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Chapter 11: Q. 7 (page 889)

What makes Definition 11.21 for curvature easy to understand? What makes it difficult to use?

Short Answer

Expert verified

The reason has been explained.

Step by step solution

01

Step 1. Given information.

We have to explain What makes Definition 11.21 for curvature easy to understand and What makes it difficult to use.

02

Step 2. Definition

The definition of curvature is, let c be the graph of a vector function r(s) defined on an interval I, parametrized by arc length s, and with unit tangent vector T.

The curvature k of c at a point on the curve is the scalar given by k=dTdsi.e. the rate of change of the tangent vector with respect to arc length is the curvature.

03

Step 3. Explanation

Let a helix is defined by r(t)=⟨sin⁡αt,cos⁡αt,βt⟩on[a,b]whereα,β,aandb

The derivative is r′(t)=⟨αcos⁡αt,−αsin⁡αt,β⟩

The arc length is :

∫ab r′(t)dt=∫ab α2cos2⁡αt+α2sin2⁡αt+β2dt=∫ab α2+β2dt=α2+β2tab=α2+β2(b−a)

Thus, the arc length of r(t) is(b−a)α2+β2

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Most popular questions from this chapter

For each of the vector-valued functions in Exercises, find the unit tangent vector and the principal unit normal vector at the specified value of t.

rt=sin2t,cos2t,t,t=π4

Find parametric equations for each of the vector-valued functions in Exercises 26–34, and sketch the graphs of the functions, indicating the direction for increasing values of t.

r(t)=(2-sint,4+cost)fort∈[0,2π]

Compute the cross product of the vector functions r1(t)=x1(t),y1(t)andr2(t)=x2(t),y2(t)by thinking of ℝ2 as the xy-plane in ℝ3.That is, let r1t=x1t,y1t,0andr2t=x2t,y2t,0,and take the cross product of these vector functions.

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r(t)=et,e−t,2tatt=0

Given a differentiable vector-valued function r(t), what is the definition of the unit tangent vector T(t)?
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