the differential equation of orthogonal trajectories of families of curves r = a(1 - cos theta) is
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Answer:
moment of inertia of the area of the upper half of the circle X square + Y square is equal to a square about the line X + Y is equal to 2 a mid density 5 integration square pin DX DY then a is equal to
Answered by
3
Answer:
A
r=
1+cosθ
2C
We have, 1−cosθ=
r
2k
or r=
1−cosθ
2k
(1)
Differentiating w.r.t. θ, we get
dθ
dr
=−
(1−cosθ)
2
2ksinθ
.(2)
Eliminating k from (2) using (1), we get
r
1
dθ
dr
=−
1−cosθ
sinθ
=−
2sin
2
2
θ
2sin
2
θ
cos
2
θ
=−
sin
2
θ
cos
2
θ
.(3)
Replacing
dθ
dr
with −r
2
dr
dθ
in (3), we get
r
dr
dθ
=
sin
2
θ
cos
2
θ
⇒
r
1
dr=
cos
2
θ
sin
2
θ
dθ
On integrating, we get
logr=−2logcos
2
θ
+logC
⇒rcos
2
2
θ
=C
⇒
2
r
(1+cosθ)=C
⇒r=
1+cosθ
2C
Step-by-step explanation:
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