MATH 241 – Multivariable Calculus Sample Exam Questions for Exam 1 – Solutions September 23, 2013

I. Consider the set Q = {(x, y, z) ∈ R^{3} | x^{2} − y^{2}/4 − z^{2} = 1}.

A) Identify the slices of Q in planes parallel to the three coordinate planes. Take x, y, z = −2, −1, 0, 1, 2 (at least). Solution: The slices by the planes x = c have equations y^{2}/4 + z^{2} = c^{2} − 1. These are ellipses if c > 1 or c < −1, single points if c = ±1 and empty if −1 < c < 1. The slices by planes y = c have equations x^{2} − z^{2} = 1 + c^{2}/4. These are hyperbolas for all c in the planes parallel to the xz-plane). The slices by planes z = c have equations x^{2} − y^{2}/4 = 1 + c^{2}. These are also hyperbolas.

B) Use that information to generate a rough sketch of Q or a verbal description. Solution: See the Maple plot posted with these solutions. A verbal description would go something like this: The surface consists of two sheets: one with x ≥ 1 and the other with x ≤ −1. There are no points on the surface with −1 < x < 1. The two sheets both have elliptical cross-sections by planes x = c. This is called a hyperboloid of two sheets because the sections by planes y = c and z = c are both hyperbolas.

C) Show that the image of the parametric curve α(t) = ((e^{t} + e^{−t})/2, 0, (e^{t} − e^{−t})/2) for t in R lies entirely in Q. Solution: Substituting x = (e^{t} + e^{−t})/2, y = 0, and z = (e^{t} − e^{−t})/2 into the equation of Q we have

x^{2} − y^{2}/4 − z^{2} = ((e^{t} + e^{−t})/2)^{2} − 0 − ((e^{t} − e^{−t})/2)^{2}
= (e^{2t} + 2 + e^{−2t})/4 − (e^{2t} − 2 + e^{−2t})/4
= 1.

Therefore all points on the image of α satisfy the equation of Q, so the curve lies entirely in the surface.

II.

A) Find the equation of the plane in R^{3} containing the point P = (1, 1, 1) and the line through Q = (0, 1, −1) with direction vector v = (−4, 1, 2). Solution: The plane contains the vector (−4, 1, 2) and P − Q = (1, 0, 2). Hence a normal vector can be found by computing the cross product

N = (−4, 1, 2) × (1, 0, 2) = (2, 10, −1)

The plane has equation

N · (x − 1, y − 1, z − 1) = 2x + 10y − z − 11 = 0.

B) The points P, Q, Q + v from part A are three corners of a parallelogram in the plane you found. Find the fourth corner and the area of that parallelogram. Solution: The fourth corner is P + v = (−3, 2, 3). The area of the parallelogram is

∥(−4, 1, 2) × (1, 0, 2)∥ = ∥(2, 10, −1)∥ = \sqrt{4 + 100 + 1} = \sqrt{105}.

C) 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 θ satisfies:

cos θ = \frac{v · w}{∥v∥∥w∥} = \frac{7}{\sqrt{5}\sqrt{29}}.

Hence θ = cos^{−1} (\frac{7}{\sqrt{5}\sqrt{29}}) ≐ 1.31 radians.

D) Show that if u = (u_{1}, u_{2}, u_{3}), v = (v_{1}, v_{2}, v_{3}), and w = (w_{1}, w_{2}, w_{3}) are general vectors in space, then u × (v + w) = u × v + u × w. Solution: We have, after applying the distributive law and splitting up the vector into a sum of two other vectors:

