Relationship between radius of curvature and slope

  • Thread starter bpcraig
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Hello all,

We know that for some well-behaved, smooth/continuous, twice differentiable function of x, f[x] there exists at each point a slope (f ' [x]) and a radius of curvature

[tex]\rho [x]=\frac{\left(1+f'[x]^2\right)^{\frac{3}{2}}}{f\text{''}[x]}[/tex]

It also seems intuitive to think that at every point on such a function, the tangent line vector and the vector from the center of curvature to this point would be normal to one another.

Is this necessarily true?
 

tiny-tim

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hello bpcraig! :wink:
It also seems intuitive to think that at every point on such a function, the tangent line vector and the vector from the center of curvature to this point would be normal to one another.

Is this necessarily true?
yes :smile:
 
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what happens when a parabola is no longer symmetrical? (cutting it half vertically that is) Some real life processes are plotted like parabolas but are not symmetrical.
 

L2E

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then this is still true. this relationship does not need a curve to be symmetrical. it even holds true for a straight line (not really twice differentiable). think about a horizontal line. The slope is 0, so the tangent will be along the original line.
The 'radius of curvature' is infinite and it comes from the bottom of the page directly below the point being observed.
This tangent and line from the radius of curvature to the point chosen intersect at 90 degrees.
 

lavinia

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Hello all,

We know that for some well-behaved, smooth/continuous, twice differentiable function of x, f[x] there exists at each point a slope (f ' [x]) and a radius of curvature

[tex]\rho [x]=\frac{\left(1+f'[x]^2\right)^{\frac{3}{2}}}{f\text{''}[x]}[/tex]

It also seems intuitive to think that at every point on such a function, the tangent line vector and the vector from the center of curvature to this point would be normal to one another.

Is this necessarily true?
For any smooth curve parametrized by arc length, the radius of curvature is the reciprocal of the length of the acceleration vector of the curve. The acceleration vector points normal to the curve.
 

tiny-tim

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ooh, so it does! …

for arc-length, v = 1, and so a = v2/r = 1/r :smile:
 

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