Substituting x 2 in the equation of each surface we

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Substituting x = 2 in the equation of each surface we get the following equations in y and z : (A) Setting x = 2 in z 2 2 = x 2 9 + y 2 4 gives z 2 2 = 4 9 + y 2 4 or equivalently y 2 4 + z 2 2 = 4 9 . This equation has no solution so it describes the empty set. In other words the surface (A) and the plane x = 2 do not intersect. 12
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(B) Setting x = 2 in z 2 4 = x 2 9 + y 2 4 1 gives z 2 4 = 4 9 + y 2 4 1 or equivalently y 2 4 z 2 4 = 5 9 which is a scaling of a standard equation of a hyperbola. (C) Setting x = 2 in z 2 4 = x 2 9 + y 2 4 gives z 2 4 = 4 9 + y 2 4 or y 2 4 z 2 4 = 4 9 which is again a scaling of a standard equation of a hyperbola. (D) Setting x = 2 in z 2 = x 2 9 y 2 4 gives z 2 = 4 9 y 2 4 or z = 8 9 y 2 2 which is the equation of a parabola. (E) Setting x = 2 in z 2 4 = x 2 9 + y 2 4 + 1 gives z 2 4 = 4 9 + y 2 4 + 1 or z 2 4 y 2 4 = 13 9 which is again a scaling of a standard equation of a hyperbola. 13
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The correct answer is (D) . square Solution of problem 1.7: Given a point Q and a vector −→ B the vector equation −→ QP · −→ B = 0 is the standard equation of the plane α that passes through Q and is perpendicular to −→ Q . The equation −→ QP · −→ B = 2 will describe a plane β which is parallel to α and thus perpendicular to −→ B . Since the right hand side in this equation is 2 negationslash = 0 we conclude that β does not pass through Q . More explicitly, if Q = ( x 0 , y 0 , z 0 ) is a fixed point and −→ B = a hatwide ı + b hatwide + c hatwide k is a given vector, then a point P = ( x, y, z ) satisfies the vector equation −→ QP · −→ B = 2 if and only if the variables x , y and z satisfy the equation a ( x x 0 ) + b ( y y 0 ) + c ( z z 0 ) = 2 . Since the coefficients of x , y and z in this equation are a , b and c , this is an equation of a plane with normal vector equal to −→ B . Moreover, since the right hand side of the equation is not zero, the point Q can not lie on this plane. Therefore the correct choice is (E) . Solution of problem 1.8: (a) The equilibria for this model are solutions of the quadratic equation 12 P P 2 H = 0 or equivalently P 2 12 P + H = 0. The discriminant of this equation is ( 12) 2 4 H = 144 4 H = 4(36 H ) . Therefore the quadratic equation will have two solutions only when H < 36. Suppose H < 36. From the quadratic formula we see that the equilibria are given by P 1 = 6 36 H and P 2 = 6 + 36 H 14
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and so dP dt = ( P P 1 )( P P 2 ) . This implies that dP/dt < 0 for P < P 1 and P > P 2 and dP/dt > 0 for P 1 < P < P 2 . In particular P is decreasing when P < P 1 , P increases when P 1 < P < P 2 , and P decreases again when P > P 2 . This shows that both P = P 1 is an unstable equilibrium and P = P 2 is a stable equilibrium . (b) In order to have a single equilibrium, we must choose H so that the quadratic equation P 2 + 12 P H = 0 has a unique solution.
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