**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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