Chapter 1

Practice Test

Instructions.  Show your work and/or explain your answers.

  1. Find the length of each vector and the angle between them.
    u = á Ö2, Ö6,  2Ö2 ñ         and        v = á0,0,1 ñ
    Solution: u·u = 2+6+8 = 16, || u|| = 4, || v|| = 1, u·v = 2Ö2, so
    cos( q) =  u·v
    || u||  || v||
    		=  2Ö2
    4
    		 =  Ö2
    2
    Thus, q = p/4.
  2. Show that if u is perpendicular to v, then
    || u|| 2+|| v|| 2 = || u-v|| 2
    Solution: || u-v||2 = ( u-v) · (u-v) = u·u + v·v - 2 u·v = || u||2 + || v||2-0
  3. Find a number k for which u = á1,2,1 ñ is perpendicular to v = ák,3,4 ñ.
    Solution: u·v = k+6+4 = k+10. Thus, u·v = 0 implies that k = -10.
  4. Find the area of the triangle whose vertices are P1(0,0,0) , P2( 1,1,0) , P3( 1,1,4) .
    Solution: The vectors are u = á1,1,0 ñ and v = á 1,1,4 ñ . Their cross product is u×v = á 4,-4,0 ñ , so that the area is
    A =  1
    2
    		 || u×v|| =   1
    2
    16+16+0
    		 = 2Ö2
  5. Find the equation of the plane through the points P1(0,0,0) , P2( 2,1,5) , and P3( -1,1,2) .
    Solution: The vectors are u = á2,1,5 ñ and v = á -1,1,2 ñ , so that u×v = á -3,-9,3 ñ . Thus, the equation of the plane is
    -3( x-0) -9( y-0) +3( z-0) = 0
    so that in functional form we have z = x+3y.
  6. Find the xy-equation and sketch the graph of the curve
    r( t) = á cos( 2t) ,sin(t) ñ ,    t in  é
    ë     		0,  p
    2
    		 ù
    û
    Solution: Since cos(2t) = 1-2sin2(t), we have x = 1-2y2. Moreover, r(0) = ácos(0) ,sin(0) ñ = (1,0) and r( p/2) = ( cos(p) ,sin(p/2) ) = (-1,1) . Thus, the parameterization is the section of the curve x = 2y2-1 from (1,0) to ( -1,1) .

  7. Find the cartesian equation of the parametric curve
    r( t) = á sin2( t) ,cos( t) ñ ,  t  in  [ 0,p]
    Then sketch the curve showing its orientation and its endpoints.
    Solution: Since cos2( t) +sin2( t) = 1, we have y2+x = 1 or x = 1-y2. Moreover, r(0) =   á 0,1 ñ and r( p) = á 0,-1 ñ .

  8. Find the velocity, speed, and acceleration of the curve with parameterization
    r( t) = á t,t2,t3 ñ
    Solution: The velocity is v( t) = á1,2t,3t2 ñ , the acceleration is a( t) = á 0,2,6t ñ , and the speed is
    v =  
    1+4t2+9t4
  9. If a rock is thrown into the air near the earth's surface with initial velocity v( 0) = á16,0,64 ñ feet per second and initial position r( 0) = á0,0,6ñ  feet, then what is the maximum height of the rock if air resistance is ignored?
    Solution: Since a( t) = á0,0,-32 ñ , we must have v( t) = òa( t) dt = á 0,0,-32t ñ+ á C1,C2,C3 ñ . Moreover, v(0) = á 16,0,64 ñ = á 0,0,0ñ+ á C1,C2,C3 ñ , so that v(t) = á 16,0,64-32t ñ . Similarly, r( t) = òv( t) dt and the initial condition lead to
    r( t) = á 16t,0,64t-16t2+6 ñ
    As a result, v·a = 0 when 64-32t = 0, or when t = 2. Thus, the maximum height is
    64·2-16·22+6 = 70  feet
  10. The acceleration due to gravity is 12.2 feet per second per second at the surface of Mars. Find the position function r(t) of an object with initial velocity v0 = á30,0,40 ñ and initial position r0 = á0,0,0 ñ .
    Solution: v( t) = ò á0,0,-12.2 ñ dt = á 0,0,-12.2t ñ +v0 = á 0,0,-12.2t ñ + á 30,0,40 ñ =   á 30,0,40-12.2t ñ . Integrating again yields
    r( t)
    =
    ó
    õ     		v( t) dt
    =
    ó
    õ     		á 30,0,40-12.2t ñ dt
    =
    á 30t,0,40t-6.1t2 ñ +r0
    =
    á 30t,0,40t-6.1t2 ñ
  11. Find the arclength and the unit tangent vector of the curve
    r( t) = á 3sin( t) ,5cos(t) ,4sin( t) ñ ,  t  in  [ 0,2p]
    Solution: v( t) = á 3cos(t) ,-5sin( t) ,4cos( t) ñ , so that the speed is
    v
    =
    9cos2( t) +25sin2( t) +16cos2( t)
    =
    25cos2( t) +25sin2( t)
    =
    5
    Thus, T(t) = á 3/5 cos(t), -sin(t) , 4/5 cos(t) ñ and the arclength is
    L = ó
    õ     		2p

