{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions3 - MATH 2007B Test 3 Solutions November 7...

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

View Full Document Right Arrow Icon
MATH 2007B Test 3 Solutions November 7, 2011 [Marks] 1. Consider the parametric curve x = t 2 1 , y = 2 t 3 + t 2 . [6] (a) Find dy dx and d 2 y dx 2 . Solution: dy dx = dy/dt dx/dt = 6 t 2 + 2 t 2 t = 3 t + 1. d 2 y dx 2 = 1 dx/dt d dt dy dx = 3 2 t . (b) Determine the point(s) where the tangent is horizontal/vertical. Solution: The tangent is horizontal when dy dx = 3 t + 1 = 0, i.e., t = 1 3 . The tangent is never vertical since dy dx = 3 t + 1 = ±∞ for any t . (c) Determine the intervals of t where the graph is concave up/down. Solution: d 2 y dx 2 = 3 2 t > 0 for t > 0 and < 0 for t < 0. Hence, the graph is concave up for t > 0 and concave down for t < 0. 2. Find the length of the parametric curve x = 3 t 2 , y = 2 t 3 , 3 t 8. [6] Solution: L = 8 3 ( dx/dt ) 2 + ( dy/dt ) 2 dt = 8 3 (6 t ) 2 + (6 t 2 ) 2 dt = 3 8 3 2 t 1 + t 2 dt = 3 9 4 u du = 2 u 3 / 2 9 4 = 2(27 8) = 38, where u = 1 + t 2 . 3. (a) Find the polar coordinates of the point ( x, y ) = ( 3 3 , 3). [2] Solution: r = x 2 + y 2 = 6 ( x, y ) = 6 3 2 , 1 2 θ = 5 π 6 ( r, θ ) = 6 , 5 π 6 .
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