LB 220 Homework 1 (due Monday, 01/14/13)

Transcription

1 LB 220 Homework 1 (due Monday, 01/14/13) Directions. Please solve the problems below. Your solutions must begin with a clear statement (or re-statement in your own words) of the problem. You solutions should be clear, legible, and demonstrate at minimum partial progress towards a complete solution to the problem. Please refer to the syllabus for the policy on grading (communication, completeness, and correctness) and late homework (homework is due at the start of class, late homework is assessed a 20% penalty if submitted within the next 48 hours) Collaboration. I encourage you to discuss the homework problems with your classmates. However, each student must write and submit his or her own homework solutions. Calculators. You can use calculators to determine a numerical approximation to an answer to an application question, but you should use exact values until the very last step in the problem. Calculators are not, however, permitted on any quizzes or exams. 1. Write an equation for the locus of points which are equidistant from the points A and B. What shape is this locus? Use the following points: A(2, 4, 1), B( 1, 0, 3). Solution: Let P (x, y, z) be a point belonging to this locus. Then the square of the distance from P to A is equal to the square of the distance from P to B: (x 2) 2 + (y 4) 2 + (z 1) 2 = (x + 1) 2 + y 2 + (z 3) 2. This simplifies to 6x + 8y 4z = 11, which is a linear equation. Therefore, the locus is a plane. An alternate solution is to observe that the midpoint M(1/2, 2, 2) of AB belongs to the locus and that if the locus is a plane, then BA = 3, 4, 2 is a normal vector to this plane. Therefore, the plane has equation 3(x 1/2) + 4(y 2) + 2(z 2) = 0, which is the same (up to a multiple of two) as the equation given above. 2. Write an equation for the locus of points whose distance to A is twice their distance to B. What is the shape of this locus? Use the following points: A(2, 4, 1), B( 1, 0, 3).

2 Solution: Let P (x, y, z) be a point belonging to this locus. Then the square of the distance from P to A is equal to four times the square of the distance from P to B: (x 2) 2 + (y 4) 2 + (z 1) 2 = 4((x + 1) 2 + y 2 + (z 3) 2 ). This simplifies to 3x 2 + 3y 2 + 3z x + 8y 22z + 19 = 0. Now divide through by 3 and complete the squares to obtain the following: (x + 2) 2 + (y )2 + (z 11 3 )2 = Therefore, the locus is a sphere of radius 116/3 centered at the point ( 2, 4/3, 11/3). 3. A team of dogs pull a sled up a snow covered path with a 5 incline with a force of 500 N at an angle of 10 above the inclined path. Determine the horizontal and vertical components of the force. Solution: The force vector is directed 15 above the horizontal. The horizontal component is therefore 500 cos N and the vertical component is 500 sin N. 4. Suppose that wind blows with a speed of 50 km/hr from the direction N45 W (meaning from a direction which is 45 to the west of the northerly direction). A pilot is piloting an aircraft in the direction N60 E at an airspeed of 250 km/hr. Determine the true course and ground speed of the aircraft. (The true course is the direction of the resultant of the velocity vectors of the wind and the aircraft; the ground speed is the magnitude of this resultant vector.) Solution: Let North be represented by the positive y-axis and East by the positive x-axis. Then the wind blows towards the fourth quadrant. Let w be the velocity vector of the wind. Its magnitude is 50 km/hr and its direction is the same as the unit vector Therefore, cos 45, sin 45 = 1 2 1, 1 = w = , , 1. 2 Similarly, if v is the velocity vector of the aircraft, then its magnitude is 250 km/hr and its direction is the same as the unit vector cos 30, sin 30 = 1 2 3, 1.

3 Therefore, v = 125 3, 1. From this, we see that the resultant vector is equal to v + w = , The ground speed is the magnitude of this vector which is approximately equal to 267 km/hr. The true course can be determined by first finding the tangent of the angle θ with which the vector makes with the positive x-axis: tan θ = Solving, we find that θ 20. Therefore, the true course is approximately N70 E. 5. A clothesline is tied between two poles, 8 m apart. The line is taught and has negligible sag. When a wet shirt with a mass of 0.8 kg is hung at the middle of the line, the midpoint is pulled down a distance of 8 cm. Determine the tension in each half of the clothesline. Solution: Let T be one of the two tension vectors directed along the clothesline from the center (where the shirt is located) towards one of the ends. Suppose that T = T x i + T y j, where T x and T y are the as yet to be determined components of the tension. The goal of the problem is compute the magnitude T = Tx 2 + Ty 2 of the tension. Since there are two tension vectors, one to the left and one to the right, which together offset the downward force of the shirt, we have that 2T y = (0.8)g = T y = 0.4g, where g 9.8 m s 2. The ratio of T y to T x is equal to the ratio of 8 cm = 0.08 m to 4 m (half the distance of the clothesline) because of the similarity of triangles. This implies that T x = 50T y. Therefore, T = Tx 2 + Ty 2 = 2500Ty 2 + Ty 2 = T y 2501 (0.4)(9.8) N

