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Total marks: 20
Lecture # 23 to 30
Due date: July 17, 2014
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+ Click Here to Search (Looking For something at vustudents.ning.com?)Arshad Bhai, In fourth line of Q3 correction is required.......
Answer # 1:
Equation of the curve in polar co-ordinates is r2=-4sin2θ
(i) Symmetry about the Initial Line.
r2=-4sin2θ
Remember that sine is an odd function
Put (r,-θ) in equation
r2=-4sin2 (-θ)
r2=-4sin (-2θ)
r2=4sin2θ
This equation is not symmetry about initial line
(ii) Symmetry about the Pole
r2=-4sin2θ
Put (-r, θ) in equation
(-r) 2 =-4sin2θ
r2=-4sin2θ
This equation is symmetry about pole
rdrdθ ko dudθ main q convert kia haiiiiiiiiiiiiii???????????????????
Q:1:- Equation of the curve in polar co-ordinates is =-4sin2 discuss the symmetry of graph of this curve about.
Equation of the curve in polar co-ordinates is
=-4sin2θ
Check it
Q:1:- Equation of the curve in polar co-ordinates is =-4sin2 discuss the symmetry of graph of this curve about.
Initial line
Pole
Equation of the curve in polar co-ordinates is
=-4sin2θ
Symmetry about the Initial Line.
=-4sin2θ
sin is odd function.
Put (r, -θ) in equation
=-4sin2 (-θ)
=-4sin (-2θ)
=4sin2θ
this equation is not symmetry about initial line
Symmetry about the Pole
=-4sin2θ
Put (-r, θ) in equation
=-4sin2θ
=-4sin2θ
This equation is symmetry about pole
MTH301_Complete_Solution
Tariq Bro, you have uploaded two solution for 3rd question which one is correct?
Doc file solution or above mentioned?
Plz ans.....
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