If x=asinthita and y=bcosthita, then d^2y_dx^2 is equal to
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Step-by-step explanation:
Correct option is
D
a
2
−b
sec
3
θ
Given, x=asinθ and y=bcosθ
On differentiating w.r.t.θ, we get
dθ
dx
=acosθ
and
dθ
dy
=−bsinθ
⇒
dx
dy
=
dx/dθ
dy/dθ
=−
a
b
tanθ
Again, differentiating w.r.t. x, we get
dx
2
d
2
y
=−
a
b
sec
2
θ⋅
dx
dθ
⇒
dx
2
d
2
y
=−
a
b
sec
2
θ⋅
acosθ
1
=−
a
2
b
sec
3
θ
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