Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 13 REVIEW ET CHAPTER 12 ¤ 129
(c) To identify the traces in y = mx we substitute y = mx into the equation of the ellipsoid:
As expected, this is a family of ellipses.
x 2
(6378.137) 2 + (mx)2
(6378.137) 2 + z 2
(6356.523) 2 =1
(1 + m 2 )x 2
(6378.137) 2 + z 2
(6356.523) 2 =1
x 2
(6378.137) 2 /(1 + m 2 ) + z 2
(6356.523) =1 2
49. If (a, b, c) satisfies z = y 2 − x 2 ,thenc = b 2 − a 2 . L 1: x = a + t, y = b + t, z = c +2(b − a)t,
L 2 : x = a + t, y = b − t, z = c − 2(b + a)t. Substitute the parametric equations of L 1 into the equation
of the hyperbolic paraboloid in order to find the points of intersection: z = y 2 − x 2
⇒
c +2(b − a)t =(b + t) 2 − (a + t) 2 = b 2 − a 2 +2(b − a)t ⇒ c = b 2 − a 2 .Asthisistrueforallvaluesoft,
L 1 lies on z = y 2 − x 2 . Performing similar operations with L 2 gives: z = y 2 − x 2
⇒
c − 2(b + a)t =(b − t) 2 − (a + t) 2 = b 2 − a 2 − 2(b + a)t ⇒ c = b 2 − a 2 . This tells us that all of L 2 also lies on
z = y 2 − x 2 .
51. The curve of intersection looks like a bent ellipse. The projection
of this curve onto the xy-plane is the set of points (x, y, 0) which
satisfy x 2 + y 2 =1− y 2 ⇔ x 2 +2y 2 =1 ⇔
x 2 +
y 2
1/
√
2
2
=1.Thisisanequationofanellipse.
13 Review ET 12
1. A scalar is a real number, while a vector is a quantity that has both a real-valued magnitude and a direction.
2. To add two vectors geometrically, we can use either the Triangle Law or the Parallelogram Law, as illustrated in Figures 3
and 4 in Section 13.2 [ ET 12.2]. Algebraically, we add the corresponding components of the vectors.
3. For c>0, c a is a vector with the same direction as a and length c times the length of a. Ifc