Question

5, (25 points 4 pages max) Suppose that γ(t) = (x(t), y(t)) is a smooth (infinitely differentiable) plane curve. For curves such that lh'(t) 0, the (signed) curvature is defined to be the quantity K(t) (a) Suppose the curve γ(t) is the graph of a function, ie x(t)-t and y(t) f(t) for some function f. Write the formula for the curve in this case. Suppose you were at a critical point of the graph of f. What does the curvature tell you in this case? (b) Suppose that γ is a unit speed curve, and that K(t)-0. Show that γ(t) must be a straight ine. (c) Suppose that γ is a unit speed curve, that K(t-k a (real) constant. i. Show that γ(t) lies on a circle ii. Find the radius of such a circle (Hint: "recall" that the solution to the equation y"(t)+w2y 0isy(t-asin(wt)+bcos(wt).)

Answer 1

a) From
r (t)
and
y(t) = f(t)
we get

$x'(t)=1,~x''(t)=0,~~y'(t)=f'(t),~y''(t)=f''(t)$

Therefore, the curvature formula for this case is

f"t

At a critical point we have , which implies

f'(t) =0

f"(t) c(t) = = f"(t)

b) For unit speed curve we have ; therefore, we get

γ"(t)

$0=\kappa(t)={\frac{x'y''-x''y'}{((x')^2+(y')^2)^{\frac 32}}}=x'y''-x''y'$

Therefore,

$0={\frac{x'y''-x''y'}{(y')^2}}={\frac{x'y''-x''y'}{(x')^2}}=d\begin{pmatrix}{\frac{y'}{x'}}\end{pmatrix}$

which gives

c +d

for some constant . This proves that $\gamma(t)=(x(t),y(t)=cx(t)+d)$ is a straight line.

c) In this case, we have $\gamma'\times \gamma''=k$ . Now, letting ${\bf n}=k\gamma'$ denote the unit normal,

${\frac d{dt}}(\gamma+{\frac 1k}{\bf n})=0$

Thus,

$\gamma+{\frac 1k}{\bf n}={\bf a}$

is a constant vector. This implies

$||\gamma-{\bf a}||={\frac 1k}$

Thus, $\gamma$ lies on a circle of radius , centered at the vector ${\bf a}$.