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# This means that the discriminant of this equation

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This means that the discriminant of this equation ought to be equal to zero, i.e. we ought have 4(36 H ) = 0. Thus the special value of H is H = 36. In this case, the differential equation becomes dP dt = 12 P P 2 36 = ( P 6) 2 . The unique equilibrium is at P = 6. If H > 36 the equation has no equilibria, and moreover we have that dP dt = 12 P P 2 H = ( P 6) 2 + (36 H ) is always negative. This shows that if we harvest the trout at rate higher than the critical harvesting rate H = 36, then the population will steadily decrease and we will eventually empty the pond completely . In contrast, if we harvest at a rate smaller than the critical harvesting rate H = 36 and if we start with enough trout in the pond, then we will always have enough fish to harvest . Solution of problem 1.9: (a) From the parametric equation vector r ( t ) = (1 + 2 t ) hatwide ı + (1 + 3 t ) hatwide + (1 + 4 t ) hatwide k we can extract a direction vector for the line. It is the vector −→ v whose components are given by the coefficients of the parameter t in the parametric equation. Thus −→ v = 2 hatwide ı +3 hatwide +4 hatwide k . Also, from the equation 2 x + 3 y 4 z = 9 of the plane we can extract a normal vector −→ n to the plane. It is the vector whose components are the coefficients of the equation of the plane. Thus −→ n = 2 hatwide ı + 3 hatwide 4 hatwide k . 15

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The line and the plane will be perpendicular when −→ v is parallel to −→ n , that is when −→ v is proportional to −→ n . But if −→ v = c · −→ n for some constant c , then we will have 2 = 2 c , 3 = 3 c , and 4 = 4 c . From he first equation we get c = 1 but from the last equation we have c = 1. This is a contradiction. Therefore the line and the plane are not perpendicular, and so (a) is False . (b) The equation x 2 = z 2 depends only on two variables so it describes a cylinder with a base in the xz -plane. Hence x 2 = z 2 can not be an ellipsoid and (b) is False . (c) If the length of vectora × vector b is zero, then the vector vectora × vector b must be the zero vector. This can happen either when one of vectora or vector b is the zero vector, or when vectora is parallel to vector b . For instance if −→ a is any vector and −→ b = −→ a we will have vectora × vector b = vectora × vectora = −→ 0 . Hence (c) is False . Solution of problem 1.10: (a) The unit normal vector is given by −→ N = d −→ T dt vextendsingle vextendsingle vextendsingle d −→ T dt vextendsingle vextendsingle vextendsingle where −→ T is the unit tangent vector. To compute −→ T we compute the velocity vector d −→ r /dt = ( 2 cos(2 t ) , 2 sin(2 t ) , 1 ) and normalize −→ T = 1 | d −→ r /dt | d −→ r /dt = 1 radicalbig 4 cos 2 (2 t ) + 4 sin 2 (2 t ) + 1 ( 2 cos(2 t ) , 2 sin(2 t ) , 1 ) = (bigg 2 3 cos(2 t ) , 2 3 sin(2 t ) , 1 3 )bigg .
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