proof of the Curvature of a plane curve. Very interesting problem!!!?
Prove that the curvature of a plane curve is K= |d(phi)/ds| where (phi) is the angle between the unit tangent vector T and L i.e the angle of inclination of T
- 1 decade agoFavorite Answer
The curvature of a plane curve is a quantity which measures the amount by which the curve differs from being a straight line. It measures the rate at which the direction of a tangent to the curve changes.
The simplest way to introduce the curvature is by first parameterizing the curve with respect to arclength. Suppose that s denotes arclength and that the curve is specified by two functions f and g of this parameter. In other words, a typical point of the curve is (f(s)g(s)) , where s must lie in some specified range. Recall that the condition that s be the arclength is that (f)2+(g)2=1 .
Let denote the angle which the tangent vector makes with the x axis. Because of the arclength condition mentioned above, we have f(s) g(s) = = cos sin The curvature is simply the derivative of this angle with respect to arclength: =dsd
It is convenient to re-express the curvature in terms of f and g . To do this, we differentiate the previous equation: f g = = ddscos=−sindsd=−sin ddssin=cosdsd=cos By eliminating , we obtain the following formulae for the curvature: = = −fg gf Because of the condition (f)2+(g)2=1 , both f and g cannot simultaneously be zero, so at least one of the above formulae must be valid at any point on the curve.
At first, it may seem odd that we have obtained two different formulae for the same quantity. The reason for this is simple. Differentiating the arclength condition (f)2+(g)2=1 gives ff+gg=0 or, dividing out, gf=−fg which explains why the two formulae for the curvature must agree. In fact, one can easily derive several other formulae for the curvature by using this identity.
For instance, one might want to obtain a formula which is explicitly invariant under rotation. Consider the following determinant: f f g g =fg−fg On the one hand, this is clearly invariant under rotation. On the other hand, we have fg−fg=(f)2+(g)2=(f)2+(g)2= hence we have the explicitly rotation-invariant formula = f f g g
Typically, when one is given a curve, it is not specified in terms of a parameterization by arclength. Since reparameterizing a curve by arclength is not always easy, it is useful to have a formula for curvature which is invariant under reparameterization since one could use such a formula with any parameterization. Such a formula can be obtained by a slight modification of the rotation-invariant formula given above.
To obtain this formula, first let us inquire into how the determinant transforms under change of parameterization. If we apply a change of parameter =(s) then, by the chain rule, dsdf=dfddsd=(s)dfd dsdg=(s)dgd Missing close brace Missing close brace Thus, we have the following transformation for the determinant: dfds d2fds2 dsdg ds2d2g =(s)3 dfd d2fd2 dgd d2gd2 Likewise, one has the following transform: dsdf2+dsdg2=(s)2dfd2+dgd2 Therefore, the following quantity is invariant under both rotation and reparameterization: dfd d2fd2 dgd d2gd2 dfd2+dgd232 In the particular case where =s , this equals the curvature; hence, by invariance, it equals the curvature for all choices of parameterization: = dfd d2fd2 dgd d2gd2 dfd2+dgd232
One special case is especially worth noting. Suppose that the curve is given as the graph of a function. That is equivalent to choosing one of f or g to be the identity function. Then the formula reduces to the following: =f1+(f)232 It is worth noting that, at points where f=0 (i.e. where the tangent to the curve is horizontal) the curvature simply equals the second derivative. This observation leads to another characterization of the curvature -- the curvature of a curve at a point can be obtained by setting up a coordinate system whose abscissa is the tangent to the curve at that point, expressing the curve as the graph of a function in this coordinate system, then taking the second derivative of this function at said point. It might also be worth pointing out the curvature of a curve at a point equals the reciprocal of the radius of the osculating circle to the curve at that point.
- davidLv 65 years ago
First draw a diagram of a curve and a point P on the curve at which the tangent makes an angle phi with the x axis. Then move a distance ds along the curve to he point Q. The tangent at point Q makes an angle phi + dphi with the x axis. Now draw a point O on the concave side of the curve. O is the centre of curvature. OPQ is a sector of a circle whose radius is r, and the arc length PQ=ds. The angle POQ is dphi. Therefore we have r dphi = ds! or dphi/ds= 1/r which is the curvature, defined as the reciprocal of the radius of curvature.
- lazareLv 43 years ago
The equation for the shortest line is f(x)=-4 abs(x-.5) +2 The equation to locate the size of the curve is the vital over the era 0 to one million of the formula (one million+ f'(x)^2)^(one million/2). Which for this reason seems to be 4.1231. the area could properly be discovered by making use of integrating f(x) over the era 0 to one million. which delivers the respond for the area to be one million.
- wghLv 51 decade ago
What is L?