Question

x =\sin \left(\frac{\theta}{2}\right), y = \cos \left(\frac{\theta}{2}\right), -\pi \leq \theta \leq \pi (a) Eliminate the parameter to find a Cartesian equation of the curve. and how does thecurve look

Integrals
ANSWERED
asked 2021-05-08
\(x =\sin \left(\frac{\theta}{2}\right), y = \cos \left(\frac{\theta}{2}\right), -\pi \leq \theta \leq \pi\) (a) Eliminate the parameter to find a Cartesian equation of the curve. and how does thecurve look

Expert Answers (2)

2021-05-09
image
17
 
Best answer
2021-09-30

Consider the following parametric equations,

\(x=\sin(\frac{\theta}{2})\) and \(y=\cos(\frac{\theta}{2}),\ -\pi\leq\theta\leq\pi\)

To eliminate \(\theta\), square x and y both sides and add them

\(x^2=\sin^2(\frac{\theta}{2})\)

\(y^2=\cos^2(\frac{\theta}{2})\)

\(x^2+y^2=\sin^2(\frac{\theta}{2})+\cos^2(\frac{\theta}{2})\)

\(x^2+y^2=1\)

The equation \(x^2+y^2=1\) represents a circle of radius. Given that the perimeter \(\theta\) varies from \(-\pi\) to \(\pi\)

Substitute \(\theta=-\pi\) in the perimetric equations

\(x=\sin(\frac{\theta}{2})\) and \(y=\cos(\frac{\theta}{2})\)

\(x=\sin(\frac{-\pi}{2})=-1\)

\(y=\cos(\frac{-\pi}{2})=0\)

So the point corresponding to \(\theta=-\pi\) is \((-1,0)\)

Substitute \(\theta=\pi\) in the perimetric equations

\(x=\sin(\frac{\theta}{2})\) and \(y=\cos(\frac{\theta}{2})\)

\(x=\sin(\frac{-\pi}{2})=-1\)

\(y=\cos(\frac{-\pi}{2})=0\)

So the point corresponding to \(\theta=-\pi\) is \((-1,0)\)

Substitute \(\theta=\pi\) in the perimetric equations

\(x=\sin(\frac{\theta}{2})\) and \(y=\cos(\frac{\theta}{2})\)

\(x=\sin(\frac{\pi}{2})=1\)

\(y=\cos(\frac{\pi}{2})=0\)

So, the point corresponding to \(\theta=\pi\) is \((1,0)\)

Hence, the curve is a senincircle with origin (0,0) radius 1.

As \(\theta\) increases from \(-\pi\) to \(\pi\), the point traces semicircles in clockwise direction, storting from the point \((-1,0)\) to the point \((1,0)\)

12

expert advice

Have a similar question?
We can deal with it in 3 hours
...