Math 32A Discussion Session Week 5 Notes November 7 and 9, 2017

This week we want to talk about and osculating circles. You might notice that these notes contain a lot of the same theory or proofs that you saw in lecture. In section we won’t focus much on this, but instead will focus on examples.

Curvature Now that we’re parametrizing curves, it makes sense to wonder how we might measure the extent to which a curve actually curves. That is, how much does our path deviate from being a straight line? When we travel along a straight line, our vector always points along our path; along a curving path, the tangent vector approximates our path, but doesn’t exactly match it.

Figure 1: Tangent vectors approximating a path.

Since the tangent vector of our parametrization gives a straight-line approximation to our curve, a reasonable first attempt at measuring curvature is to ask how quickly our tangent vector is changing. That is, if we have a parametrization r(t) for our curve, we might consider kr00(t)k, the speed with which r0(t) is changing. To see why this might measure curvature, have a look at Figure1. Near points where our path makes sharp turns, the tangent vectors change quickly, but along straighter parts the tangent vector doesn’t change much. Unfortunately this approach has a big problem — our tangent vector r0(t) depends on how we parametrize the curve. To see what we mean, consider the following parametrizations:

r1(t) = hcos(t), sin(t)i, 0 ≤ t ≤ 2π and r2(t) = hcos(πt/2), sin(πt/2)i, 0 ≤ t ≤ 4.

Both of these parametrizations trace out the unit circle, but they do so at different speeds. As a result, their tangent vectors have different lengths, as can be seen in Figure2. It’s unsurprising, then, that these parametrizations give us different vectors:

π2 r00(t) = h− cos(t), − sin(t)i and r00(t) = h− cos(πt/2), − sin(πt/2)i. 1 2 4 Both parametrizations would tell us that the circle has constant curvature (which seems accurate), 2 but r1 would call this curvature 1, while r2 would call it π /4: π2 kr00(t)k ≡ 1 and kr00(t)k ≡ . 1 2 4

1 Figure 2: Tangent vectors for r1 (in blue) and r2 (in orange).

As a means of standardizing our curvature measurements, we can replace the idea of measuring how quickly the tangent vector is changing with measuring how quickly the unit tangent vector is changing. That is, as long as our tangent vector r0(t) is never zero, we can scale it to be unit length and call this vector T(t) — our unit tangent vector. We could then measure how quickly this vector changes. Notice that for our parametrizations of the circle we have

0 r1(t) T1(t) = 0 = h− sin(t), cos(t)i kr1(t)k and 0 r2(t) 2 D π π E T2(t) = 0 = − sin(πt/2), cos(πt/2) = h− sin(πt/2), cos(πt/2)i. kr2(t)k π 2 2

In Figure2 this corresponds to scaling our orange vectors down to unit length, so that if r1(t1) = r2(t2) for some parameter values t1, t2, then T(t1) = T(t2). Our second attempt at curvature is to measure how quickly this unit tangent vector changes, so we compute

0 kT1(t)k = kh− cos(t), − sin(t)ik = 1 and D π π E π kT0 (t)k = − cos(πt/2), − sin(πt/2) = . 2 2 2 2 So our new measure of curvature still has the problem that it depends on how we parametrize our curves.

The problem with asking how quickly the unit tangent vector changes is that we need to be 0 consistent with the value against which we measure this change. When we compute kT1(t)k we’re 0 measuring the change in our unit tangent vector against the parameter for r1, and kT2(t)k mea- sures this change against the parameter for r2. What we really want to know is how T changes with respect to arclength. We want to ask how T will be different one unit down the curve. The following definition is made with this consideration in mind.

2 Curvature. Suppose r(s) is an arclength parametrization for the curve C. The the curvature of C at the point r(s) is given by

dT κ(s) = (s) , ds where T(s) is the unit tangent vector to C at r(s).

Example 1. The parametrization r1(t) of the unit circle given earlier is an arclength parametriza- 0 0 tion, since kr1(t)k ≡ 1. It follows that the curvature of the circle is kT1(t)k ≡ 1. So the circle has constant curvature. This is consistent with our intuition, since the circle is always deviating from its tangent vector in the same way.

The definition of curvature we finally settled on is on a good theoretical footing, since it prevents us from finding different curvature values by using different parametrizations for our curve. But it has a practical problem: finding arclength parametrizations for curves can be very difficult. We’d like to find a way of computing curvature from parametrizations that aren’t necessarily arclength parametrizations.

