Hw16-2b - Section 16.2 Solutions 16.2.42 (a) First we find...

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

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

Unformatted text preview: Section 16.2 Solutions 16.2.42 (a) First we find a tangent vector to the circle. To be tangent, we require that 2x + 2yy = 0 ⇒ y = −x/y to be the slope of the tangent line. Thus a tangent vector pointing in the clockwise direction is a = yi − xj . To make it a yi − xj unit vector we simply divide by the magnitude: G = . Note that G points in the clockwise direction. x2 + y 2 (b) G = −F 16.2.44 y (a) The slope of the line through (x, y ) and the origin is x . This implies that v = −xi − yj is a vector parallel to that line and pointing toward the origin. Note that |v | = x2 + y 2 which is the distance from the origin. Thus F = v . (b) Now we want the magnitude to be √ 1 . x2 +y 2 1 C ∗ |v | = ⇒C= We thus want to solve: x2 (where C = 0) + y2 1 x2 ( + y 2 )( x2 + y2 ) = 1 x2 + y 2 −xi − yj has magnitude inversely proportional to the distance from (x, y ) to the origin. In fact, any x2 + y 2 nonzero multiple of F also has this property. Thus F = 16.2.46 r = xi + yj = xi + f (x)j ⇒ dr = i + f (x)j dx k (xi + yj ) has magnitude k and points away from the origin. x2 + y 2 dr kx k ∗ y ∗ f (x) kx + kf (x)f (x) d F· =2 + = =k x2 + |f (x)|2 (chain rule) 2 2 + y2 2 + [f (x)]2 dx x +y x x dx F= ⇒ F · Tds = C F· c =k dr dx = dx b k a b2 + [f (b)]2 − 1 d dx x2 + [f (x)]2 dx = k x2 + |f (x)|2 a2 + [f (a)]2 , as claimed. b a ...
View Full Document

This note was uploaded on 06/11/2011 for the course MATH 1920 at Cornell University (Engineering School).

Ask a homework question - tutors are online