u × (v + w) = (u_{1}, u_{2}, u_{3}) × (v_{1} + w_{1}, v_{2} + w_{2}, v_{3} + w_{3})
= (u_{2}(v_{3} + w_{3}) − u_{3}(v_{2} + w_{2}), −(u_{1}(v_{3} + w_{3}) − u_{3}(v_{1} + w_{1})),
u_{1}(v_{2} + w_{2}) − u_{2}(v_{1} + w_{1}))
= (u_{2}v_{3} − u_{3}v_{2} + u_{2}w_{3} − u_{3}w_{2}, −(u_{1}v_{3} − u_{3}v_{1}) − (u_{1}w_{3} − u_{3}w_{1}),
u_{1}v_{2} − u_{2}v_{1} + u_{1}w_{2} − u_{2}w_{1})
= (u_{2}v_{3} − u_{3}v_{2}, −(u_{1}v_{3} − u_{3}v_{1}), u_{1}v_{2} − u_{2}v_{1})+
(u_{2}w_{3} − u_{3}w_{2}, −(u_{1}w_{3} − u_{3}w_{1}), u_{1}w_{2} − u_{2}w_{1})
= u × v + u × w.

III. A) Sketch the parametric curve α(t) = (4 cos(πt), 3 sin(πt)) for 0 ≤ t ≤ 1. Show the

starting and final points and indicate the direction the curve is traced with an arrow. Solution: This is the top half of the ellipse \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1, starting at α(0) = (4, 0) and ending at α(1) = (−4, 0). The curve is traced in the counter-clockwise direction.

B) Give a parametrization of the circle with center (1, 3) and radius r = 4 so that one full circuit of the circle is traced out clockwise for 0 ≤ t ≤ 4π. Solution: β(t) = (4 cos(t/2) + 1, −4 sin(t/2) + 3) is one such parametrization. starting from β(0) = (5, 3).) Another curve satisfying the given conditions is γ(t) = (4 sin(t/2) + 1, 4 cos(t/2) + 3). This one starts from γ(0) = (1, 7).

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

α(t) = ((1 + cos(t)) sin(t), −(1 + cos(t)) cos(t)),

called a cardioid (see the Maple plot).

A) Show that for all t, the distance from the origin to the point α(t) is given by 1+cos(t). Solution: The distance is the same as the magnitude ∥α(t)∥. Computing this, by means of the basic trig identity, we find:

∥α(t)∥ = \sqrt{(1 + cos(t))^{2} sin^{2}(t) + (1 + cos(t))^{2} cos^{2}(t)}
= \sqrt{(1 + cos(t))^{2}(sin^{2}(t) + cos^{2}(t))}
= \sqrt{(1 + cos(t))^{2}}
= |1 + cos(t)|
= 1 + cos(t),

since cos(t) ≥ −1 for all t, so 1 + cos(t) ≥ 0.

B) At how many different times t in [0, 2π] is the y-coordinate of α(t) equal to zero? Find them. Solution: The y-coordinate is −(1+cos(t)) cos(t). This is only = 0 if either cos(t) = −1 or cos(t) = 0. There are three such t values in the given interval: t = π/2, π, 3π/2.

C) Find a parametrization of the tangent line to the cardioid at the point α(π/4). Solution: The point on the cardioid is

α(π/4) = ((1 + \sqrt{2}/2)\sqrt{2}/2, −(1 + \sqrt{2}/2)\sqrt{2}/2) = (\frac{1 + \sqrt{2}}{2}, −\frac{1 + \sqrt{2}}{2}) .

The tangent vector at a general α(t) is

α′(t) = (− sin^{2}(t) + cos^{2}(t) + cos(t), 2 sin(t) cos(t) + sin(t)).

So α′(π/4) = (\sqrt{2}/2, 1 + \sqrt{2}/2). The tangent line is

λ(u) = α(π/4) + uα′(π/4) = (\frac{1 + \sqrt{2}}{2}, −\frac{1 + \sqrt{2}}{2}) + u (\frac{\sqrt{2}}{2}, 1 + \frac{\sqrt{2}}{2}) .

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, every point on the cardioid is distance at most 2 from the origin. So if we take any radius r > 2 and center (0, 0), the cardioid will be contained in the interior region of the circle: δ(t) = (3 cos(t), 3 sin(t)) gives one such circle if we take t ∈ [0, 2π].