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Explain why the parametrization x=sint+1,y=sin2t-4for t∈(-∞,∞)repeatedly traces the same small portion of the graph of the function y=x2-2x-3.

Short Answer

Expert verified

The parameterization of the curves sint=x-1,y=sin2t-4 traces exactly the same curve y=x2-2x-3as that of the equations x=2t+1,y=4t2-4.

Step by step solution

01

Step: 1 Given information

Consider the parametric curvesx=sint+1,y=sin2t-4,t∈(-∞,∞).

02

Calculation

The goal is to figure out how to parametrize the curves.

Remove the parameter tfrom the parametric equations to parameterize the curves.

Take x=sint+1.

Add -1to both sides of the equation.

x-1=sint+1-1x-1=sint+1-1x-1=sint

Now substitute sint=x-1in the equation y=sin2t-4.

Then,

y=(x-1)2-4y=x2-2x+1-4y=x2-2x-3

03

Further simplification

Here the sintfunction is periodic and lies in between -1,1. And x, y are also bounded.

Here the values that satisfy the equationy=x2-2x-3 are

Thus, the parameterization of the curves sint=x-1,y=sin2t-4 traces exactly the same curve y=x2-2x-3 as that of the equations x=2t+1,y=4t2-4.

Hence the explanation.

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