if x=a cos^3thita and y = a sin^3 thita = pie / 3
Answers
Step-by-step explanation:
We have,
x=acos
3
θ ……. (1)
y=asin
3
θ ……… (2)
On differentiating to equation (1) w.r.t θ, we get
dθ
dx
=a(3cos
2
θ)(−sinθ)
dθ
dx
=−3asinθcos
2
θ
On differentiating both sides w.r.t θ, we have
dθ
2
d
2
x
=−3a(cos
2
θcosθ+sinθ(2cosθ(−sinθ)))
dθ
2
d
2
x
=−3a(cos
3
θ−2sin
2
θcosθ)
Similarly,
On differentiating to equation (2) w.r.t θ, we get
dθ
dy
=a(3sin
2
θ)(cosθ)
dθ
dy
=3asin
2
θcosθ
On differentiating both sides w.r.t θ, we have
dθ
2
d
2
y
=3a(sin
2
θ(−sinθ)+cosθ(2sinθ(cosθ)))
dθ
2
d
2
y
=3a(−sin
3
θ+2cos
2
θsinθ)
Therefore,
dθ
2
d
2
x
dθ
2
d
2
y
=
−3a(cos
3
θ−2sin
2
θcosθ)
3a(−sin
3
θ+2sinθcos
2
θ)
dx
2
d
2
y
=
(−cos
3
θ+2sin
2
θcosθ)
(−sin
3
θ+2sinθcos
2
θ)
Put θ=
6
π
,we get
dx
2
d
2
y
∣
∣
∣
∣
∣
θ=
6
π
=
(−cos
3
(
6
π
)+2sin
2
(
6
π
)cos(
6
π
))
(−sin
3
(
6
π
)+2sin(
6
π
)cos
2
(
6
π
))
dx
2
d
2
y
∣
∣
∣
∣
∣
θ=
6
π
=
⎝
⎛
−(
2
3
)
3
+2(
2
1
)
2
×(
2
3
)
⎠
⎞
⎝
⎛
−(
2
1
)
3
+2×
2
1
×(
2
3
)
2
⎠
⎞
dx
2
d
2
y
∣
∣
∣
∣
∣
θ=
6
π
=
(−
8
3
3
+
4
3
)
(−
8
1
+
4
3
)
dx
2
d
2
y
∣
∣
∣
∣
∣
θ=
6
π
=
(
8
−3
3
+2
3
)
(−
8
1+6
)
dx
2
d
2
y
∣
∣
∣
∣
∣
θ=
6
π
=
3
7