   Chapter 12.2, Problem 68E

Chapter
Section
Textbook Problem

# ProofIn Exercises 61–68, prove the property. In each case, assume r , u , and v are differentiable vector-valued functions of t in space, w is a differentiable real-valued function of t , and c is a scalar. d d t [ r ( t ) ± u ( t ) ] = r ′ ( t ) ± u ′ ( t )

To determine

To prove: The expression ddt[r(t)±u(t)]=r(t)±u(t)

Explanation

Given:

Assume r, u, and v are differentiable vector-valued functions of t in space, w is a differentiable real-valued function of t and c is a scalar.

Formula used:

If f(t)=f1(t)i+f2(t)j+f3(t)k, then ddt[f(t)]=f1(t)+f2(t)+f3(t).

Proof:

Let r(t)=f1(t)i+f2(t)j+f3(t)k and u(t)=g1(t)i+g2(t)j+g3(t)k.

Where f1,f2,f3,g1,g2,g3 are real valued differentiable functions.

Now,

r(t)±u(t)=[f1(t)i+f2(t)j+f3(t)k]±[g1(t)i+g2(t)j+g3(t)k]=[f1(t)±g1(t)]i+[f2(t)±g2(t)]j+[f3(t)±g3(t)]k

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