{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm 2 - SOLUTION TO 18.02 MIDTERM#2 BJORN POONEN...

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

View Full Document Right Arrow Icon
SOLUTION TO 18.02 MIDTERM #2 BJORN POONEN October 22, 2009, 1:05–1:55pm (50 minutes) 1) A particle is moving in the xy -plane so that relative to the origin it is rotating coun- terclockwise at a rate of 2 radians per second while its distance to the origin is increasing at a rate of 10 meters per second. At a time when the particle is at ( - 4 , 3), what is dy dt ? Solution: At the given time r = p ( - 4) 2 + 3 2 = 5 and (cos θ, sin θ ) = ( - 4 / 5 , 3 / 5). We have y = r sin θ , so the chain rule yields dy dt = r cos θ dt + dr dt sin θ = 5( - 4 / 5)2 + 10(3 / 5) = - 8 + 6 = - 2 m / s . 2) Let f ( x, y, z ) = 2 xy 2 + z 3 . (a) (10 pts.) Find the unit vector u in R 3 that minimizes the directional derivative D u f at the point (0 , 2 , - 1). (b) (5 pts.) Find the value of D u f at (0 , 2 , - 1) in that direction u . Solution: (a) We have f = h 2 y 2 , 4 xy, 3 z 2 i , which is h 4 , 0 , 3 i at ( x, y, z ) = (0 , 2 , - 1). The length of f there is 5, so the unit vector in the direction of f is h 4 / 5 , 0 , 3 / 5 i . The directional derivative D u f = ( f ) · u is minimized by choosing u to be in the opposite direction, i.e., u = h- 4 / 5 , 0 , - 3 /
Image of page 1

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

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