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Mathematics
Mathematics
Problem
If
x
=
a
sin
2
θ
(
1
+
cos
2
θ
)
,
y
=
b
cos
2
θ
(
1
−
cos
2
θ
)
x = a \sin 2\theta(1 + \cos 2\theta), y = b \cos 2\theta(1 - \cos 2\theta)
x
=
a
sin
2
θ
(
1
+
cos
2
θ
)
,
y
=
b
cos
2
θ
(
1
−
cos
2
θ
)
, then
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
(a)
b
tan
θ
a
\frac{b \tan \theta}{a}
a
b
t
a
n
θ
(b)
a
tan
θ
b
\frac{a \tan \theta}{b}
b
a
t
a
n
θ
(c)
a
b
tan
θ
\frac{a}{b \tan \theta}
b
t
a
n
θ
a
(d)
b
a
tan
θ
\frac{b}{a \tan \theta}
a
t
a
n
θ
b
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