DEPARTMENT OF MATHEMATICS AND STATISTICS UNIVERSITY OF MASSACHUSETTS. MATH 233 SOME SOLUTIONS TO EXAM 1 Fall 2018

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1 DEPARTMENT OF MATHEMATICS AND STATISTICS UNIVERSITY OF MASSACHUSETTS MATH SOME SOLUTIONS TO EXAM 1 Fall 018 Version A refers to the regular exam and Version B to the make-up 1. Version A. Find the center and radius of the sphere given by the equation x 4x + y + 6y + z z = 11. Solution. By completing squares we can rewrite the equation as follows (x 4x + ( ) ) + (y + 6y + ) + (z z + ( 1) ) = 11 + ( ) + + ( 1) = 5 (x ) + (y + ) + (z 1) = 5. Hence, the center is C = (,, 1) and the radius is r = 5. Version B. Find the center and radius of the sphere given by the equation x + y 6y + z + z = 1. Solution. By completing squares we can rewrite the equation as follows x + (y 6y + ( ) ) + (z + z + 1 ) = 1 + ( ) + 1 = 9 x + (y ) + (z + 1) =. Hence, the center is C = (0,, 1) and the radius is r =.. Version A. (a) Let a be the vector with initial point P (0,, ) and terminal point Q(1, 1, 7) and let b be the vector with initial point R(, 0, 1) and terminal point S(, 1, ). Find the unit vector in the direction of b + a. Solution. a = P Q = 1 0, 1, 7 = 1,, 9 b = RS = ( ), 1 0, 1 = 1, 1,

2 b + a = 1, 1, + 1,, 9 =,, 6 + 1,, 9 = 4, 0, b + a = (4) + (0) + ( ) = 5 = 5 So the unit vector in the direction of b + a is u = b+a b+a = 4 5, 0, 5 (b) Let a = 1,, 9 and b = 1, 1,. The vector b a lies in either the xy-, xz-, or yz-plane. Which one is it? Explain your reasoning. Solution. b a = 1, 1, 1,, 9 =,, 6 1,, 9 = 4, 0, Since the y component is 0, it lies in the xz-plane. Version B. (a) Let a = 1,, 9 and b = 1, 1,. Find the vector in the direction of b a whose length is. Solution. b a = 1, 1, 1,, 9 =,, 6 1,, 9 = 4, 0, b a = (4) + (0) + ( ) = 5 = 5 The unit vector in the direction of b a is u = b+a b+a = 4 5, 0, 5 Thus the vector in the direction of b a with length is u = 8 5, 0, 6 5 (b) Let a be the vector with initial point P (0, 5, 0) and terminal point Q(, 1, 7) and let b be the vector with initial point R(1, 1, 7) and terminal point S(, 5, 0). Are a and b equivalent vectors? Why or why not? Solution. Yes, a and b are equivalent vectors, since: a = P Q = 0, 1 5, 7 0 =, 6, 7 b = RS = 1, 5 1, 0 ( 7) =, 6, 7. Version A. (a). Consider the points P = (1,, 4), Q = (, 1, ), and R = (0, 1, ). Find the vector projection of the vector P Q onto the vector P R. Solution. P Q = Q P = 1, 4, 1, P R = R P = 1,, 1.

3 proj P R ( P Q P R P Q) = ( ) P R P R = 1( 1) + ( 4)( ) + ( 1)( 1) ( 1) + ( ) + ( 1) 1,, 1 = 4, 8, 4 (b). A particle moves in a straight line from the point A = (1,, ) to the point B = ( 1, 0, 7) with the help of a force F of magnitude 100 newtons that points in the direction 1, 1, 1. Find the work done by the force F on the particle as it moves from A to B. (Distance is measured in meters). Solution. F = 100 1, 1,1, D = B A =,, 4. Work = F D = 100 (( 1)( ) + ( 1)( ) + 1(4)) = 800 joules (or newton-meters). Version B. (a). Consider the points P = ( 1,, 5), Q = (, 1, ), and R = (1,, 1). Find the scalar projection of the vector P Q onto the vector P R. Solution. P Q = Q P = 4, 4,, P R = R P =, 0, 4. comp P R ( P Q P R P Q) = ( ) P R 4() + ( 4)(0) + ( )( 4) = () + (0) + ( 4) = (b). A force F = i + j k, measured in newtons, moves a particle in a straight line from the origin for 50 meters in the direction (1, 1, 1). Find the work done by the force on the particle. Solution. F =,, 1, D = 50 1,1,1. Work = F D = 50 (()(1) + ()(1) + ( 1)(1)) = 00 joules (or newton-meters). 4. Version A. Consider the lines l 1 : l : x 1 x 4 = y 6 = y = z 4 = z 1 (a) Show that the lines l 1 and l are parallel. Solution.

