Question

find the image of the set s under the given transformation. s = (u, v) ; x = 2u 3v, y = u − v

Answer 1

To find the image of set S under the given transformation, substitute the values of u and v from set S into the transformation equations. Find the corresponding values of x' and y' for each point.

Step-by-step explanation:

To find the image of set S under the given transformation, we substitute the values of u and v from set S into the transformation equations. The given transformation equations are:

x' = 2u + 3v

y' = u - v

We substitute the values of u and v from set S and find the corresponding values of x' and y'.

For example, when u = 0 and v = 0:

x' = 2(0) + 3(0) = 0

y' = 0 - 0 = 0

Similarly, we can find the image of each point in set S under the given transformation by substituting the corresponding values of u and v into the transformation equations.

Answer 2

Under the given transformation, the image of the set S corresponds to the range
`\(0 \leq x \leq 37\) and \(0 \leq y \leq 1\)`.

The set S is defined as
`\(S = \{(u, v) \mid 0 \leq u \leq 8, 0 \leq v \leq 7\}\)`.

The transformation given is
`\(x = 2u + 3v\) and \(y = u - v\)`.

Let's find the range of values for x and y that correspond to the set S.

For x:

Minimum x value:

When u = 0 and v = 0,
`\(x = 2 * 0 + 3 * 0 = 0\)`.

Maximum x value:

When u = 8 and v = 7,
`\(x = 2 * 8 + 3 * 7 = 16 + 21 = 37\)`.

So, the range for x is
`\(0 \leq x \leq 37\)`

For y:

Minimum y value:

When u = 0 and v = 0, y = 0 - 0 = 0.

Maximum y value:

When u = 8 and v = 7, y = 8 - 7 = 1.

So, the range for
`\(y\) is \(0 \leq y \leq 1\)`.