!!How do we change this parametric equation to cartesian equation?? (1 Viewer)

fabl · Apr 27, 2015

r(t) = 2tan(t)i + 2cos^2(t)j
I guess

x(t) = 2tan(t)
and
y(t) =2cos^2(t)

??

InteGrand · Apr 27, 2015

fabl said:
r(t) = 2tan(t)i + 2cos^2(t)j
I guess

x(t) = 2tan(t)
and
y(t) =2cos^2(t)

??

Those are still parametric equations, but they're the right parametric equations.

To eliminate t:

$\frac{1}{y}=\frac{1}{2}\sec ^2 t= \frac{1}{2}(\tan^2 t + 1) = \frac{1}{2}\left(\left(\frac{x}{2}\right) ^2 + 1\right)$

So $y=\frac{2}{\left(\frac{x}{2}\right) ^2 + 1} = \frac{2}{\frac{x^2}{4} + 1}=\frac{8}{x^2 + 4}$ .

braintic · Apr 27, 2015

Square x, turn tan^2 into sec^2, rearrange to make cos^2 the subject, and you almost have y.

!!How do we change this parametric equation to cartesian equation?? (1 Viewer)

fabl

Member

InteGrand

Well-Known Member

braintic

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)