Quiz 10 Key PDF

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Quiz 10 - Takehome 11/19/14

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The point value for each problem is listed next to it. 1. (10 pts) Given the following equation for a hyperbola, (a) list both asymptotes, (b) foci, and (c) vertices: (y − 2)2 (x + 4)2 − =1 9 16

Solution: To find the asymptotes, change the 1 to a 0 and solve for y. 3(x + 4) x+4 y−2 (x + 4)2 (y − 2)2 (x + 4)2 (y − 2)2 =⇒ y = 2 ± =± =⇒ − = = 0 =⇒ 4 3 16 9 9 4 16     3 3 3 3 So y = 2 − x + 3 = x − 1 and y = 2 + x + 3 = x + 5 are the asymptotes. 4 4 4 4 The center is at (−4, 2), and the foci and vertices are in the y. a2 = 9 so a = 3. Then the p direction of √ vertices are at (−4, 2 ± 3), or (−4, −1) and (−4, 5). c = a2 + b2 = 9 + 16 = 5. Then the foci are at (−4, 2 ± 5), or (−4, −3) and (−4, 7) 2. (10 pts) Given the following equation for a conic with one focus at the origin, (a) identify the conic (Hyperbola, parabola, ellipse), (b) the eccentricity e, and (c) the equation of the directrix corresponding to the focus at the origin. 3 r= 2 − cos θ Solution: This is not currently in standard form. Divide the top and bottom by 2 to get r = 3/2 , so e = 1/2. Therefore the conic is an ellipse. We know ek = 3/2, so (1/2)k = 3/2 =⇒ 1 − (1/2) cos θ k = 3. Since the sign in front of cos is negative, then the directrix is x = −3. 3. (10 pts) Given the following parametric equations, (a) identify the conic by finding the equation for the conic in Cartesian coordinates (as in Section 9.6), (b) sketch the graph of the parametric equation, including the direction of travel. p x = t, y = 4 − t2 , −2 ≤ t ≤ 2

p 2 2 2 2 2 2 Solution: y = 4 − px =⇒ y = 4 − x =⇒ x + y = 4 = 2 . This is the equation for a circle of radius 2. Since y = 4 − x2 , then y ≥ 0, and −2 ≤ x ≤ 2, so it is only the top half of the circle. The graph goes from x = −2 to x = 2, so clockwise along the top half of the circle.

4. (10 pts) Given the following vectors, find the requested quantities: u =< 0, 5, 12 >

v = 3i − 2j + 6k

w =< −4, 2, −3 >

(a) 3u − 2v

Solution: 3 < 0, 5, 12 > −2 < 3, −2, 0 >=< 0, 15, 36 > − < 6, −4, 0 >=< −6, 19, 36 >

(b) −2w

Solution: −2 < −4, 2, −3 >=< 8, −4, 6 >

(c) |v| Solution: (d) |w − u|

p

32 + (−2)2 + 62 =

√ √ 9 + 4 + 36 = 49 = 7

Solution: | < −4, √2, −3 >√− < 0, 5, 12√> | = | < −4, −3, −15 > | = √ 16 + 9 + 225 = 250 = 25 · 10 = 5 10

p

(−4)2 + (−3)2 + (−15)2 =

(e) Find the direction vector of u. p u . |u| = 02 + 52 + 122 = Solution: This is a unit vector pointing in the direction of u, |u| √ √ u 5 12 5 12 >. 25 + 144 = 169 = 13, so = (5j + 12k)/13 = j + k or < 0, , 13 13 13 13 |u| 5. (10 pts) Give the equation(s) that define the given region or object in three dimensions. (a) The plane through (2, −4, 1) perpendicular to the y-axis.

Solution: All planes perpendicular to y-axis are y = k for k ∈ R. The only such plane with (2, −4, 1) in it is y = −4.

(b) The sphere of radius 3 centered at (1, 2, 3). Solution: (x − 1)2 + (y − 2)2 + (z − 3)2 = 32 (c) The line passing through the points (5, −2, 3) and (−3, −2, 3).

Solution: We need nonparallel planes that both contain these two points. Notice that y = −2 and z = 3 satisfy these requirements. So the line is defined by the two equations y = −2 and z=3

(d) The plane through (2, −4, 1) parallel to the xy-plane.

Solution: The xy-plane is z = 0. The parallel plane containing (2, −4, 1) is z = 1.

(e) The line passing through the point (−2, 0, 3) parallel to the z-axis. Solution: The z-axis is defined by the equations x = 0, y = 0, which are two planes. Then two planes parallel to these, which would make a line parallel to the z-axis, and crossing the point (−2, 0, 3) are x = −2, y = 0....