Plane Curve with Given Curvature

Question

Answer 1

There is probably a better way to do this. Review your notes from lecture and the examples in your book. I'm starting with on a basis from vector calculus, so your class might have learned some additional material for this problem.

Remember the curvature of a planar curve is

k(x) = y''(x)/(1+y'(x)²)^3/2

Let

k(x) = 1/(1+s(x)²)

where s(x) is the length of the curve starting at a defined as,

s(x) = ∫_a^x (1+y'(t)²)^1/2 dt.

From the fundamental theorem of calculus

s'(x) = (1+y'(x)²)^1/2.

Then

y'(x) = (s'(x)²-1)^1/2

y''(x) = s''(x)/(s'(x)²-1)^1/2

Plugging these in and making the curvature requirement gives

k(x) = s''(x)/(s'(x)²-1)^1/2/s'(x)² = 1/(1+s(x)²).

s''(x)/(s'(x)²-1)^1/2/s'(x) = s'(x)/(1+s(x)²)

Integrating both sides (and ignoring the integration constant) gives

-arctan(√s'(x)²-1) = arctan(s(x))

√ s'(x)²-1 = s(x)

s'(x)²-1 = s(x)²

Solving the first order differential equation gives

s(x) = sinh(x)

s'(x) = cosh(x) = (1+y'(x)²)^1/2

y'(x) = sinh(x)

y(x) = cosh(x) + C

Check

|y''(x)| = cosh(x)

y'(x) = sinh(x)

(1+y'(x)²)^3/2 = (1 + sinh²(x))^3/2 = cosh³(x)

k(x) = 1/cosh²(x) = 1/(1+sinh(x)²) = 1/(1+s²)