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Mathematics
Mathematics
Problem
If
x
=
3
a
t
1
+
t
3
,
y
=
3
a
t
2
1
+
t
3
x = \frac{3at}{1+t^3}, y = \frac{3at^2}{1+t^3}
x
=
1
+
t
3
3
a
t
,
y
=
1
+
t
3
3
a
t
2
, then
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
(a)
t
(
2
+
t
3
)
1
−
2
t
3
\frac{t(2+t^3)}{1-2t^3}
1
−
2
t
3
t
(
2
+
t
3
)
(b)
t
(
2
−
t
3
)
1
−
2
t
3
\frac{t(2-t^3)}{1-2t^3}
1
−
2
t
3
t
(
2
−
t
3
)
(c)
t
(
2
+
t
3
)
1
+
2
t
3
\frac{t(2+t^3)}{1+2t^3}
1
+
2
t
3
t
(
2
+
t
3
)
(d)
t
(
2
−
t
3
)
1
+
2
t
3
\frac{t(2-t^3)}{1+2t^3}
1
+
2
t
3
t
(
2
−
t
3
)
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