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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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140 D CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE<br />

(c) To identify the traces in y = mx we substitute y = mx into the equation of the ellipsoid:<br />

As expect<strong>ed</strong>, this is a family of ellipses.<br />

x 2 (mx? . z 2<br />

(6378.137) 2 + (6378.137)2 + (6356.523) 2 = 1<br />

(1 + m 2 )x 2 z 2<br />

(6378.137) 2 + (6356.523)2 = 1<br />

x2<br />

(6378.137) 2 /(1 + m2) + (6356.523)2 = 1<br />

z2<br />

49. If (a, b, c) satisfies z = y 2 - x 2 , then c = b 2 - a 2 . L 1 : x = a + t, y = b + t, z = c + 2( b - a )t,<br />

£2: x =a+ t, y = b- t, z = c- 2(b + a)t. Substitute the parametric equations of £1 into the equation<br />

of the hyperbolic paraboloid in order to find the points of intersection: z = y 2 - x 2 =><br />

. '<br />

I<br />

c + 2(b - a)t = (b + t) 2 - (a+ t) 2 = b 2 - a 2 + 2(b- a)t => c = b 2 - a 2 . As this is true for· all values oft,<br />

L 1 lies on z = y 2 - x 2 . Performing similar operations with £2 gives: z = y 2 - x 2 =><br />

c- 2(b + a)t = (b- t) 2 - (a+ t? = b 2 - a 2 - 2(b + a)t => c = b 2 - a 2 . This tells us that all of £2 also lies on<br />

51.<br />

The curve of intersection looks like a bent ellipse. The projection<br />

of this curve onto the xy-plane is the set of points (x, y, 0) which<br />

satisfy x 2 + y 2 = 1 - y 2 x 2 + 2y 2 = 1 <br />

y2<br />

x 2 + = 1. This is an equation of an ellipse.<br />

{1//2)2<br />

12 Review<br />

CONCEPT CH ECK<br />

1. A scalar is a real number, while a vector is a quantity that has both a real-valu<strong>ed</strong> magnitude and a direction.<br />

2. To add two vectors geometrically, we can use either.the Triangle Law or the Parallelogram Law, as illustrat<strong>ed</strong> in Figures 3<br />

and 4 in ·section 12.2. Algebraically, we add the corresponding components of the vectors.<br />

3. For c > 0, c a is a vector with. the same direction as a and length c times the length of a. If c < 0, ca points in the opposite<br />

direction as a and has length JcJ times the length of a. (See Figures 7 and IS in Section 12.2.) Algebraically, to find c a we<br />

multiply each component of a by c.<br />

4. See (I) in Section 12.2.<br />

5. See Theorem 12.3.3 and Definition 12.3.1.<br />

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