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Show your work and give reasons for you answers on all of the following problems.
No calculators, books or notes are allowed.

1. The position of a particle at time t is given by

   r = < 3t5,4t7/2,3t4 > ,       -¥ < t < ¥

   1. Find the velocity and acceleration of the particle at time t.
   2. Find parametric equations for the tangent line to the path of the particle at t = 1.

   Answer:

   1. The velocity v(t) is the rate of change of the position r(t) with respect to time. Therefore, to obtain the velocity we differentiate the position vector r(t) with respect to t and obtain

      v(t) = < 15t4,(28/2)t5/2,12t3 >

      The acceleration a(t) is the rate of change of the velocity v(t) with respect to time. Thus, to obtain the acceleration we differentiate the velocity vector v(t) with respect to t and obtain

      a(t) = < 60t3,(140/4)t3/2,36t2 >

   2. In order to find parametric equation for a line we need a point P_o(x_o,y_o,z_o) on the line and a direction vector v = < a,b,c > for the line. The point P_o is the point where the particle is at at time t = 1. That is < x_o,y_o,z_o > = r(1) = < 3,4,3 > . The velocity vector v(1) is along the tangent line which we are interested in. Therefore, we take can v(1) = < 15,(28/2),12 > , as the direction vector of the tangent line. Now we have a point on the line and a direction vector for the line, and we can write parametric equations for it:

      x = 3+15t,  y = 4+(28/2)t,  z = 3+12t,      -¥ < t < ¥

2. A curve has the parametric equations

   r = < 5t4,4t2,4t4 > ,       -¥ < t < ¥

   1. Find parametric equations for the tangent line to the curve when t = 1.
   2. Find a unit tangent vector to the curve when t = 1.

   Answer:

   1. In order to find parametric equations for a line we need a point P_o(x_o,y_o,z_o) on the line and a direction vector v = < a,b,c > for the line. The point P_o is r(1). That is < x_o,y_o,z_o > = r(1) = < 5,4,4 > . The velocity vector v(1) = r¢(1) is along the tangent line which we are interested in. Therefore, we can take v(1) = < 20,(8),16 > , as the direction vector of the tangent line. Now we have a point on the line and a direction vector for the line, and we can write parametric equations for it:

      x = 5+20t,  y = 4+(8)t,  z = 4+16t,      -¥ < t < ¥

   2. We computed a tangent vector to the curve when t = 1. Namely, r¢(1). Therefore, a unit tangent vector u when t = 1 is r¢(1)/|r¢(1)|. That is, we devide r¢(1) by its length |r¢(1)| and obtain.

      u = < 20/   ___ Ö720   ,8/   ___ Ö720   ,16/   ___ Ö720   >

3. The position of a particle of unit mass at time t is given by

   r(t) = 4costi+8sint j+6 t k

   1. The motion of this particle is confined to a surface. Find an equation for this surface, sketch it and describe it in words. Sketch the path of the particle.
   2. What is the force that is acting on this particle and confines its motion to this particular surface?
   3. Find parametric equations for the tangent line to the path of this particle when t = p/3.

   Answer:

   1. Recall that cos²t+sin²t = 1. Also notice that x = 4cost and y = 4sint. Therefore,

      x2/16+y2/64 = 1

      In \mathbbR², i.e. in the xy-plane, this is an equation of an ellipse. In \mathbbR³, i.e. in the 3 dimensional space, this is an equation of a vertical elliptic cylinder with its axis along the z it-axis. By elliptical we mean that if we take a horizontal cross section (i.e. if we cut the cylinder with any horizontal plane z = c) we obtain an ellipse whose equation is the above equation.
   2. Newton's equations says that

      (mass)(acceleration) = force

      The particle has unit mass. Thus acceleration = force = F. The acceleration a = v¢(t) = r¢¢(t). Thus

      v(t) = -4sint i+8cost j+6k

      F(t) = r(t) = -4costi-8sintj

   3. In order to find parametric equations for a line we need a point P_o(x_o,y_o,z_o) on the line and a direction vector v = < a,b,c > for the line. The point P_o is r(p/3). Recall that

      cos(p/3) = 1/2,        sin(p/3) = Ö3/2

      Thus < x_o,y_o,z_o > = r(p/3) = (4/2) i+(8Ö3/2) j+(6p/3)k. The velocity vector v(p/3) = r¢(p/3) is along the tangent line which we are interested in. Therefore, we can take v(p/3) = -(4Ö3/2)i+(8/2)j+6k, as the direction vector of the tangent line. Now we have a point on the line and a direction vector for the line, and we can write parametric equations for it:

      x = 4/2-(4Ö3/2)t,  y = 8Ö3/2 +(8/2)t,  z = 6p/3+6t,      -¥ < t < ¥

4. Find the arc length of the parametric curve

   r(t) = cos4ti+2t3/2j+ sin4tk       1 £ t £ 2

   Answer: The arc length of a parametric curve

   r(t) = x(t)i+y(t)j+z(t)k,        a £ t £ b

   is given by the integral

   L = ó õ b a  ||r¢(t)|| dt = ó õ b a    æ Ö (x¢(t))2+ (y¢(t))2+ (z¢(t))2     dt

   (1)

   For our curve

   r¢(t) = -4cos4ti+(6/2)t1/2j+4sin4tk

   Recall that cos² q+ sin²q = 1. It follows that the arc length of our curve is given by the integral

   ó õ 2 1    _________ Ö16+(36/4)t

   In order to evaluate this integral we make a substitution

   u = 16+(36/4)t

   It follows that

   du = (36/4)dt

   Now the integral becomes

   ó õ (4/36)u1/2du = (4/36)(2/3)u3/2

   Now, the arc length is

   L = (8/108)[16+(36/4)t]3/2\mid12

   = (8/108){[16+(36/2)]3/2- [16+(36/4)]3/2}

5. Find the arc length of the parametric curve

   r(t) = e2ti+ (Ö8  t+2)j+ e-2tk,       1 £ t £ 2

   Answer: The arc length of a parametric curve

   r(t) = x(t)i+y(t)j+z(t)k,        a £ t £ b

   is given by the integral

   L = ó õ b a  ||r¢(t)|| dt = ó õ b a    æ Ö (x¢(t))2+ (y¢(t))2+ (z¢(t))2     dt

   (2)

   For our curve

   r¢(t) = 2e2ti+ (Ö8)j-2e-2tk,

   It follows that the arc length is

   L = ó õ 2 1    æ Ö 4e4t+8+4e-4t     dt = ó õ 2 1    æ Ö [2e2t +2e-2t ]2     dt = ó õ 2 1  [2e2t +2e-2t   ]dt

   = [e2t-e-2t]\mid12 = e4 -e-4-e2+e-2