Question

.
18. Show that for the curve y= log x, the least value of radius of curvature is
3√3/2​

Answer 1

Answer:

At the displacement  

Δ

s

along the arc of the curve, the point  

M

moves to the point  

M

1

.

The position of the tangent line also changes: the angle of inclination of the tangent to the positive  

x

−

axis

at the point  

M

1

will be  

α

+

Δ

α

.

Thus, as the point moves by the distance  

Δ

s

,

the tangent rotates by the angle  

Δ

α

.

(The angle  

α

is supposed to be increasing when rotating counterclockwise.)

The absolute value of the ratio  

Δ

α

Δ

s

is called the mean curvature of the arc  

M

M

1

.

In the limit as  

Δ

s

→

0

,

we obtain the curvature of the curve at the point  

M

:

K

=

lim

Δ

s

→

0

 

∣

∣

∣

Δ

α

Δ

s

∣

∣

∣

.

From this definition it follows that the curvature at a point of a curve characterizes the speed of rotation of the tangent of the curve at this point.

For a plane curve given by the equation  

y

=

f

(

x

)

,

the curvature at a point  

M

(

x

,

y

)

is expressed in terms of the first and second derivatives of the function  

f

(

x

)

by the formula

K

=

|

y

′

′

(

x

)

|

[

1

+

(

y

′

(

x

)

)

2

]

3

2

.

If a curve is defined in parametric form by the equations  

x

=

x

(

t

)

,

 

y

=

y

(

t

)

,

then its curvature at any point  

M

(

x

,

y

)

is given by

K

=

|

x

′

y

′

′

−

y

′

x

′

′

|

[

(

x

′

)

2

+

(

y

′

)

2

]

3

2

.

If a curve is given by the polar equation  

r

=

r

(

θ

)

,

the curvature is calculated by the formula

K

=

∣

∣

r

2

+

2

(

r

′

)

2

−

r

r

′

′

∣

∣

[

r

2

+

(

r

′

)

2

]

3

2

.

The radius of curvature of a curve at a point  

M

(

x

,

y

)

is called the inverse of the curvature  

K

of the curve at this point:

R

=

1

K

.

Hence for plane curves given by the explicit equation  

y

=

f

(

x

)

,

the radius of curvature at a point  

M

(

x

,

y

)

is given by the following expression:

R

=

[

1

+

(

y

′

(

x

)

)

2

]

3

2

|

y

′

′

(

x

)

|

.

At the displacement  

Δ

s

along the arc of the curve, the point  

M

moves to the point  

M

1

.

The position of the tangent line also changes: the angle of inclination of the tangent to the positive  

x

−

axis

at the point  

M

1

will be  

α

+

Δ

α

.

Thus, as the point moves by the distance  

Δ

s

,

the tangent rotates by the angle  

Δ

α

.

(The angle  

α

is supposed to be increasing when rotating counterclockwise.)

The absolute value of the ratio  

Δ

α

Δ

s

is called the mean curvature of the arc  

M

M

1

.

In the limit as  

Δ

s

→

0

,

we obtain the curvature of the curve at the point  

M

:

K

=

lim

Δ

s

→

0

 

∣

∣

∣

Δ

α

Δ

s

∣

∣

∣

.

From this definition it follows that the curvature at a point of a curve characterizes the speed of rotation of the tangent of the curve at this point.

For a plane curve given by the equation  

y

=

f

(

x

)

,

the curvature at a point  

M

(

x

,

y

)

is expressed in terms of the first and second derivatives of the function  

f

(

x

)

by the formula

K

=

|

y

′

′

(

x

)

|

[

1

+

(

y

′

(

x

)

)

2

]

3

2

.

If a curve is defined in parametric form by the equations  

x

=

x

(

t

)

,

 

y

=

y

(

t

)

,

then its curvature at any point  

M

(

x

,

y

)

is given by

K

=

|

x

′

y

′

′

−

y

′

x

′

′

|

[

(

x

′

)

2

+

(

y

′

)

2

]

3

2

.

If a curve is given by the polar equation  

r

=

r

(

θ

)

,

the curvature is calculated by the formula

K

=

∣

∣

r

2

+

2

(

r

′

)

2

−

r

r

′

′

∣

∣

[

r

2

+

(

r

′

)

2

]

3

2

.

The radius of curvature of a curve at a point  

M

(

x

,

y

)

is called the inverse of the curvature  

K

of the curve at this point:

R

=

1

K

.

Hence for plane curves given by the explicit equation  

y

=

f

(

x

)

,

the radius of curvature at a point  

M

(

x

,

y

)

is given by the following expression:

R

=

[

1

+

(

y

′

(

x

)

)

2

]

3

2

|

y

′

′

(

x

)

|

.

Explanation:

Answer 2

Answer:

Lets say that a curve, r(t) = x(t) , y(t), z(t) in $R^3$ that has a curvature as $\frac{|r2(t) * r2(t)|}{|r1(t)|^3}$

So we can say that, our curve here is r(t) = <t. Ln(t),0>

So, r1(t) = <1,1/t,0> and r2(t) =<0,$-t^2$,0>

Since the value of x is equal to t, x=t, the curvature of ln(x) at any value of x in the given domain will be equal to :

$\frac{| < 1, 1/x, 0 > * < 0, -x^-2,0 > }{| < 1,1/x,0|^3}$

$\frac{| < 0,0,-x^-2 > |}{(\sqrt{1^2+(1/x)^2+0^2})^3 }$

$\frac{x^-2}{(1+1/x^2)^3/2}$

$\frac{1}{\frac{(x^2+1)^3/2}{x} }$

$\frac{x}{(x^2+1)^3/^2}$

The goal is to maximise this function. So we solve while setting the derivative equal to 0.

$\frac{(x^2+1)^3^/^2 - 2x^2 \frac{3}{2} (x^2+1)^1^/^2}{(x^2+1)^3}$ = 0

$(X^2+1)^3^/^2 - 2x^2 \frac{3}{2} (x^2+1)^1^/^2 = 0$

$(x^2 +1) ^1^/^2$ is absolutely positive so we know that we may divide by this to yield :

$(x^2+1) -3x^2 = 0$

$1= 2x^2$

Therefore this gives us :

$x =$± $\frac{1}{\sqrt{2} }$

Since the negative option is out of the question, we rule it out. If we were being cautious, we would make sure that we have a maximum and not a minimum or a saddle by checking the second derivative of the curvature function. However, examining the graph of In (x). Since the curvature appears to reach its maximum in this area, we can tell that we have reached a maximum.

Additionally, inserting $x = 1/\sqrt{2}$into the calculated curvature function results in a maximum curvature of

$\frac{1/\sqrt{2} }{(3/2)^3^/^2} =\frac{2}{3\sqrt{3} }$ ≈$0.3849$

Henceforth proved.

#SPJ3