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Hw16-2b

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

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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 ...
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