Suppose r(t) parametrizes the curve C and that r0(t) is never zero, but is not necessarily an arclength parametrization. Then r0(t) T(t) = , kr0(t)k so the chain rule tells us that dT dT ds dT d Z 1  = = kr0(u)kdu . dt ds dt ds dt 0 We then apply the fundamental theorem of calculus to find dT dT = kr0(t)k. dt ds So we can compute the arclength derivative of T according to dT 1 dT = . ds kr0(t)k dt

Since the curvature is the magnitude of this vector, we have

1 dT κ(t) = . kr0(t)k dt

0 Example 2. We’ve already computed kT2(t)k = π/2 from our parametrization r2(t) of the unit 0 circle. Since kr2(t)k = π/2, we see that

1 0 π/2 κ(t) = 0 kT2(t)k = = 1, kr2(t)k π/2 and this agrees with our computation of curvature from the parametrization r1(t).

3 Before computing curvature for a curve other than the circle, we’ll come up with one more formula for curvature that starts from an arbitrary1 parametrization r(t). First we need a lemma. Lemma 1. Let T(t) be a parametrization of the unit tangent vector to a curve C. Then T(t) is perpendicular to T0(t). Proof. Let f(t) = kT(t)k2. Because T(t) always has unit length, f is the constant function 1. At the same time, d f 0(t) = (T(t) · T(t)) = 2T(t) · T0(t). dt Because f is constant, f 0 is always 0, so 2T(t)·T0(t) = 0 for all t, meaning that T(t) is perpendicular to T0(t) for all t.

This observation may seem irrelevant, but it will be helpful in the computation that follows. Suppose r(t) is a parametrization for the curve C so that r0(t) is never zero. Then for all t,

r0(t) = kr0(t)kT(t).

From this we can compute r00(t): d r00(t) = (kr0(t)k)T(t) + kr0(t)kT0(t). dt Now (for no immediately apparent reason) we’ll compute the cross product2 r0(t) × r00(t):

 d  r0(t) × r00(t) = (kr0(t)kT(t)) × (kr0(t)k)T(t) + kr0(t)kT0(t) dt d = kr0(t)k (kr0(t)k)(T(t) × T(t)) + kr0(t)k2(T(t) × T0(t)) dt = kr0(t)k2(T(t) × T0(t)).

The first term on the second-to-last line vanishes because we’re crossing a vector with itself. We now have that

kr0(t) × r00(t)k = kr0(t)k2kT(t) × T0(t)k = kr0(t)k2 sin θkT(t)kkT0(t)k, where θ is the angle between T(t) and T0(t). But our lemma says that this angle must always be π/2. Because sin(π/2) = 1 and kT(t)k = 1, we see that

kr0(t) × r00(t)k = kr0(t)k2kT0(t)k.

This is very nearly the curvature formula — we need only divide by kr0(t)k3. So we see that if our curve is parametrized by r(t) and r0(t) never vanishes, then

kr0(t) × r00(t)k κ(t) = . kr0(t)k3

1Not quite arbitrary: we do require that r0(t) not vanish. 2Computing the cross product assumes that r0(t) and r00(t) are vectors in three dimensions. In the case that we’re working in two dimensions, we simply assume the z-coordinate to be 0.

4 Figure 3: Unit tangent vectors for an .

Example 3. Let C be the ellipse defined by x2 y2 + = 1. 25 4 Find the extreme values for the curvature of C.

(Solution) Since C has major and minor axes of length 5 and 2, respectively, we can parametrize C by r(t) = h5 cos(t), 2 sin(t)i, 0 ≤ t ≤ 2π. Then r0(t) = h−5 sin(t), 2 cos(t)i and r00(t) = h−5 cos(t), −2 sin(t)i, so

i j k 0 00 2 2 r (t) × r (t) = −5 sin(t) 2 cos(t) 0 = h0, 0, 10 sin (t) + 10 cos (t)i = h0, 0, 10i.

−5 cos(t) −2 sin(t) 0 Then kr0(t) × r00(t)k = 10. We also have q q kr0(t)k = 25 sin2(t) + 4 cos2(t) = 21 sin2(t) + 4, so kr0(t) × r00(t)k 10 κ(t) = = . kr0(t)k3 (21 sin2(t) + 4)3/2 We’re interested in minimizing and maximizing κ(t), so we compute 630 cos(t) sin(t) κ0(t) = − . (21 sin2(t) + 4)5/2 Then κ0(t) will vanish whenever either cos(t) or sin(t) is 0, so we have critical points at t = 0, π/2, π, and 3π/2. We calculate the curvature at each of these points and find t 0 π/2 π 3π/2 κ(t) 5/4 2/25 5/4 2/25 So C achieves its maximum curvature of 5/4 at r(0) = (5, 0) and r(π) = (−5, 0) and achieves its minimum curvature of 2/25 at r(π/2) = (0, 2) and r(3π/2) = (0, −2). Notice that this fits with the plot of C in Figure3. The path is curving most sharply at the two horizontal edges, where our unit tangent vectors are changing rapidly. At the vertical extrema of our path, the curve is much straighter, and this is reflected in its smaller curvature. ♦

