Arc Length and Curvature

Transcription

1 Arc Length and Curvature. Last time, we saw that r(t) = cos t, sin t, t parameteried the pictured curve. (a) Find the arc length of the curve between (, 0, 0) and (, 0, π). (b) Find the unit tangent vector at the point (, 0, π). (c) Find the curvature at the point (, 0, π). (d) Find the unit normal vector at the point (, 0, π).

2 . Suppose that r(t), 0 t 3, parameteries the following curve in space, with r(0) = 0, 3, 0 and r(0) = 0, 0,. The curve lies entirel in the plane = 0, and the right picture shows just that plane. We are told that the arc length of the curve is approimatel Find each of the following, or eplain wh there is not enough information to do so. (a) A sketch of the arc length function s(t). (b) The unit tangent vector T at the point (0, 0, ). (c) The unit tangent vector T (). (d) The osculating plane at (0, 0, ). (e) The unit normal vector N at the point (0, 0, ). (f) The unit normal vector N(). (g) The binormal vector B at the point (0, 0, ). (h) The normal plane at (0, 0, ). (i) Which of the following is the best estimate for the curvature of the curve at (0, 3, 0)? 0 0

3 Arc Length and Curvature. Last time, we saw that r(t) = cos t, sin t, t parameteried the pictured curve. (a) Find the arc length of the curve between (, 0, 0) and (, 0, π). Solution. We d like to find the arc length of the curve parameteried b r(t) between t = 0 (the point (, 0, 0)) and t = π (the point (, 0, π)). If we imagine r(t) as describing the position of a particle in space, the particle s velocit at time t is r (t) = sin t, cos t,, so the particle s speed at time t is r (t) =. To find the distance traveled b the particle between times 0 and π, we π simpl integrate the speed between t = 0 and t = π: the arc length is r (t) dt = π. (b) Find the unit tangent vector at the point (, 0, π). Solution. As we computed alread, r (t) = sin t, cos t, and r (t) =, so T (t) = r (t) sin t, cos t,. The point (, 0, π) corresponds to t = π, and T (π) = 0 0,,. r (t) = (c) Find the curvature at the point (, 0, π). Solution. We ve alread found that T (t) = sin t, cos t,, so, T (t) = cos t, sin t, 0 and T (t) =. We found in (a) that r (t) =, so κ(t) =. This means that the curvature is everwhere along the curve, so the curvature at the point (, 0, π) is. (d) Find the unit normal vector at the point (, 0, π). Solution. We know that the unit normal vector N(t) T is equal to (t) T. Using the calculations (t) from the previous part, this is equal to cos t, sin t, 0. In particular, N(π) =, 0, 0.. Suppose that r(t), 0 t 3, parameteries the following curve in space, with r(0) = 0, 3, 0 and r(0) = 0, 0,. The curve lies entirel in the plane = 0, and the right picture shows just that plane. We are told that the arc length of the curve is approimatel 5.3.

4 3 Find each of the following, or eplain wh there is not enough information to do so. (a) A sketch of the arc length function s(t). Solution. There is not enough information. If we are imagining r(t) as describing the position of a particle in space, s(t) describes the distance the particle has traveled b time t. However, we don t know how fast the particle is moving along its path, so it is impossible for us to know what this function looks like. We do, however, know that the particle has traveled nowhere at time t = 0, so s(0) = 0, and that the particle has traveled approimatel 5.3 units at time t = 3, so s(3) 5.3. In addition, the distance the particle has traveled certainl increases with time, so s(t) is an increasing function. (b) The unit tangent vector T at the point (0, 0, ). Solution. Since the curve lies entirel in the plane = 0, we know the unit tangent vector must also lie in the plane = 0. From looking at the picture, we can see that the tangent line at (0, 0, ) is flat in the -plane. Since the particle is moving counter-clockwise around the curve, the unit tangent vector is therefore 0,, 0. (c) The unit tangent vector T (). Solution. This is asking us to find the unit tangent vector at time t =, but since we don t know the eact parameteriation, we have no idea where the particle is at time t =. Therefore, there is not enough information to sa what the unit tangent vector is. (d) The osculating plane at (0, 0, ). Solution. The osculating plane at (0, 0, ) is the plane that comes closest to containing the curve near (0, 0, ). Of course, our curve sits entirel in the plane = 0, so that must be the osculating plane. (e) The unit normal vector N at the point (0, 0, ). Solution. Remember that the unit normal vector sits in the osculating plane, is perpendicular to the unit tangent vector, has length, and points in the direction that the curve is turning. Therefore, the unit normal vector in this case must be 0, 0,. (f) The unit normal vector N().

5 Solution. For the same reason as in (c), there is not enough information to determine this. (g) The binormal vector B at the point (0, 0, ). Solution. We know B is T N; from parts (b) and (e), we know that T = 0,, 0 and N = 0, 0, at the point (0, 0, ), so B =, 0, 0. (h) The normal plane at (0, 0, ). Solution. This is the plane which is normal to the unit tangent vector at (0, 0, ) and which contains the point (0, 0, ). We found in (b) that the unit tangent vector at (0, 0, ) is 0,, 0. The plane normal to 0,, 0 which contains the point (0, 0, ) is = 0. (i) Which of the following is the best estimate for the curvature of the curve at (0, 3, 0)? 0 0 Solution. Remember that the curvature of a circle of radius a is a. So, if we can find the circle which best matches the curve near ( 3, 0, 0), the curvature should be over the radius of that circle. If we draw in such a circle, its diameter looks a little less than, so its radius should be slightl less than. Therefore, its curvature should be slightl more than. 3 (In fact, the actual curvature is 7 3.) 3