Finding maximum curvature on lnx

In summary: First, f '' (x) should be f " (x), and second, the derivatives should be dy/dx, not dx/dy. Here are the corrected formulas:k(x) = 1/x2 / { [1 + (1/x)2 ] 3/2 }k(x) = 1/x2 / { [ x2 +1 /x2 ] 3/2 }
  • #1
tensor0910
Gold Member
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Homework Statement



Determine the point on a plane curve f(x) = ln x where the curvature is maximum.[/B]

Homework Equations



k(x) = || T ' (x) || / || r ' (x ) ||

k (x) = f '' (x) / [ 1 + ( f '' (x))2 ] 3/2[/B]

The Attempt at a Solution



f ' (x) = 1/x
f " (x) = -1/x2

k(x) = 1/x2 / { [1 + (1/x)2 ] 3/2 }

k(x) = 1/x2 / { [ x2 +1 /x2 ] 3/2 }

iirc we can use the first derivative and try to find the local max...but how do we even start it with this mess? quotient rule? Maybe there's something about the problem that tells us its impossible...? I'm lost here...
 
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  • #2
tensor0910 said:

Homework Statement



Determine the point on a plane curve f(x) = ln x where the curvature is maximum.[/B]

Homework Equations



k(x) = ιι T ' (x) ιι / ιι r ' (x ) [/B]
There's a lot of extra stuff that I can't read in your formula above. What are these characters? ιι
tensor0910 said:

k (x) = ι f '' (x) ι / [ 1 + ( f '' (x))2 ] 3/2
And what's this one? ι
tensor0910 said:

The Attempt at a Solution



f ' (x) = 1/x
f " (x) = 1/x2
You lost a sign here (above).
tensor0910 said:
k(x) = 1/x2 / { [1 + (1/x)2 ] 3/2 }

k(x) = 1/x2 / { [ x2 +1 /x2 ] 3/2 }

iirc we can use the first derivative and try to find the local max...but how do we even start it with this mess? quotient rule? Maybe there's something about the problem that tells us its impossible...? I'm lost here...
Sure, quotient rule would work, provided you are careful and methodical. It would be helpful to simplify the expression as much as possible for attempting to differentiate it.
 
  • #3
I cleaned it up a bit. Sorry!
 
  • #4
tensor0910 said:

Homework Statement



Determine the point on a plane curve f(x) = ln x where the curvature is maximum.[/B]

Homework Equations



k(x) = || T ' (x) || / || r ' (x ) ||

k (x) = f '' (x) / [ 1 + ( f '' (x))2 ] 3/2[/B]

The Attempt at a Solution



f ' (x) = 1/x
f " (x) = -1/x2

k(x) = 1/x2 / { [1 + (1/x)2 ] 3/2 }

k(x) = 1/x2 / { [ x2 +1 /x2 ] 3/2 }

iirc we can use the first derivative and try to find the local max...but how do we even start it with this mess? quotient rule? Maybe there's something about the problem that tells us its impossible...? I'm lost here...
There are some mistakes in your formulas.
 

What is the purpose of finding maximum curvature on lnx?

The purpose of finding maximum curvature on lnx is to determine the point where the graph of the function has the sharpest change in direction. This can help in analyzing the behavior of the function and its critical points.

How do you find the maximum curvature on lnx?

To find the maximum curvature on lnx, you need to calculate the second derivative of the function and set it equal to zero. Then solve for x to get the critical points. After that, you can use the first derivative test or the second derivative test to determine if the critical points are points of maximum curvature.

What is the formula for calculating maximum curvature on lnx?

The formula for calculating maximum curvature on lnx is:
K = |(f''(x))/(1+(f'(x))^2)^(3/2)|
where f''(x) is the second derivative of the function and f'(x) is the first derivative of the function.

Can the maximum curvature on lnx be negative?

Yes, the maximum curvature on lnx can be negative. This means that the graph of the function has a point of inflection where the concavity changes from upwards to downwards or vice versa.

What are the real-world applications of finding maximum curvature on lnx?

Finding maximum curvature on lnx can be useful in various fields such as engineering, physics, and economics. It can help in optimizing the shape of objects, analyzing the trajectory of objects in motion, and understanding the behavior of economic models.

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