14.3%20Prel%201-4,%20Ex%201-6

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
486 CHAPTER 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 ) = h x 1 ( t ), y 1 ( t ), z 1 ( t ) i and r 2 ( t ) = h x 2 ( t ), y 2 ( t ), z 2 ( t ) i be differentiable vector-valued func- tions. Then, Z r 1 ( t ) · r 0 2 ( t ) dt = r 1 ( t ) · r 2 ( t ) Z r 0 1 ( t ) · r 2 ( t ) To verify this formula we compute the dot product on the left-hand side and use componentwise integration. This gives: Z r 1 ( t ) · r 0 2 ( t ) = Z h x 1 ( t ), y 1 ( t ), z 1 ( t ) i · - x 0 2 ( t ), y 0 2 ( t ), z 0 2 ( t ) ® = Z x 1 ( t ) x 0 2 ( t ) + y 1 ( t ) y 0 2 ( t ) + z 1 ( t ) z 0 2 ( t ) = Z x 1 ( t ) x 0 2 ( t ) + Z y 1 ( t ) y 0 2 ( t ) + Z z 1 ( t ) z 0 2 ( t ) We now use Integration by Parts for scalar-valued functions for each of the three integrals. We obtain: Z r 1 ( t ) · r 0 2 ( t ) = x 1 ( t ) x 2 ( t ) Z x 0 1 ( t ) x 2 ( t ) + y 1 ( t ) y 2 ( t ) Z y 0 1 ( t ) y 2 ( t ) + z 1 ( t ) z 2 ( t ) Z z 0 1 ( t ) z 2 ( t ) = x 1 ( t ) x 2 ( t ) + y 1 ( t ) y 2 ( t ) + z 1 ( t ) z 2 ( t ) ±Z x 0 1 ( t ) x 2 ( t ) + Z y 0 1 ( t ) y 2 ( t ) + Z z 0 1 ( t ) z 2 ( t ) We rewrite this as: Z r 1 ( t ) · r 0 2 ( t ) = r 1 ( t ) · r 2 ( t ) Z - x 0 1 ( t ), y 0 1 ( t ), z 0 1 ( t ) ® · h x 2 ( t ), y 2 ( t ), z 2 ( t ) i = r 1 ( t ) · r 2 ( t ) Z r 0 1 ( t ) · r 2 ( t ) 71. Show that if k r ( t ) k≤ K for t ∈[ a , b ] ,then ° ° ° ° ° Z b a r ( t ) ° ° ° ° ° K ( b a ) Think of r ( t ) as a velocity vector. Then, Z b a r ( t ) gives the displacement vector from the location at time t = a to the time t = b ,andso ° ° ° ° ° Z b a r ( t ) ° ° ° ° ° gives the length of this displacement vector. But, since speed is k r ( t ) k 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 ° ° ° ° ° Z b a r ( t ) ° ° ° ° ° K ( b a ) ,as desired.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online