). These equations transform coordinates from the \((u, v)\) plane to the \((x, y)\) plane.
Understanding how to efficiently move between these two coordinate systems requires analyzing and possibly inverting these equations.
Coordinate transformation involves a set of equations that relate the coordinates of one system to another. In the given exercise, the transformation \( T \) is defined by the equations \( x = \sqrt{v-u} \) and \( y = v+u \). These equations transform coordinates from the \((u, v)\) plane to the \((x, y)\) plane.
Understanding how to efficiently move between these two coordinate systems requires analyzing and possibly inverting these equations.
Inverse Functions
Inverse functions reverse the effect of the original function. By finding the inverse of a function, one can map output values back to the corresponding input values. When dealing with coordinate transformations, finding the inverse is crucial to revert back to the original coordinate system.
For the transformation \( T \) defined by \( x = \sqrt{v-u} \) and \( y = v+u \), the inverse transformation \( T^{-1} \) allows us to determine \( (u, v) \) from \( (x, y) \).
For the transformation \( T \) defined by \( x = \sqrt{v-u} \) and \( y = v+u \), the inverse transformation \( T^{-1} \) allows us to determine \( (u, v) \) from \( (x, y) \).
- By rearranging \( x = \sqrt{v-u} \), we derive \( v-u = x^2 \).
- From \( y = v+u \), two equations emerge: \( v-u = x^2 \) and \( v+u = y \).
- Adding these equations gives \( 2v = x^2 + y \), leading to \( v = \frac{x^2 + y}{2} \).
- Subtracting them provides \( 2u = y - x^2 \), yielding \( u = \frac{y - x^2}{2} \).
Mapping in the Coordinate Plane
Mapping in the coordinate plane refers to the process of translating points from one set of coordinates to another. This is a common task in many fields including physics, engineering, and computer graphics.
In terms of the exercise, we need to find the images of specific points \((x, y)\) in the \(xy\)-plane using the inverse transformation \( T^{-1} \).
In terms of the exercise, we need to find the images of specific points \((x, y)\) in the \(xy\)-plane using the inverse transformation \( T^{-1} \).
- For the point \((1, 1)\), we found \( T^{-1}(1, 1) = (0, 1) \).
- For the point \((1, 3)\), the image is \( T^{-1}(1, 3) = (1, 2) \).
- Finally, for the point \((\sqrt{2}, 2)\), we calculated \( T^{-1}(\sqrt{2}, 2) = (0, 2) \).
Other exercises in this chapter
Problem 1
Graph the curve traced by the given vector function. $$ \mathbf{r}(t)=2 \sin t \mathbf{i}+4 \cos t \mathbf{j}+t \mathbf{k} ; t \geq 0 $$
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$$ \text { Verify the divergence theorem. } $$ $$ \begin{aligned} &\mathbf{F}=6 x y \mathbf{i}+4 y z \mathbf{j}+x e^{-y} \mathbf{k} ; D \text { the region bound
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Evaluate the given iterated integral. $$ \int_{1}^{3} \int_{1}^{x} \int_{2}^{x y} 24 x y d z d y d x $$
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In Problems, use a double integral in polar coordinates to find the area of the region bounded by the graphs of the given polar equations. $$ r \quad 2+\cos \th
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