notes001 (44)

# notes001 (44) - The acceleration is a t = d v dt =& 4...

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

MATH 2203 °Quiz 3 (Version 1) Solution September 24, 2010 NAME_____________________________ The motion of a particle is given by x = sin (2 t ) y = ° cos (2 t ) . Find ± the position function: r ( t ) ± the velocity: v ( t ) ( 4 points if you get to here ) ± the speed: v ( t ) = j v ( t ) j ± the acceleration: a ( t ) ( 10 points if you get to here ) ± the unit tangent vector: T ( t ) ± the curvature: ° ( t ) ( 16 points if you get to here ) ± the tangential component of acceleration: a T ( t ) ± the normal component of acceleration: a N ( t ) ( 20 points if you get to here ) Solution: The vector°valued function for the position of the particle is r ( t ) = sin (2 t ) i ° cos (2 t ) j . The velocity is v ( t ) = d r dt = 2 cos (2 t ) i + 2 sin (2 t ) j and the speed is v ( t ) = j v ( t ) j = q (2 cos (2 t )) 2 + (2 sin (2 t )) 2 = 2 . 1

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

View Full Document
The acceleration is
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The acceleration is a ( t ) = d v dt = & 4 sin (2 t ) i + 4 cos (2 t ) j . The unit tangent vector is T ( t ) = 1 v ( t ) v ( t ) = 1 2 (2 cos (2 t ) i + 2 sin (2 t ) j ) = cos (2 t ) i + sin (2 t ) j . Also d T dt = & 2 sin (2 t ) i + 2 cos (2 t ) j and & & & & d T dt & & & & = q ( & 2 sin (2 t )) 2 + (2 cos (2 t )) 2 = 2 so the curvature is & ( t ) = 1 v ( t ) & & & & d T dt & & & & = 1 2 (2) = 1 . The tangential component of acceleration is a T ( t ) = dv dt = d dt (2) = 0 and the normal component of acceleration is a N ( t ) = & ( t ) ( v ( t )) 2 = (1) (2) 2 = 4 . 2...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern