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Chapter 11: Q. 31 (page 889)

In Exercises 31鈥�35 find the curvature of the given function at the indicated value of x. Then sketch the curve and the osculating circle at the indicated point.

y=ex,x=0

Short Answer

Expert verified

The curvature atx=0isk=24.

The graph is

Step by step solution

01

Step 1. Given information.

The given function and point isy=ex,x=0.

02

Step 2. Curvature at the point.

The curvature is given by,

k=f'(x)1+f'(x)232f'(x)=exf''(x)=exk=ex1+ex232k=ex1+e2x32Atx=0,k=e01+e2(0)32k=1232k=212232路212k=222k=24

03

Step 3. Graph.

The graph will be,

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Most popular questions from this chapter

Compute the cross product of the vector functions r1(t)=x1(t),y1(t)andr2(t)=x2(t),y2(t)by thinking of 鈩�2 as the xy-plane in 鈩�3.That is, let r1t=x1t,y1t,0andr2t=x2t,y2t,0,and take the cross product of these vector functions.

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

r(t)=(cos2t,sin2t)fort鈭�[0,2蟺]

For each of the vector-valued functions, find the unit tangent vector.

r(t)=(sin2t,cos2t,t)

Find the unit tangent vector and the principal unit normal vector at the specified value of t.

r(t)=(asinht,bcosht)Whereaandbarepositive,t=0

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=cos3t,sin3t,t=蟺4
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