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Sketch the image of the unit circle under the line transformation

T(x)=(5002)x

Short Answer

Expert verified

The image of circle under T(x)=(5002)xis an ellipse.

Step by step solution

01

Compute the functions 

Consider the matrix.

T(x)=(5002)x

The vector that has points on the unit circle is,

C={(costsint):0t2}

Let the vector be,

x=(costsint)

The linear transformation is,

T(x)=(5002)(costsint)T(x)=(5cost2sint)T(x)=cost(50)+sint(02)

02

Consider the transformation 

The transformation of the circle is,

T(C)={cost(50)+sint(02):0t2}

The above equation represents the ellipse with the axes given as,

e1=[50],e2=[02]

As, e1e2=[50][02]=0, thus, these are perpendicular to each other.

03

Sketch the graph 

Consider the vectors.

e1=[50],e2=[02]

Consider the linear transformations.

T(C)={cost(50)+sint(02):0t2}

Consider the circle.

C={costsint:0t2}

The graph of the same is,

Hence, the image of circle underT(x)=(5002)xis an ellipse.

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