Envelope of the family of curves

fitzgilbert

New member
Joined
Feb 4, 2008
Messages
1
Find envelope of the family of curves x^2cos? + y^2sin? = a^2 where ? is the parameter

I differentiated eq1 with respect to theta and got

-x^2sin theta + y^2cos theta

which is eq 2 but i just cant figure out how to isolate The theta in such a way that i can sub it back into eq 1

thanks
Fitzgilbert
 
Hello, fitzgilbert!

\(\displaystyle \text{Find envelope of the family of curves: }x^2\cos\theta + y^2\sin\theta \;=\;a^2\text{, where }\theta\text{ is the parameter}\)

\(\displaystyle \text{I differentiated eq1 with respect to }\theta\text{ and got: }-x^2\sin\theta + y^2\cos\theta \:=\:0\)

\(\displaystyle \text{We have: }\;x^2\sin\theta \:=\:y^2\cos\theta\quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:\frac{y^2}{x^2}\quad\Rightarrow\quad \tan\theta \:=\:\frac{y^2}{x^2}\)

\(\displaystyle \text{Then: }\:\tan\theta \:=\:\frac{y^2}{x^2} \:=\:\frac{opp}{adj}\)

\(\displaystyle \theta\text{ is in a right triangle with: }\:eek:pp = y^2,\;adj = x^2\)

. . \(\displaystyle \text{Then: }\;hyp \:=\:\sqrt{x^4+y^4}\)

\(\displaystyle \text{Hence: }\;\sin\theta \:=\:\frac{y^2}{\sqrt{x^4+y^4}}\qquad \cos\theta \:=\:\frac{x^2}{\sqrt{x^4+y^4}}\)

\(\displaystyle \text{Now substitute those into the equation . . . }\)

 
Top