Math, asked by jangirhimani8, 2 months ago

the center of curvature of y =x^2 at origin is

Answers

Answered by shilpisanjanakumari
0

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

.

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