4 4 The direction vector of l 1 is v 1 =,,. The direction vector of l is v = 4, 6, 4 = v 1. Therefore these two lines are parallel. (b) Find the equation of the plane which passes through l 1 and l. Solution. We pick a point p 1 on l 1, for example p 1 = (1,, 1). Similarly we pick a point p on l, for example p = (0,, ). Consider the vector p 1 p = 1, 0, 1. We compute the cross product n = v 1 p 1 p =, 4,. The plane P with normal vector n which passes through p 1 is the desired plane. The vector equation of the plane P is Version B. Consider the lines, 4, x 1, y, z 1 = 0 (x 1) 4(y ) + (z 1) = x 4y + z + 6 = 0. l 1 : x = t + 1, y = t +, z = t + 1 l : x = 4t, y = 6t +, z = 4t + (a) Show that the point p = (0,, ) is both on l 1 and l. Solution. For the parameter t = 1 on l 1 we get the point p. For the parameter t = 0 on l we get the point p. (b) Show that l 1 and l are not the same lines. Solution. The point p 1 = (1,, 1) is on l 1 but not l. Indeed if (1,, 1) is on l 1, for some t we should have 1 = 4t, = 6t +, 1 = 4t + which is not possible. (c) Find the vector equation of the plane which contains both lines l 1 and l. Solution. The direction vector of l 1 is v 1 = 1, 1, 1. The direction vector of l is v = 4, 6, 4. We compute the cross product n = v 1 v =, 8, 10. The plane P with normal vector n which passes through p 1 is the desired plane. The vector equation of the plane P is, 8, 10 x, y, z = 0 (x) + 8(y ) 10(z ) = x + 8y 10z + 4 = Version A. Consider the points below P (0, 1, ), Q(, 4, 5), R( 1, 0, 1) and S(6, 1, 4). (a) Find a unit vector orthogonal to the plane through the points P, Q, and R. Solution. The vector P Q P R is orthogonal to both P Q and P R and is therefore to the plane through the points P, Q, and R. Since P Q = i+j+k and P R = i j k we obtain P Q i j k P R = = = 1 1 i 1 1 j k = = j + k,

5 5 and P Q P R = j + k = 0 + ( 1) + 1 =. Thus, the unit vector orthogonal to both is u = 1 ( j + k) = j + k. (b) Find the area of the triangle P QR. Solution. The area of the parallelogram with adjacent sides is P Q P R =. Therefore, the are of the triangle P QR is half the area of this parallelogram, that is,. (c) Find the volume of the parallelepiped with adjacent edges P Q, P R, and P S. Do the points P, Q, R and S lie in the same plan? Justify your answer. Solution. We have that P S = 6i j + k and P Q ( P R P S) = Thus, the volume of the parallelepiped V = P Q ( P R P S) = 4 = 4. = 4. Since the volume of the parallelepiped is different from 0 then the points P, Q, R and S do not lie in the same plan. Version B. Consider the points below A = (1, 1, 1), B = (, 0, ), C = (4, 1, 7) and D = (, 1, ). (a) Find two vectors perpendicular to the plane through the points A, B, and C. Solution. Since AB = i j + k and AC = i + 6k, we obtain AB AC = i j k = i 1 6 j k = 6i + k. 0 Thus, the two vectors orthogonal to both are 6i + k and 6i k. (b) Find the area of the parallelogram determined by AB and AC. Solution. The area of the parallelogram with adjacent sides is AB AC = 6i + k = ( 6) = 45. (c) Find the volume V of the parallelepiped such that the A, B, C and D are vertices and the vertices B, C, D are all adjacent to the vertex A. Are the vectors AB, AC and AD coplanar? Justify your answer.

