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# B determine the special value of h for which the

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(b) Determine the special value of H for which the population growth has a single equilibrium. What will happen if we start harvesting at a rate higher than this special value? Solution Key: 2.8 Solution: 3.8 Question 9: True or false. Explain your reasoning. (a) The line vector r ( t ) = (1 + 2 t ) hatwide ı + (1 + 3 t ) hatwide + (1 + 4 t ) hatwide k is perpendicular to the plane 2 x + 3 y 4 z = 9. (b) The equation x 2 = z 2 in three dimensons, describes an ellipsoid. (c) | vectora × vector b | = 0 implies that either vectora = 0 or vector b = 0. Solution Key: 2.9 Solution: 3.9 Question 10: A bug is crawling along a helix and his position at ttime t is given by −→ r ( t ) = ( sin(2 t ) , cos(2 t ) , t ) . Which of the following statements are true and which are false? Explain and justify your reasoning. (a) The unit normal vector always points toward the z -axis. (b) The bug travels upward at a constant rate, i.e. the unit tangent vector has a constant z -component at any moment of time. (c) The unit binormal vector always points straight up or straight down. 6

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Solution Key: 2.10 Solution: 3.10 7
2 Solution key (1) (D) (2) (E) (3) (B) (4) (F) (5) (D) (6) (D) (7) (E) (8) (a) H < 36, one unstable and one stable equilibrium; (b) to have one equilibrium we must harvest at a rate H = 36. If H > 36, then the trout population will go extinct. (9) (a), (b), and (c) are false (10) (a) is true , (b) is true , and (c) is false . 8

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3 Solutions Solution of problem 1.1: The point P on the curve corresponding to the value of the parameter t = 0 has position vector −→ r (0) = ( 2 , 1 , 0 ) . In other words P has coordinates P (2 , 1 , 0). The tangent vector to the curve at P is the vector d −→ r dt (0) = parenleftbigg d dt (big t 2 + 3 t + 2 , e t cos t, ln( t + 1) )big parenrightbigg | t =0 = parenleftbigg(bigg 2 t + 3 , e t cos t e t sin t, 1 t + 1 )biggparenrightbigg | t =0 = ( 3 , 1 , 1 ) . The tangent line L at t = 0 is the line passing through P and having the tangent vector d −→ r /dt (0) as a direction vector. Thus L is given by the parametric equations x = 2 + 3 s, y = 1 + s, z = s. To intersect the line L with the plane x + y + z = 8 we substitute the parametric expressions for x , y , and z in the equation of the plane and solve for s . We get (2+3 s )+(1+ s )+ s = 8, i.e. s = 1. Thus the point of intersection is the point on the line which corresponds to the value of the parameter s = 1, i.e. the point (5 , 2 , 1) . The correct answer is (D) . Solution of problem 1.2: A vector is perpendicular to a plane if and only if it is parallel to a normal vector for the plane. To find a vector −→ n which is normal to the plane containing P (2 , 1 , 5), Q ( 1 , 3 , 4), and R (3 , 0 , 6) we need to find two non-parallel vectors in the plane and compute their cross products. The vectors −→ PQ = (− 1 2 , 3 1 , 4 5 ) = (− 3 , 2 , 1 ) −→ PR = ( 3 2 , 0 1 , 6 5 ) = ( 1 , 1 , 1 ) 9
are not parallel and belong to the plane, so we have −→ n = −→ PQ × −→ PR = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle hatwide ı hatwide hatwide k 3 2 1 1 1 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (2 1) hatwide ı + (3 1) hatwide + (3 2) hatwide k = hatwide ı + 2 hatwide +

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• Fall '07
• Temkin
• Math, Calculus, Quadratic equation, Parametric equation

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