1points are (-3√3a4 , a4) ;( 3√3a4 , a4);(0,2a)
@sibal explain the process how u did the graph ..........
sketch the graph for
x = a sin 2θ(1 + cos2θ)
y = a cos 2θ(1 - cos2θ)
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6 Answers
1is der ne technique to draw the graphs of given equationss.........bhaiiya can u sggest sum trickss or general approach to 'em.....
1
1see, to get a rough graph, what sibal has done, presume is,
that he has first found the roots for x=0 andy=0, that happen to be
(0,2a),(a,0),(-a,0).
This is achieved by getting values of theta for x=0, and putting in y, and vise-versa.
now, we analyse dy/d(2theta)
= dy/d(2theta)
= a(cos2@ + cos4@)
and similarly, dx/d(2@) = a(sin4@-sin2@)
to get max and min attainable y, the former is 0, and for max and min attainable x the latter should be 0.
that way you get the boundary points as (-3√3a/4 , a/4) ;( 3√3a/4 , a/4);(0,2a)
between 0 and pi/2, dy/dx does not ever become 0, hence its safe to assume the function is increasing from x=0 to a, and similarly decreasing from -a to 0. also we can similarly arrive at the conclusion that it decreases from -3√3a/4 again to 0.
but this ofcourse doesnot give u the clearest picture of the graph.
you need to investigate d2y/dx2 to get the clearest graph.
cheers!!
