14.3%20Prel%201-4,%20Ex%201-6 - 486 C H A P T E R 14 C A L C U L U S O F VE C T O R VA L U E D F U N C T I O N S(ET CHAPTER 13 70 Formulate and verify a

# 14.3%20Prel%201-4,%20Ex%201-6 - 486 C H A P T E R 14 C A L...

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486 C H A P T E R 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) 70. Formulate and verify a version of Integration by Parts for vector-valued integrals. SOLUTION Let r 1 ( t ) = x 1 ( t ), y 1 ( t ), z 1 ( t ) and r 2 ( t ) = x 2 ( t ), y 2 ( t ), z 2 ( t ) be differentiable vector-valued func- tions. Then, r 1 ( t ) · r 2 ( t ) dt = r 1 ( t ) · r 2 ( t ) r 1 ( t ) · r 2 ( t ) dt To verify this formula we compute the dot product on the left-hand side and use componentwise integration. This gives: r 1 ( t ) · r 2 ( t ) dt = x 1 ( t ), y 1 ( t ), z 1 ( t ) · x 2 ( t ), y 2 ( t ), z 2 ( t ) dt = x 1 ( t ) x 2 ( t ) + y 1 ( t ) y 2 ( t ) + z 1 ( t ) z 2 ( t ) dt = x 1 ( t ) x 2 ( t ) dt + y 1 ( t ) y 2 ( t ) dt + z 1 ( t ) z 2 ( t ) dt We now use Integration by Parts for scalar-valued functions for each of the three integrals. We obtain: r 1 ( t ) · r 2 ( t ) dt = x 1 ( t ) x 2 ( t ) x 1 ( t ) x 2 ( t ) dt + y 1 ( t ) y 2 ( t ) y 1 ( t ) y 2 ( t ) dt + z 1 ( t ) z 2 ( t ) z 1 ( t ) z 2 ( t ) dt = x 1 ( t ) x 2 ( t ) + y 1 ( t ) y 2 ( t ) + z 1 ( t ) z 2 ( t ) x 1 ( t ) x 2 ( t ) dt + y 1 ( t ) y 2 ( t ) dt + z 1 ( t ) z 2 ( t ) dt We rewrite this as: r 1 ( t ) · r 2 ( t ) dt = r 1 ( t ) · r 2 ( t ) x 1 ( t ), y 1 ( t ), z 1 ( t ) · x 2 ( t ), y 2 ( t ), z 2 ( t ) dt = r 1 ( t ) · r 2 ( t ) r 1 ( t ) · r 2 ( t ) dt 71. Show that if r ( t ) K for t ∈ [ a , b ] , then b a r ( t ) dt K ( b a ) SOLUTION Think of r ( t ) as a velocity vector. Then, b a r ( t ) dt gives the displacement vector from the location at time t = a to the time t = b , and so b a r ( t ) dt gives the length of this displacement vector. But, since speed is r ( t ) which is less than or equal to K , then in the interval a t b , the object can move a total distance not more than K ( b a ) . Thus, the length of the displacement vector is K ( b a ) , which gives us b a r ( t ) dt K ( b a ) , as desired.