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# Quiz 7 - Version OO4/AAABA w DGQO7C — Gilbert —(78759 1...

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Unformatted text preview: Version OO4/AAABA w DGQO7C — Gilbert — (78759) 1 This print—out should have 3 questions. after differentiating with respect to y. Conse— Muitiple—choice questions may continue on quentiy, the next column or page - ﬁnd all choices before ansWering. fm “ 131(531!) +4 . y “7 ' ' III-I": . 002 10-0 pants an. BACK NEXT Which of the following surfaces couid have m- CB contour map 001 10.0 points Determine fxy when flay) ll my 1n(a:y) + 2233;. 1. fxy x ln(5ry) + 4 correct 2. fan; as 2(1n(my)+:cy)-3 3. fay = (ln(:ry) ~+~ my) ~l~ at 4. fm :2 111(23y) —- 4 5- fey m 2011(339) _ my) +3 1, 913,39 6- fey = 2111359) "” 3 2. hyperbolic paraboioid Explanation: Since , 3. cone correct ln(.:ry) = inst + lny, 4. paraboloid we see that 5. parabolic cylinder ﬂag, y) m \$90115; + in y) + 3339, _ Explanation: The graphs in the contour map show that But then, the horizontal cross—sections of the surface are an all circles as in cones and paraboloicis. On the fm m 1,011 5;; + la y) + “L;— + 2y other hand, the values of the contours and the 3; grid show that these cross—sections increase at = yang; +1119) + By, a constant rate. This occurs £01" a cone, but not for a paraboloid. in which case Consequently, of the surfaces listed, the 1.! only one having the given contour map is a .fxy = (3119: +lny) + — +3 9 = (lnm+lny)+4, - Version OSe/AAABA — DGQ07c ~ Giibert — (78759) keywords: contour map, surface, quadric sur- face, plane, ' ' 003 10.0 points Find an equation in parametric form for the tangent iine to the graph of r(s) m_ (35, 33, 36) at the point (1, 1, 1). 1. 2:03) m 1+4t, y(t) m 1+2t, z(t) = 1+5t 2. 16(1)) 2 5:91, y(t) = 3t_1, 20:) 2: 6t~1 3. 5130:) = 1+5t, y(t) =1+3t, z(t) = 1 +6t correct 4. m(t)=4t-1,y(t) =2tw1, 5t“ 1 5. \$(t) m 5-54, y(t) 2 353, 2(t) = 62:5 6. 33(t) = 1—-5t, y(t) = 1+3t, z(t)m1~—6t Explanation: The graph of r(s) has tangent vector r’(s) m {534, 332, 635). On the other hand, (1, 1, 1) = r(1). Thus, at (1, 1, 1) the equation of the tangent line in vector form is ‘ ‘ L(t) = r(1)+tr’(1) = (1+5t,1+3t, 1+6t), which in parametric form becomes 23(t) m 1 +515, y(t) 2= 1 +3t, 2(t) = 1 +613 keywords: tangent iine, tangent vector, vector function, derivative, vector form, parametric form, ...
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Quiz 7 - Version OO4/AAABA w DGQO7C — Gilbert —(78759 1...

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