Engineering Calculus Notes 314

Engineering Calculus Notes 314 - 3.5.7 We apply Lemma 3.5.8...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
302 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Proof of Lemma 3.5.8 . For any θ , bardbl (cos θ ) −→ v + (sin θ ) −→ w bardbl 2 = ((cos θ ) −→ v + (sin θ ) −→ w ) · ((cos θ ) −→ v + (sin θ ) −→ w ) = bardbl −→ v bardbl 2 (cos 2 θ ) + 2 −→ v · −→ w cos( θ sin θ ) + bardbl −→ w bardbl 2 (sin 2 θ ) = 1 2 bardbl −→ v bardbl (1 + cos 2 θ ) + −→ v · −→ w sin 2 θ + 1 2 bardbl −→ w bardbl (1 cos 2 θ ); a standard calculation (Exercise 10 ) shows that the extreme values of this function of θ occur when tan 2 θ = 2 −→ v · −→ w bardbl −→ v bardbl 2 −bardbl −→ w bardbl 2 ; denote by θ 0 the value where the minimum occurs. It is clear that we can express θ 0 as a function of −→ v and −→ w ; let K ( −→ v, −→ w ) = bardbl (cos θ 0 ) −→ v + (sin θ 0 ) −→ w bardbl . Since −→ v and −→ w are linearly independent, we automatically have
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.5.7 . We apply Lemma 3.5.8 to the vectors −→ v = ∂ −→ p ∂s −→ w = ∂ −→ p ∂t to ±nd a positive, continuous function K ( s,t ) de±ned on the domain of −→ p such that for every θ the vector −→ v ( s,t,θ ) = (cos θ ) ∂ −→ p ∂s ( s,t ) + (sin θ ) ∂ −→ p ∂t ( s,t ) has b −→ v ( s,t,θ ) b ≥ K ( s,t ) . In particular, given s , t , and an angle θ , we know that some component of the vector −→ v ( s,t,θ ) must have absolute value exceeding K ( s,t ) / 2: | v j ( s,t,θ ) | > K ( s,t ) 2 ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern