# REVIEW 2, MATH 3020 AND MATH 3030

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1 REVIEW, MATH 300 AND MATH Let P = (0, 1, ), Q = (1,1,0), R(0,1, 1), S = (1,, 4). (a) Find u = PQ and v = PR. (b) Find the angle between u and v. (c) Find a symmetric equation of the plane σ that contains the points P, Q, and R. (d) Find parametric equations of the line that is perpendicular to the plane π and that goes through the point S. (e) Find the area of the triangle PQR. (f) Find the distance between the point S and the plane PQR.. Prove that for every two vectors u and v the following inequality holds: u v u v. 3. The velocity vector of a particle at time t is given by v = 3cos(t) i + 4 j + 3sin(t) k where at time t = 0 the particle is at the point (0,, 1/). (a) Find the position of the particle at time t. (b) Find the acceleration of the particle at time t. (c) Find the speed of the particle at time t. (d) Find the distance traveled by the particle at time t. 4. Find the arc-length function of the curve given by x = t 3, y = 3t, 1 t 3.. Consider the curve x = t cos(t), y = sint, 0 t < +. Find all points on the curve where the tangent is vertical, and all points where it is horizontal. 6. Find the parametric equation of the surface obtained by rotating the curve z = 4y around the z axis. (a) Find parametric equations of the surface in cylindrical coordinates. (b) Find symmetric equation of of the surface. 7. Let A(1,0,1), B(1,1,1), and C( 1,,0) be three points in the space. (a) Find the angle between the vectors AB and AC. (b) Find a symmetric equation of the plane that contains the point A and is perpendicular to the line connecting B and C. 8. (a) Find parametric equations of the curve γ obtained as the intersection of the surface with the plane x + 3z =. (b) Find the lowest point on γ. x + y + 16z = (a) Find the symmetric equation of the surface α obtained by rotating the hyperbola y z = 1 around the z axis. (b) Prove that the surface α contains the entire curve whose parametrization is: x(t) = t + 1, y(t) = t, z(t) = t, < t < +. (c) Prove that the parametrization r(t) = x(t), y(t), z(t) of every curve that lies on the surface α satisfies the following condition: x (t),y (t),z (t) x(t),y(t), z(t). 1

2 10. Let A(3,1, 1), B(4,, 1), and C(6,1,) be three points in the space. (a) Find the angle BAC. (b) Find the equation of the plane that contains the points A, B, and C. 11. Let L 1 be the line obtained as the intersection of the planes 3x y z = 3 and x +y z = 1. Let L be the line given by 1,0,1 + t,1,1. (a) Find the parametric equations of the line L 1. (b) Prove that the lines L 1 and L are neither parallel nor intersecting. 1. The acceleration vector of the curve is given by a (t) = 1t,, sint (t 0), the initial position is r (0) = 3, 1,, and the initial velocity is v (0) = 1,0,1. (a) Find the parametric equations of the curve. (b) Find the point P at which the curve intersects the xz plane. (c) Find the tangent vector of the curve at the point P. 13. Find the parametric equation of the curve that is obtained as the intersection of the paraboloid z = 9x + 4y and the cylinder x + y = A plane is given by its symmetric equation 8x 7y + z = 1. Find the point on the plane that is closest to the point P with coordinates (3, 1, ). 1. Initially, a paint nozzle is placed at the point (100,0,0) and the nozzle is moving along the line parallel to the z-axis. The position of the nozzle at time t is (100,0,t ). The paraboloid z = x +y is rotating around z-axis with constant angular speed. As time goes from 0 to 1, the paraboloid completes one rotation around the z-axis and ends up in the original position. If the nozzle is spraying the paint towards the paraboloid the entire time, set up the integral (but do not evaluate) for the length of the curve that is painted on the surface during the time interval from t = 0 to t = 1. There is no gravity nor air resistance that interfere with the painting.