5 Figure 4: A pair of osculating circles on the curve parametrized by r(t) = h3 cos t, 7 sin t, 4 cos 2ti.

The Osculating Circle In lecture you’ve talked about the unit normal vector to r(t), which can be computed by

T0(t) N(t) = kT0(t)k or using the equation T0(t) = kr0(t)kκ(t)N(t). One important use for the normal vector is to construct the Frenet frame {T, N, B} for our 3 cure. The Frenet frame provides an orthonormal basis for R at each point of our curve, and the derivatives of this frame can be computed within the frame itself :

T0(s) = κN(s) N0(s) = −κT(s) + τB(s), B0(s) = −τN(s) where s is an arclength parameter. This may not sound that impressive at first, but it means that the Frenet frame does a good job of capturing the rotating and twisting done by the curve. We won’t talk much about the Frenet frame, but you should check it out. The above equations are called the Frenet formulas.

Today we’ll use the normal vector to construct a sort of circle-of-best-fit for curves. Suppose (x, y) is a point on the curve C, and that C has nonzero curvature at (x, y). Then the osculating circle at this point is the unique circle which passes through (x, y), has the same curvature at C at (x, y), and whose center lies in the direction of the standard unit normal vector to C. The plane in which the circle lies is called the osculating plane of C at (x, y). See Figure4 for a couple of examples of osculating circles. Notice that the circle is smallest when the curvature is greatest.

6 Example 4. Let E be the ellipse in the xy-plane determined by the equation x2 y2 + = 1, a2 b2 where a and b are positive. Using the parametrization

r(t) = ha cos t, b sin ti, 0 ≤ t ≤ 2π for E, give a parametrization c(t) of the center of the osculating circle to E at r(t).

(Solution) If we can find a parametrization N(t) for the standard unit normal to E at r(t) and also a parametrization R(t) for the radius of the osculating circle at r(t), then the center of this circle will be given by c(t) = r(t) + R(t)N(t). Recall that the curvature of a circle is constant, and in particular is equal to 1/R, where R is the radius of the circle. Since we want C to have the same curvature as E, we see that R(t) = 1/κ(t), where κ(t) is the curvature of E at r(t). In the interest of finding κ(t) we compute

r0(t) = h−a sin t, b cos ti and r00(t) = h−a cos t, −b sin ti.

Then we have

i j k 0 00 2 2 r (t) × r (t) = −a sin t b cos t 0 = h0, 0, ab(sin t + cos t)i = h0, 0, abi,

−a cos t −b sin t 0 so kr0(t) × r00(t)k = ab. We also see that

kr0(t)k = (a2 sin2 t + b2 cos2 t)1/2, so kr0(t) × r00(t)k ab κ(t) = = . kr0(t)k3 (a2 sin2 t + b2 cos2 t)3/2 Then the radius R(t) of the osculating circle C at r(t) is given by

(a2 sin2 t + b2 cos2 t)3/2 R(t) = . ab Next we must find N(t). First we have the unit tangent vector r0(t) h−a sin t, b cos ti T(t) = = . kr0(t)k (a2 sin2 t + b2 cos2 t)1/2 Now the unit normal vector is defined by N(t) = T0(t)/kT0(t)k. Rather than compute this relatively messy derivative, we’ll cheat a little bit. Because of our familiarity with the parametrization r(t), we know that the unit tangent vector T(t) is always turning to the left. In particular, T0(t) is always to the left of T(t), and thus we can obtain N(t) by rotating T(t) 90◦ counterclockwise. We know how to make this rotation component-wise, so we have h−b cos t, −a sin ti N(t) = . (a2 sin2 t + b2 cos2 t)1/2

7 Finally we have

c(t) = r(t) + R(t)n(t) (a2 sin2 t + b2 cos2 t)3/2 h−b cos t, −a sin ti = ha cos t, b sin ti + ab (a2 sin2 t + b2 cos2 t)1/2 a2 sin2 t + b2 cos2 t = ha cos t, b sin ti + h−b cos t, −a sin ti ab a2 b2 a2 sin2 t + b2 cos2 t a2 sin2 t + b2 cos2 t  = h cos t, sin ti − cos t, sin t a b a b a2(1 − sin2 t) − b2 cos2 t b2(1 − cos2 t) − a2 sin2 t  = cos t, sin t a b a2 − b2 b2 − a2  = cos3 t, sin3 t . a b

To see our parametrization of the center in action, have a look at this animation. Letting a = 5 and b = 2 gives us the ellipse considered in Example3, and the center of the osculating circle of that ellipse is parametrized by

21 −21  c(t) = cos3 t, sin3 t . 5 2

♦

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