the center of curvature of y =x^2 at origin is
Answers
Step-by-step explanation:
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
.
Example 3.
Find the curvature and radius of curvature of the curve
y
=
cos
m
x
at a maximum point.
Solution.
This function reaches a maximum at the points
x
=
2
π
n
m
,
n
∈
Z
.
By the periodicity, the curvature at all maximum points is the same, so it is sufficient to consider only the point
x
=
0.
Write the derivatives:
y
′
=
(
cos
m
x
)
′
=
−
m
sin
m
x
,
y
′
′
=
(
−
m
sin
m
x
)
′
=
−
m
2
cos
m
x
.
The curvature of this curve is given by
K
=
|
y
′
′
|
[
1
+
(
y
′
)
2
]
3
2
=
∣
∣
−
m
2
cos
m
x
∣
∣
[
1
+
(
−
m
sin
m
x
)
2
]
3
2
=
∣
∣
−
m
2
cos
m
x
∣
∣
(
1
+
m
2
sin
2
m
x
)
3
2
.
At the maximum point
x
=
0
,
the curvature and radius of curvature, respectively, are equal to
K
(
x
=
0
)
=
m
2
(
1
+
m
2
sin
2
0
)
3
2
=
m
2
,
R
=
1
K
=
1
m
2
.