6 6 Solution. We have that AD = i j k and AB ( AC AD) = Thus, the volume of the parallelepiped = 1. V = P Q ( P R P S) = 1 = 1. Since the volume of the parallelepiped is different from 0 the vectors AB, AC and AD are non coplanar. 6. Version A. Sketch the surface in R with equation z = 4 x. Label the coordinate axes, and include and label the trace at y =. Solution. The equation z = 4 x factors as z = ( x)( + x), and thus gives a parabolic cylinder, the z = 0 trace of which is a downward opening parabola with roots x = ±. The y = trace appears as an identical downward opening parabola in the plane z =. Figure 1: The trace at y = is the light blue parabolic curve. Version B. Sketch the surface with equation y + z = 9. Label the coordinate axes and include and label the trace at x = 4. Solution. The equation describes a right circular cylinder, with central axis the x-axis, and with radius. The trace at x = 4 is a circle of radius.

7 7 Figure : The trace at x = 4 is the red circle. 7. Version A. The path r(t) of a particle satisfies dr dt = 8, 5 t, 4t. If r(0) = 1, 6, 0, where is the particle located when t = 4? Solution. The general solution is r(t) = 8, 5 t, 4t dt = 8t, 5t t, 4 t + C r(0) = 8(0), 5(0) (0), = 0, 0, 0 + C 4 (0) + C Since r(0) = 1, 6, 0, we have C = 1, 6, 0. Then r(t) = 8t, 5t t, 4 t + 1, 6, 0 = 8t + 1, 5t t + 6, Therefore, when t = 4, the position of the particle is 4 t r(4) = 8(4) + 1, 5(4) (4) + 6, =,, 56 4 (4)

8 8 Version B. The path r(t) of a particle satisfies dr dt where is the particle located when t =? = 8t, 5 8 sin(t). If r(0) = 0, 10, Solution. The general solution is r(t) = 8t, 5 8 sin(t) dt = 4t, 5t + 4 cos(t) + C r(0) = 4(0), 5(0) + 4 cos(0) + C = 0, 4 + C Since r(0) = 0, 10, we have C = 0, 6. Then r(t) = 4t, 5t + 4 cos(t) + 0, 6 = 4t, 5t + 4 cos(t) + 6 Therefore, when t =, the position of the particle is r() = 4(), 5() + 4 cos(6) + 6 = 6, cos(6) 8. Version A. (a) Find a vector function that represents the curve of intersection of the cylinder x + y = 9 and the surface z = xy. Solution. Let x = cos(t) and y = sin(t), with 0 t π. Then Thus the vector equation is z = 18 cos(t) sin(t) = 9 sin(t). r = cos(t), sin(t), 9 sin(t). (b) Two particles are traveling along the space curves: r 1 = t, t, t and r = 1 + t/, + t, 4 + t. Do the particles collide? If they do then find the position at which they collide. Solution. The particles collide if we can find t such that r 1 (t) = r (t). That means: t = 1 + t/, t = + t and t = 4 + t. There is only one solution t =, thus two particles will collide at t =. The position of collision is (, 4, 8). (c) Find the limit lim ( t t t4 + 1, t + 1, arctan t) e t Solution. By taking limit for each component, as t, the vector tends to ( 1, 0, π ). Version B. (a) Find a vector function that represents the curve of intersection of the paraboloid z = 4x + y and the parabolic cylinder y = x.

9 9 Solution. Since y = x, we have that z = 4x + x 4. Thus let x = t with < t < we have that y = t and z = 4t + t 4. The vector equation is: r = t, t, 4t + t 4. (b) Two particles travelling along the space curves: r 1 = (t, 7t, t 1) and r = (1 + t, 6 + 5t, 4 + t). Do the particles collide? If they do then find the position at which they collide. Solution. The particles collide if we can find t such that r 1 (t) = r (t). That means: t = 1 + t, 7t = 6 + 5t and t 1 = 4 + t. There is only one solution to the first two equations: t =. However, t = does not satisfy the third equation, thus two particles will not collide. (c) Find the limit t + t lim t (te t, t 1, t sin 1 t ) Solution. By taking limit of each component, as t, the vector tends to (0, 1, 1).

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