Center of curvature for curve x^3+y^3=2 at (1,1)
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Answer:
Solution.
Write the derivatives of the quadratic function:
y
′
=
(
x
2
)
′
=
2
x
;
y
′
′
=
(
2
x
)
′
=
2.
Then the curvature of the parabola is defined by the following formula:
K
=
y
′
′
[
1
+
(
y
′
)
2
]
3
2
=
2
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
.
At the origin (at
x
=
0
), the curvature and radius of curvature, respectively, are
K
(
x
=
0
)
=
2
(
1
+
4
⋅
0
2
)
3
2
=
2
,
R
=
1
K
=
1
2
.
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