HW09-solutions

E by setting respectively x a y a and z a in the

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Unformatted text preview: + 2t, 2 + t, −4 − 2t 4. r(t) = 4 − 2t, 2 − t, 4 − 2t 5. r(t) = 2 + 4t, 1 + 2t, 2 − 4t 6. r(t) = 4 + 2t, −2 + t, 4 − 2t correct Explanation: A line passing through a point P (a, b, c) and having direction vector v is given in vector form by r(t) = a + tv , a= 3. 2x2 + y 2 + 3z 2 = 1 Explanation: The graphs of each of the given quadratic relations is a quadric surface in standard position. We have to decide which quadric surface goes with which equation - a good way of doing that is by taking plane slices parallel to the coordinate planes, i.e., by setting respectively x = a, y = a and z = a in the equations once we’ve decided what the graphs of those plane slices should be. Now as slicing of a, b, c . Now for the given line, its direction vector will be parallel to the normal to the plane 2x + y − 2z = 6 , so v = 2, 1, 2 ; shows, slices of this hyperbolic paraboloid by z = a are hyperbolas, while slices by x = a and y = a are parabolas, opening in diﬀerent directions. Only the graph of on the other hand, a = 4, 2, 4 , since the line passes through P (4, −2, 4). Consequently, in vector form the line is given by r(t) = 4 + 2t, −2 + t, 4 − 2t . z = y 2 − x2 has these properties. geyer (dag2798) – HW09 – rusin – (55735) 7 z 013 10.0 points x Which one of the following equations has graph But this circle has radius 1 because the cylinder has radius 1, and so its equation is x2 + (z + 1)2 = 1 when the circular cylinder...
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