Mathematics 41, section 1 -- Multivariable Calculus

Solutions for Sample Exam 1

September 22, 1999

Sample Exam

I. Consider the set Q = {(x,y,z) : x² - y²/4 + z² = 1}. Identify the slices of Q in planes parallel to the three coordinate planes, and use that information to generate a rough sketch of Q.

Solution: The slice in the plane z = c is given by x² - y²/4 = 1 - c², a hyperbola if c <> 1, -1, and a union of two lines y = 2x, y = -2x, z = c if c = 1, -1. The slice in the plane y = b is given by x² + z² = 1 + b²/4, a circle with radius r = sqrt(1 + b²/4). The slice in the plane x = a is given by - y²/4 + z² = 1 - a², a hyperbola if a <> +/- 1, and a union of two lines y = 2z, y = -2z, x = a if a = -1, 1. Here is a Maple plot of this surface (a hyperboloid of one sheet):

II.

• A) Find the equation of the plane in R³ containing the point P = (1,1,1) and the line through Q = (0,1,-1) with direction vector v = (-4,1,2).

  Solution: Let v = (-4,1,2) -- the direction vector of the line -- and u = P - Q = (1,0,2). Then N = v x u = (2,10,-1) is a normal vector to the plane and the equation is 0 = N . ((x,y,z) - (1,1,1)) = (2,10,-1) . (x-1,y-1,z-1) = 2x + 10y - z - 11, or 2x + 10y - z = 11. (Note: In the review session, I think I used a different point Q by accident, so the result was different. The method is exactly the same, though!)

• B) Two lines with direction vectors v = (0,1,2) and w = (4,3,2) meet at (0,0,0). Find the acute angle between the two lines.

  Solution: The angle between the lines is the same as the angle between their direction vectors: theta = arccos((0,1,2).(4,3,2)/(sqrt(5)sqrt(29)) = arccos(7/sqrt(145)) which is about 58 degrees (or 1.01 radians).

• C) Show that if u = (u₁,u₂,u₃), v = (v₁,v₂,v₃), and w = (w₁,w₂,w₃) are general vectors in space, then u x (v + w) = u x v + u x w.

  Solution: Starting on the left-hand side,
  u x (v + w) = (u₁,u₂,u₃) x (v₁+w₁,v₂+w₂,v₃+w₃)
  = (u₂(v₃+w₃) - u₃(v₂+w₂), -(u₁(v₃+w₃) - u₃(v₁+w₁)), u₁(v₂+w₂) - u₂(v₁+w₁)).
  Expanding in each component using the distributive law, then rearranging, we get:
  = ((u₂v₃-u₃v₂)+ (u₂w₃-u₃w₂), -((u₁v₃-u₃v₁)+ (u₁w₃-u₃w₁)), (u₁v₂-u₂v₁)+ (u₁w₂-u₂w₁))
  = u x v + u x w.

III. All parts of this problem refer to the parametric curve

• A) Show that for all t, the distance from the origin to the point alpha(t) is given by 1 + cos(t).

  Solution: The distance is the magnitude of the vector alpha(t):
  ||alpha(t)|| = sqrt[(1+cos(t))²sin²(t) + (1+cos(t))²cos²(t))]
  = sqrt[(1+cos(t))²(sin²(t) + cos²(t))]
  = sqrt[(1+cos(t))²]
  = 1 + cos(t),
  (since 1 + cos(t) >= 0 for all t.)

• B) At how many different times t in [0,2 pi] is the y-coordinate of alpha(t) equal to zero? Find them.

  Solution: Three different t's: t = Pi/2, Pi, 3Pi/2. (The first and third are the roots of the cos(t) factor; the second is the root of the 1 + cos(t) factor.)

• C) Find a parametrization of the tangent line to the cardioid at the point alpha(pi/4).

  Solution: The point alpha(Pi/4) = ( (1 + sqrt(2)/2)sqrt(2)/2,-(1 + sqrt(2)/2)sqrt(2)/2 ) = (sqrt(2)/2 + 1/2, -sqrt(2)/2 - 1/2. The direction vector for the tangent line is alpha'(Pi/4). By the product rule
  alpha'(t) = ( (1+cos(t))cos(t) - sin²(t), (1+cos(t))sin(t) + sin(t)cos(t) )
  So,
  alpha'(Pi/4) = ( (1 + sqrt(2)/2)sqrt(2)/2 - 1/2, (1 + sqrt(2)/2)sqrt(2)/2 + 1/2 )
  = (sqrt(2)/2, sqrt(2)/2 + 1/2)
  The parametrization of the tangent line is
  alpha(Pi/4) + t alpha'(Pi/4) = ( sqrt(2)/2 + 1/2, -sqrt(2)/2 - 1/2 ) + t ( sqrt(2)/2, sqrt(2)/2 + 1/2 )
  for t in R

• D) Give a parametrization for any one circle whose interior region completely contains the cardioid. Explain how you determined your center and radius.

  Solution: By part A, the distance from (0,0) to alpha(t) is 1 + cos(t), which is between 0 and 2 for all t. If we take any radius > 2, then the circle will contain the entire curve. For example, (2.1 cos(t), 2.1 sin(t)), t in [0, 2 Pi] works.