3 1. Let P = (0, 1, ), Q = (1,1,0), R(0,1, 1), S = (1,, 4). (a) Find u = PQ and v = PR. (b) Find the angle between u and v. SOLUTIONS (c) Find a symmetric equation of the plane σ that contains the points P, Q, and R. (d) Find parametric equations of the line that is perpendicular to the plane π and that goes through the point S. (e) Find the area of the triangle PQR. (f) Find the distance between the point S and the plane PQR. (a) u = PQ = 1,1,0 0, 1, = 1,, and v = PR = 0,1, 1 0, 1, = 0,,1. (b) If α is the angle between u and v we have: u v = u v cosα. Thus and α = arccos. cosα = u v u v = 1,, 0,, =, (c) In order to find the symmetric equation we need to find first a normal vector and a point in the plane. We have that P = P(0, 1, ) is a point in the plane. A normal vector can be obtained as a cross product of any two vectors parallel to π. The vectors u and v are parallel to π hence a normal vector is: N = u v = det i j k = i j + k =, 1,. Therefore a symmetric equation is: (x 0) (y + 1) + (z + ) = 0. (d) Since the vector of the line is N =, 1, the required parametric equations are: x = 1 t y = t z = 4 + t < t < +. (e) The area of PQR is equal to the one half of the area of the rectangle determined by vectors PQ and PR, thus A( PQR) = 1 u v == 1 ( ) + ( 1) + = 3. (f) d(s,σ) = (1 0) 1( + 1) + ( 4 + ) = 3 4 = 3. ( ) + ( 1) + 3. If u and v are any two vectors, prove that u v u v. If α is the angle between u and v we have: u v = u v cosα. Since 1 cosα 1 we know that cosα 1 hence u v u v. 3. The velocity vector of a particle at time t is given by v = 3cos(t) i + 4 j + 3sin(t) k where at time t = 0 the particle is at the point (0,, 1/). (a) Find the position of the particle at time t. (b) Find the acceleration of the particle at time t. 3

4 (c) Find the speed of the particle at time t. (d) Find the distance traveled by the particle at time t. (a) We have that v(t) = r (t) = 3cos(t),4,3sin(t) hence r(t) = 3 sin(t) +C 1,4t +C, 3 cos(t) +C 3, where C 1, C, and C 3 are three real numbers. We can obtain these numbers from r(0) = 0,, 1/. Therefore r(0) = 3 sin( 0) +C 1,4 0 + C, 3 cos( 0) +C 3 = 0,, 1 and C 1,C, 3 +C 3 = 0,, 1. This implies that C 1 = 0, C =, and C 3 = 1. Hence (b) a(t) = v (t) = 6sin(t),0,6cos(t). (c) v(t) = v(t) = (9cos (t) sin (t) = =. (d) s(t) = t 0 v(τ)dτ = t 0 dτ = t. 3 r(t) = sin(t),4t +, 3 cos(t) Find the arc-length function of the curve given by x = t 3, y = 3t, 1 t 3. From r(t) = t 3,3t we obtain that v(t) = 6t,6t and v(t) = v(t) = 36t t = 6t t + 1. Hence s(t) = t 1 6τ τ + 1dτ. We use the substitution z = τ +1. Then dz = τ dτ, and the bounds of integration become z t +1. Hence s(t) = 6 t +1 dz t+1 z = 3 zdz = 3 t +1 3 z3/ z= = ( t ).. Consider the curve x = t cos(t), y = sint, 0 t < +. Find all points on the curve where the tangent is vertical, and all points where it is horizontal. The vector equation of the curve is r(t) = t cos(t),sint. Hence v(t) = r (t) = + sin(t),cost. The tangent line is parallel to the velocity. The tangent is horizontal if it is parallel to 1,0. That happens if and only if cost = 0. This happens if and only if t = π + kπ for some k N 0. The tangent is vertical if and only if it is parallel to 0,1. This happens if and only if + sin(t) = 0, which never happens since sin(t) [ 1,1]. 6. Find the parametric equation of the surface obtained by rotating the curve z = 4y around the z axis. (a) Find parametric equations of the surface in cylindrical coordinates. (b) Find symmetric equation of of the surface. (a) The curve z = 4y belongs to the yz-plane. Each point Q of the curve has coordinates (t,4t ) for some real number y. This point rotates and traces the circle whose z coordinate is consent and is equal to 4t. Thus the parametrization of the surface is: x = t sinθ y = t cosθ z = 4t 0 θ π < t +. (b) From the parametric equation we obtain x + y = t, hence the symmetric equation is z = 4(x + y ). 7. Let A(1,0,1), B(1,1,1), and C( 1,,0) be three points in the space. (a) Find the angle between the vectors AB and AC. (b) Find a symmetric equation of the plane that contains the point A and is perpendicular to the line connecting B and C. 4