4 Remark: In the English or Imperial system, units of pounds are used. There are two kinds of pounds however. There is the pound-mass which is approximately equal to kg or 1 kg 2.2 lb. And there is the pound-force which is approximately equal to 4.45 N. There is a simple relation: suppose I have kg of vegetables, i.e. one pound-mass of vegetables; if I place one pound-mass of vegetables on a scale at the grocery store, then a force of 4.45N, i.e. one pound-force, is exerted on the scale. So, one pound-mass exerts a force of one pound-force. To distinguish between the two pound-units, the pound-mass may be denoted by lb m and the pound-force by lb f. To obtain the extimates given above, use the definition of the newton and the measurement of the acceleration of gravity near the earth s surface: 1 N = (1 kg)(1 m s 2 ), g 9.8 m ft 32 s2 s 2, 1 lb f (1 lb m )(32 ft s 2 ) (0.453 kg)(9.8 m ) 4.45 N. s2 The answer to the problem above was 196 N 44 lb f. So, the force of tension is about the same as the force exerted by 44 pounds of vegetables on a scale in the grocery store. You can check this is correct: 44/ kg and (20)(9.8) 196 N. The pound-mass of the wet shirt was given to be 0.8 kg 1.76 lb m. The wet shirt exerts a force of approximately 1.76 pound-force which is approximately (1.76)(4.45) (0.8)(9.8) 7.8 N of force. So, most of the tension in the clothesline is due to the fact that the clothesline has negligible sag. Another units: the poundal (unit of force) is 1 pdl = 1 lbm ft. So s 2 1 lb f 32 pdl. And the slug (unit of mass) is 1 lb f = (1 slug)(1 ft ). s 2 Crazy. What s the average mass of a slug in slugs? Here s what you should keep in mind: the force in newtons exerted by 1 pound-mass of vegetables on a scale due to the acceleration of gravity (i.e. 1 pound-force) is approximately equal to 4.45 N. In reverse, to exert one newton of force on a scale, you should place approximately pound-mass of vegetables on a scale. The following link is useful for specific examples of forces we might observe, measured in newtons:

5 6. Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and is half its length. Solution: Let A, B, and C be the vertices of a triangle. Let M be the midpoint of AB and N the midpoint of AC. The following vector equations hold: AB + BC + CA = 0, AB = 2MB, BC = 2BN. Consider the triangle M BN. The following vector equation holds: MB + BN + NM = 0. (I am using the fact that vector addition v + w can be computed by placing the tail of w at the head of v and representing v + w by the directed line segment from the tail of v to the head of w.) Multiply the above equation by two and re-write in terms of AB, and BC: 2 MB + 2 BN + 2 NM = AB + BC + 2 NM = 0. Combined with the very first equation we deduce that 2 NM = CA. Thus, NM is parallel to CA. And therefore, AC is parallel to MN. In fact, we have proved more: the length of AC is twice that of MN.

6 7. Determine the measure of the acute angles at the points of intersection of the curves y = x 2 and y = x 3. Solution: x 2 = x 3 = x 2 (x 1) = 0 = x = 0, 1. So, the points of intersection are O(0, 0) and P (1, 1). At O, the slope of the tangent line to y = x 2 is y (0) = 2(0) = 0. The same is true for the tangent line to y = x 3 at O. Therefore, the angle between the curves at O is equal to zero. At P, the slope of the tangent line to y = x 2 is 2 and the slope of the tangent line to y = x 3 is y (0) = 3(1) 2 = 3. The vector a = 1, 2 has slope 2 and the vector b = 1, 3 has slope 3. The cosine of the angle between these two vectors is equal to (a b)/( a b ) = 7/( 5 10) = θ. Therefore, the angle between the two curves at P is equal to arccos θ Determine the two unit vectors orthogonal to both 1, 2, 3 and 1, 0, 1. Solution: Let v be the vector product of the above two vectors: i j k v = = 2, 4, The length of v is equal to 24 = 2 6. Therefore, the unit vectors 1 24 v = 1 6 1, 2, 1 and its negative are the two unit vectors orthogonal to both of the two given vectors. 9. Use the triple scalar product to determine whether the points A(1, 3, 2), B(3, 1, 6), C(5, 2, 0), and D(3, 6, 4) are coplanar. Solution: AB = 2, 4, 4, AC = 4, 1, 2, AD = 2, 3, 6. The points are coplanar if and only if the following determinant (representing the volume of the parallelepiped spanned by the above three vectors) is equal to zero = ( 4) = 2(6 + 6) + 4( ) + 4(12 + 2) = = 0 So, the points are coplanar.

7 10. A wrench 30 cm long lies along the positive y-axis and grips a bolt at the origin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find the magnitude of the force needed to supply 100 N m of torque to the bolt. Let τ be the torque vector so that τ = r F, where r = 0, 30/100, 0 (in meters) is the position vector of the point where the force is applied (relative to the point of rotation placed at the origin) and where F is the force vector. We are given that F has the same direction as v = 0, 3, 4. Therefore, F = F v v. We are to determine the value of α = F so that τ = 100. From the above equation we have that F = α 5 v. The easiest way to solve the problem is to make a sketch and recognize that the vectors are parallel to the yz-plane and so the problem can be reduced to a two-dimensional problem using a (3, 4, 5)-triangle. The problem can be solved using the vector product, but it is somewhat easier to use the scalar product and the identity sin θ = 1 cos 2 θ for 0 θ π. For, cos θ = r F r F = 3 5. (You will see that the factor of α cancels in the above computation.) From this we have that sin θ = 4 5. And therefore, τ = 100 = r F sin θ = 3 10 α4 5 and so α = = N of force needs to be applied.