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Chapter 11: Q. 61 (page 873)

Let r(t)be a vector-valued function whose graph is a curve C, and let a(t)be the acceleration vector. Prove that if a(t)is always zero, then C is a straight line.

Short Answer

Expert verified

Ans: If a(t)=鉄�0,0,0鉄�then r(t)=c1t伪,c2t尾,c3t+纬the graph is a straight line.

Step by step solution

01

Step 1. Given information:

The graph of the vectors valued function r(t)is on curve C. a(t)be the acceleration vector.

02

Step 2. Proving C is a straight line:

Let a(t)=鉄�0,0,0鉄�

The velocity vector, v(t)=鈭�a(t)dt

=鈭�鉄�0,0,0鉄�dt

=c1,c2,c3wherec1,c2andc3are constantsr(t)=鈭�v(t)dt=鈭�c1,c2,c3dt=鈭�c1dti+鈭�c2dtj鈭�c3dtk

=c1t+伪i+c2t+尾j+c3t+纬kwhere 伪,尾,纬are constants.

r(t)=c1t伪,c2t尾,c3t+纬, the graph is a straight line because it is in linear form.

Thus if the acceleration vector ofr(t)is always zero, then the graph ofr(t),c, is a straight line.

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Most popular questions from this chapter

Let r(t)=xt,yt,t鈭�[a,鈭�), be a vector-valued function, where a is a real number. Explain why the graph of r may or may not be contained in a circle centered at the origin. (Hint: Graph the functions r1t=1t,1tand r2t=t,t,both with domain 摆1,鈭�).)

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)=(3+t)i+(3-1t)jfort>0

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

r(t)=(t,t2),t=1

Carefully outline the steps you would use to find the equation of the osculating plane at a point rt0on the graph of a vector function.

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=cos伪t,sin伪t,where伪>0,t=蟺
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