5 This is the same as problem (a) Find parametric equations of the curve γ obtained as the intersection of the surface with the plane x + 3z =. (b) Find the lowest point on γ. (a) In order to find the intersection we solve the system: x + y + 16z = 17 x + y + 16z = 17 x + 3z =. From the second equation we get x = 3z. Placing this in the first equation yields: y +z 30z = 8, which is equivalent to y + (z 3) = 1. We can now take y = cosθ and z 3 = sinθ. Then we have: x = sinθ y = cos θ z = 3 + sinθ 0 θ π. (b) The lowest point is the point on the surface that corresponds to the minimal z-value. The minimal z occurs when sinθ is minimal, and that happens for θ = 3π and sinθ = 1. Then we have z =, x = 19, and y = 0. Thus the lowest point is ( 19,0, ). 9. (a) Find the symmetric equation of the surface α obtained by rotating the hyperbola y z = 1 around the z axis. (b) Prove that the surface α contains the entire curve whose parametrization is: x(t) = t + 1, y(t) = t, z(t) = t, < t < +. (c) Prove that the parametrization r(t) = x(t), y(t), z(t) of every curve that lies on the surface α satisfies the following condition: x (t),y (t),z (t) x(t),y(t), z(t). (a) The equation of the surface α is x + y z = 1. (b) We need to check that x = t + 1, y = t, z = t x + y z = satisfy the equation from part (a). This follows from: ( ) ( t t + 1 +t = t4 4 t + 1 +t t4 4 = 1. ) (c) Since (x(t),y(t),z(t)) α we have x(t) + y(t) z(t) = 1. Differentiating with respect to t yields: x(t)x (t) + y(t)y (t) z(t)z (t) = 0 hence x(t),y(t),z(t) x (t),y (t),z (t) = 0. This means that the two vectors are orthogonal. 10. Let A(3,1, 1), B(4,, 1), and C(6,1,) be three points in the space.

6 (a) Find the angle BAC. (b) Find the equation of the plane that contains the points A, B, and C. (a) Let us denote θ = BAC. We have AB = 1,1,0, AC = 3,0,3. Then we have AB = = and AC = = 3. From AB AC = AB AC cosθ we obtain: 3 = 3 cosθ hence cosθ = 1 and θ = π 3. (b) The normal vector to the plane can be obtained as n = AB AC. Thus i j k n = = 3 i 3 j 3 k. The symmetric equation of the plane is: 3(x 3) 3(y 1) 3(z + 1) = Let L 1 be the line obtained as the intersection of the planes 3x y z = 3 and x +y z = 1. Let L be the line given by 1,0,1 + t,1,1. (a) Find the parametric equations of the line L 1. (b) Prove that the lines L 1 and L are skew lines. (a) The parametric equations of L 1 are obtained by solving the system 3x y z = 3 x + y z = 1. Adding the two equations yields 4x z = 4 and if we choose x = t we get z = t and y = 1+z x = 1+t t = t 1. Hence the parametric equations of L 1 are: (b) The parametric equations of L can be written as x = t y = t 1 z = t. x = 1 + s y = s If the lines L 1 and L intersect, then the system of equations z = 1 + s. t = 1 + s t 1 = s t = 1 + s would have a solution. However, from the first two equations we get s = and t = 3, which does not satisfy t = 1 + s. Since the lines are not parallel ( 1,1, is not parallel to,1,1 ), they are skew. 1. The acceleration vector of the curve is given by a (t) = 1t,, sint (t 0), the initial position is r (0) = 3, 1,, and the initial velocity is v (0) = 1,0,1. (a) Find the parametric equations of the curve. 6

