Calculus III - Change of Variables

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Section 15.8 : Change of Variables

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5. If \(R\) is the parallelogram with vertices \(\left( {1,0} \right)\), \(\left( {4,3} \right)\), \(\left( {1,6} \right)\) and \(\left( { - 2,3} \right)\) determine the region we would get applying the transformation \(x = \frac{1}{2}\left( {v - u} \right)\), \(y = \frac{1}{2}\left( {v + u} \right)\) to \(R\).

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Start Solution

Let’s start off with a sketch of \(R\).

The equations of each of the boundaries of the region are given in the sketch. Given that we know the coordinates of each of the vertices its simple algebra to determine the equation of a line given two points on the line so we’ll leave it to you to verify the equations.

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Okay, now all we need to do is apply the transformation to each of the boundary equations. To do this we simply plug the transformation into each of the boundary equations.

Let’s start with \(y = x + 5\). Applying the transformation to this equation gives,

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = \frac{1}{2}\left( {v - u} \right) + 5\\ v + u & = v - u + 10\\ 2u &= 10\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {u = 5} \end{align*}\]

So, this boundary will transform into the equation \(u = 5\).

Show Step 3

Let’s now apply the transformation to \(y = - x + 7\).

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = - \frac{1}{2}\left( {v - u} \right) + 7\\ v + u & = - v + u + 14\\ 2v & = 14\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {v = 7} \end{align*}\]

This boundary transforms in the equation \(v = 7\).

Show Step 4

Next, let’s apply the transformation to \(y = x - 1\).

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = \frac{1}{2}\left( {v - u} \right) - 1\\ v + u & = v - u - 2\\ 2u & = - 2\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {u = - 1} \end{align*}\]

So, this boundary will transform into the equation \(u = - 1\).

Show Step 5

Finally let’s apply the transformation to \(y = - x + 1\).

\[\begin{align*}\frac{1}{2}\left( {v + u} \right) & = - \frac{1}{2}\left( {v - u} \right) + 1\\ v + u & = - v + u + 2\\ 2v & = 2\hspace{0.5in} \Rightarrow \hspace{0.5in}\underline {v = 1} \end{align*}\]

This boundary transforms in the equation \(v = 1\).

Show Step 6

Sketching the transformed equations gives the following region.

So, we transform the diamond shaped region into a rectangle under the transformation.

Note that we chose \(u\) to be the horizontal axis and \(v\) to be vertical axis in the transformed region for this problem. There is no real reason for doing that other than it is just what we’ve always done. Regardless of your choices here make sure to label the axes to make it clear!