Problem 6

Question

Find the image of the set $S$ under the given transformation. $$ \text { S: } 1 \leq u \leq 2,1 \leq v \leq 2 ; x=u v, y=v^{2} $$

Step-by-Step Solution

Verified

Answer

The image of S is bounded by $ x = \sqrt{y} $, $ x = 2\sqrt{y} $ from $ y = 1 $ to $ y = 4 $.

1Step 1: Identify the Transformation Equations

The transformation is given by the equations: $ x = uv $ and $ y = v^2 $. These equations will allow us to convert the set $ S $ from $ (u, v) $ space to $ (x, y) $ space.

2Step 2: Determine Boundary Curves for S

The set $ S $ is defined by the inequalities $ 1 \leq u \leq 2 $ and $ 1 \leq v \leq 2 $. The boundaries occur at constant values of $ u $ and $ v $: $ u = 1 $, $ u = 2 $, $ v = 1 $, and $ v = 2 $.

3Step 3: Transform the Boundaries into (x, y) Space

- For $ u = 1 $: - $ x = 1v = v $, $ y = v^2 $, describing a curve $ x = \sqrt{y} $ for $ 1 \leq v \leq 2 $.- For $ u = 2 $: - $ x = 2v $, $ y = v^2 $, describing a curve $ x = 2\sqrt{y} $ for $ 1 \leq v \leq 2 $.- For $ v = 1 $: - $ x = u $, $ y = 1 $, describing a line segment from $ x = 1 $ to $ x = 2 $.- For $ v = 2 $: - $ x = 2u $, $ y = 4 $, describing a line segment from $ x = 2 $ to $ x = 4 $.

4Step 4: Sketch Region in (x, y) Space

Using the transformed boundaries:- Curve from $ y = 1 $ to $ y = 4 $ bounded by the curves $ x = \sqrt{y} $ and $ x = 2\sqrt{y} $.- Horizontal lines at $ y = 1 $ from $ x = 1 $ to $ x = 2 $ and at $ y = 4 $ from $ x = 2 $ to $ x = 4 $. This forms a region in $ (x, y) $ space.

Key Concepts

Boundary CurvesCoordinate TransformationParametric Equations

Boundary Curves

When we talk about boundary curves in transformations in geometry, we are dealing with the limits or edges of a shape or region in space. For the set $ S $ defined by $ 1 \leq u \leq 2 $ and $ 1 \leq v \leq 2 $, the boundaries are established at these constant values of $ u $ and $ v $. This means that the edges of $ S $ are:

$ u = 1 $
$ u = 2 $
$ v = 1 $
$ v = 2 $

By transforming these boundaries through given equations like $ x = uv $ and $ y = v^2 $, we translate the boundary from $(u, v)$ space into $(x, y)$ space. This helps in visualizing how the original shape or region changes under the transformation.
In our solution, four lines in $(u, v)$ space were converted into curves in $(x, y)$ space, giving us a clear view of the transformed boundaries.

Coordinate Transformation

Coordinate transformation is about changing from one coordinate system to another. It's like changing the way you describe the location of a point or the shape of a region. In the given exercise, transforming from coordinate $(u, v)$ to $(x, y)$ happens via the equations $ x = uv $ and $ y = v^2 $.

For $ u = 1 $, the transformation gives us a curve described by $ x = v $ and $ y = v^2 $.
For $ u = 2 $, we get $ x = 2v $ and $ y = v^2 $, forming another curved boundary.
For $ v = 1 $, it's a straight line segment with $ x = u $ and $ y = 1 $.
Finally, $ v = 2 $ transforms to $ x = 2u $ and $ y = 4 $.

Through this process, the original square-like shape in $(u, v)$ space gets re-shaped in $(x, y)$ space, making the region look different visually. Understanding this concept allows you to predict and comprehend geometric transformations' effects.

Parametric Equations

Parametric equations give a way to define a curve with a set of equations. Instead of describing a curve as $ y=f(x) $, parametric equations express both x and y in terms of one or more parameters. In the context of our problem, the parameters $ u $ and $ v $ serve to express the transformation equations:

$ x = uv $
$ y = v^2 $

By plugging in different values for $ u $ and $ v $, we get different points $(x, y)$ that make up the boundary and the interior of the region.
The elegance of parametric equations is in their ability to describe complex curves and shapes succinctly, even when a simple $ y=f(x) $ form might not be possible. In the exercise, defining curves like $ x = \sqrt{y} $ and $ x = 2\sqrt{y} $ in parametric form simplifies understanding and handling transformations.

Problem 5

Problem 6

Other exercises in this chapter

Problem 5

Find the gradient of the given function at the indicated point. $$ f(x, y)=x^{2}-4 y^{2} ;(2,4) $$

View solution

Problem 5

Graph the curve traced by the given vector function. $$ \mathbf{r}(t)=\left\langle e^{t}, e^{2 t}\right\rangle $$

View solution

Problem 6

Use the divergence theorem to find the outward flux $\iint_{S}(\mathbf{F} \cdot \mathbf{n}) d S$ of the given vector field $\mathbf{F}$. $$ \begin{aligned}

View solution

Problem 6

Evaluate the given iterated integral. $$ \int_{0}^{\sqrt{2}} \int_{\sqrt{y}}^{2} \int_{0}^{e^{x}} x d z d x d y $$

View solution