7 (b) Find the point P at which the curve intersects the xz plane. (c) Find the tangent vector of the curve at the point P. (a) The velocity vector satisfies v (t) = 6t +C 1,t +C,cost +C 3. From the initial velocity v (0) = 1,0,1 we obtain that C 1 = 1, C = 0, and C 3 = 0. Thus v (t) = 6t + 1,t,cost. Therefore r (t) = t 3 +t + D 1,t + D,sint + D 3, and from the initial position r (0) = 3, 1, we obtain D 1 = 3, D = 1, and D 3 =. Therefore r (t) = t 3 +t + 3,t 1,sint +. (b) The xz plane has the equation y = 0. The intersection is the point for which t = 1, hence P = (6,0,sin1 +. (c) The velocity vector at P is v (1) = 7,,cos1 and the tangent vector is T (1) = v (1) v (1) = cos 1 7,,cos Find the parametric equation of the curve that is obtained as the intersection of the paraboloid z = 9x + 4y and the cylinder x + y = 16. The parametric equations of the cylinder are x,y,z = 4cosθ,4sinθ,z, 0 θ < π, < z < +. Points of the cylinder that belong to the paraboloid must satisfy: x = 4cosθ y = 4sinθ z = 144cos θ + 64sin θ. 14. A plane is given by its symmetric equation 8x 7y + z = 1. Find the point on the plane that is closest to the point P with coordinates (3, 1, ). Let us denote by α the given plane. The required point is the intersection of α with the line through P orthogonal to α. The normal vector to α is 8, 7,. The equation of the line through P orthogonal to α is: x = 3 + 8t y = 1 7t z = + t. The intersetion of the line with α is the point (3+8t,1 7t, +t) for which 8(3+8t) 7(1 7t)+( +t) = 1. Solving for t yields: (64+49+)t = 1, which implies that t = = Thus the required point is Q( ,1 69, + 69). 1. Initially, a paint nozzle is placed at the point (100,0,0) and the nozzle is moving along the line parallel to the z-axis. The position of the nozzle at time t is (100,0,t ). The paraboloid z = x +y is rotating around z-axis with constant angular speed. As time goes from 0 to 1, the paraboloid completes one rotation around the z-axis and ends up in the original position. If the nozzle is spraying the paint towards the paraboloid the entire time, set up the integral (but do not evaluate) for the length of the curve that is painted on the surface during the time interval from t = 0 to t = 1. There is no gravity nor air resistance that interfere with the painting. 7

8 The length of the curve is the same as the length of the curve obtained by nozzle that rotates around the paraboloid in unit time. The equation of the rotating nozzle is: x(t) = 100 cos(πt) y(t) = 100 sin(πt) z(t) = t 0 t 1. When the cylindrical coordinates of the nozzle are (θ,r,z) the paint on paraboloid will appear at the point whose cylindrical coordinates are (θ, z,z). Hence, the parametric equation of the curve is: x(t) = t cos(πt) y(t) = t sin(πt) z(t) = t 0 t 1. The velocity vector is v(t) = cos(πt) πt sin(πt), sin(πt) + πt cos(πt), t, and the speed is v(t) = 1 + 4t + 4π t = 1 + (4 + 4π )t. The required length is l = (4 + 4π )t dt. 8