2.1 DIFFERENTIATION
Show that the radius of curvature at the point(x, y)on the ellipse
x^2
a^2
+
y^2
b^2
=1
has magnitude(a^4 y^2 +b^4 x^2 )^3 /^2 /(a^4 b^4 )and the opposite sign toy. Check the special case
b=a, for which the ellipse becomes a circle.
Differentiating the equation of the ellipse with respect toxgives
2 x
a^2
+
2 y
b^2
dy
dx
=0
and so
dy
dx
=−
b^2 x
a^2 y
.
A second differentiation, using (2.13), then yields
d^2 y
dx^2
=−
b^2
a^2
(
y−xy′
y^2
)
=−
b^4
a^2 y^3
(
y^2
b^2
+
x^2
a^2
)
=−
b^4
a^2 y^3
,
where we have used the fact that (x, y) lies on the ellipse. We note thatd^2 y/dx^2 , and hence
ρ, has the opposite sign toy^3 and hence toy. Substituting in (2.19) gives for the magnitude
of the radius of curvature
|ρ|=
∣
∣
∣∣
∣
[
1+b^4 x^2 /(a^4 y^2 )
] 3 / 2
−b^4 /(a^2 y^3 )
∣
∣
∣∣
∣
=
(a^4 y^2 +b^4 x^2 )^3 /^2
a^4 b^4
.
For the special caseb=a,|ρ|reduces toa−^2 (y^2 +x^2 )^3 /^2 and, sincex^2 +y^2 =a^2 ,thisin
turn gives|ρ|=a, as expected.
The discussion in this section has been confined to the behaviour of curves
that lie in one plane; examples of the application of curvature to the bending of
loaded beams and to particle orbits under the influence of a central forces can be
found in the exercises at the ends of later chapters. A more general treatment of
curvature in three dimensions is given in section 10.3, where a vector approach is
adopted.
2.1.10 Theorems of differentiation
Rolle’s theorem
Rolle’s theorem (figure 2.5) states that if a functionf(x) is continuous in the
rangea≤x≤c, is differentiable in the rangea<x<cand satisfiesf(a)=f(c)
then for at least one pointx=b,wherea<b<c,f′(b) = 0. Thus Rolle’s
theorem states that for a well-behaved (continuous and differentiable) function
that has the same value at two points either there is at least one stationary point
between those points or the function is a constant between them. The validity of
the theorem is immediately apparent from figure 2.5 and a full analytic proof will
not be given. The theorem is used in deriving the mean value theorem, which we
now discuss.