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Chapter 11: Q. 30 (page 889)

In Exercises 28鈥�30 show that the given vector-valued functions are not arc length parametrizations. Then find an arc length parametrization for the curves defined by those functions.

r(t)=鉄�cost,sint,t鉄�

Short Answer

Expert verified

The arc-length parametrization isr(s)=coss2,sins2,s2.

Step by step solution

01

Step 1. Given information.

The given function isr(t)=鉄�cost,sint,t鉄�.

02

Step 2. Required answer.

Now,

r'(t)=鉄�-sint,cost,1鉄�r'(t)=(-sint)2+(cost)2+1=2Consider鈭�cdr'(t)dt=鈭�cd2dt=2tcd=2(d-c)鈮�d-c

Hence the given vector-valued functions are not arc length parameterizations on the interval [0,1]the arc length of the curve is 2(1-0)=2

Now, replace t withs2inr(t),we get,

r(s)=coss2,sins2,s2

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Most popular questions from this chapter

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

r(t)=(3sint,5cost,4sint)

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

r(t2)=(t2+5,5t,4t3)

For constants 伪,尾, and 纬, the graph of a vector-valued function of the form

r(t)=伪e纬t(cost)i+伪e纬t(sint)j+尾e纬tk,t鈮�0

Find and graph the vector function determined r(t)=鉄�x(t),y(t)鉄�by the differential equation

x鈥�(t)=x,y鈥�(t)=x2,x(0)=1,y(0)=2. (HINT: Start by solving the initial-value problem x鈥�(t)=x,x(0)=1.)

Find and graph the vector function determined r(t)=鉄�x(t),y(t)鉄�by the differential equation

role="math" localid="1649566464308" x鈥�(t)=1+x2,y鈥�(t)=x2,x(0)=0,y(0)=1. ( HINT: Start by solving the initial-value problemrole="math" localid="1649566360577" x鈥�(t)=1+x2,x(0)=0.)
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