    0 
    		 vdt = ó
    õ     		2p

    0 
    		 5dt = 10p
  12. Find the unit normal N for the curve
    r( t) = á sin( t3) ,t3,cos( t3) ñ
    Solution: To begin with, v( t) =   á3t2cos( t3) ,3t2,3t2sin( t3) ñ , so that the speed is
    v =
    Ö
     
    9t4cos2( t3) +9t4+9t4sin2(t3)
     
    		 = 3Ö2t2
    Thus, the unit tangent vector is T( t) = ácos( t3) /Ö2, 1/Ö2, sin( t3) /Ö2 ñ and
     dT
    dt
    		 =
    		 -3t2
    Ö2
    		 sin(t3) ,0,  -3t2
    Ö2
    		 cos( t3)
    from which we find that the normal vector is N( t) = á -sin( t3) ,0,-cos( t3) ñ .
  13. Find the arclength of the curve
    r( t) = á e2t,t,2et ñ ,   t  in  [ 0,1]
    Solution: v( t) = á2e2t,1,2et ñ , so v = ( 4e4t+4e2t+1)1/2 = ( [ 2e2t+1] 2) 1/2 = 2e2t+1. Thus,
    L = ó
    õ     		1

    0 
    		 vdt = ó
    õ     		1

    0 
    		 ( 2e2t+1) dt = e2
  14. Find the length of the astroid
    r( t) = á cos3( t) ,sin3( t) ñ ,    t  in  [ 0,2p]
    Solution: The velocity is v = á -3cos2( t) sin( t) ,3sin2( t) cos( t) ñ , so that the speed is
    v = 
    9cos4( t) sin2( t) +9sin4(t) cos2( t)
    		 =  
    9cos2( t) sin2( t) ( cos2( t) +sin2(t) )
    Thus, the speed is v = 3| cos( t) sin( t)| , so that the arclength is
    L = ó
    õ     		2p

    0 
    		 vdt = 4 ó
    õ     		p/2

    0 
    		 3sin( t) cos(t) dt = 6
  15. Find the curvature of the curve
    r( t) = á sin( t) ,cos(t) ,ln| sec( t) | ñ
    Solution: The velocity is v( t) = ácos( t) ,-sin( t) ,tan( t) ñ , which implies that the speed is
    v =  
    cos2( t) +sin2( t) +tan2(t)
    		 = 
    1+tan2( t)
    		 =  
    sec2( t)
    Thus, v = sec( t) , which implies that the unit tangent vector is
    T( t)
    =
    cos( t) á cos(t) ,-sin( t) ,tan( t) ñ
    =
    á cos2( t) ,-sin( t) cos(t) ,sin( t) ñ
    It then follows that the derivative of the unit tangent is
     dT
    dt
    		 = á 2cos( t) sin(t) ,sin2( t) -cos2( t) ,cos(t) ñ = á sin( 2t) ,-cos(2t) ,cos( t) ñ
    As a result, the curvature is
    k =  1
    sec( t)
    sin2( 2t) +cos2( 2t) +cos2( t)
    		 = cos(t)
    1+cos2( t)