Mathematics 241, section 1 – Multivariable Calculus Solutions for Exam 1 October 1, 2010

I. All parts of this question refer to the three points P = (0, 0, 1), Q = (1, 2, −4), and R = (1, −1, 0) in R^{3}.

A) (15) Find the equation of the plane containing P, Q, R.

Solution: The vector from P to Q is Q − P = (1, 2, −5) and the vector from P to R is R − P = (1, −1, −1). Hence N = (Q − P) × (R − P) = (−7, −4, −3) is a normal vector for the plane we want. Using N and the point P, equation is (−7, −4, −3) · (x − 0, y − 0, z − 1) = 0, or

7x + 4y + 3z = 3.

B) (10) Give a parametrization of the line segment from R to Q (in that direction), including the proper range of t-values.

Solution: The direction vector we want is Q − R = (0, 3, −4). The line segment is

(1, −1, 0) + t(0, 3, −4) = (1, −1 + 3t, −4t) where 0 ≤ t ≤ 1.

C) (10) Which t-value gives the midpoint of the line segment from part B (the point equidistant from R and Q) in your parametrization?

Solution: Since the line segment is traversed at constant speed in this parametrization between t = 0 and t = 1, the midpoint will be reached at t = \frac{1}{2}. The midpoint is M = (1, \frac{1}{2}, −2). It can be checked that ∥M − Q∥ = ∥M − R∥ = \frac{5}{2}.

D) (10) Compute the angle between the vectors P⃗ Q and P⃗ R.

Solution: Q − P = (1, 2, −5) and R − P = (1, −1, −1). The angle satisfies

cos(θ) = \frac{(1, 2, −5) · (1, −1, −1)}{\sqrt{30}\sqrt{3}} = \frac{4}{3\sqrt{10}} = \frac{4\sqrt{10}}{30}.

The angle is

θ = cos^{−1} (\frac{4\sqrt{10}}{30}) .

II. Let u, v, w be vectors in R^{3} with tails at the origin. A) (10) Show that

w × (u + v) = w × u + w × v.

Solution: Let u = (u_{1}, u_{2}, u_{3}), v = (v_{1}, v_{2}, v_{3}), and w = (w_{1}, w_{2}, w_{3}). Then

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

B) (5) Show that if w is in the plane spanned by u and v, then (u × v) · w = 0.

Solution: If w is in the plane spanned by u and v, then there are scalars s, t such that w = su + tv. Then by another vector identity,

(u × v) · w = (u × v) · (su + tv) = s(u × v) · u + t(u × v) · v.

Since u × v is orthogonal to both u and v, both dot products here are zero. Hence (u × v) · w = 0.

III. All parts of this question refer to

α(t) = (cos(3t) cos(t), cos(3t) sin(t))

called a 3-leaved rose curve. A) (15) Starting from t = 0, what is the first t with α(t) = (0, 0)? At how many different

t is α(t) = (0, 0) in the range 0 ≤ t < π?

Solution: Starting from t = 0, the first t with α(t) = (0, 0) is t = π/6. There are three t in the range 0 ≤ t < π where α(t) = (0, 0): t = π/6, π/2, 5π/6. These all come from zeroes of the function cos(3t).

B) (15) Find a parametrization of the tangent line to the rose curve at t = π/3.

Solution: We have

α(π/3) = (cos(π) cos(π/3), cos(π) sin(π/3)) = (−\frac{1}{2}, −\frac{\sqrt{3}}{2}) .

Then by the product rule in each component,

α′(t) = (−3 sin(3t) cos(t) − cos(3t) sin(t), −3 sin(3t) sin(t) + cos(3t) cos(t)).

Hence

α′(π/3) = (0 + \frac{\sqrt{3}}{2}, 0 − \frac{1}{2}) = (\frac{\sqrt{3}}{2}, −\frac{1}{2}) .

The tangent line is

(−\frac{1}{2}, −\frac{\sqrt{3}}{2}) + s (\frac{\sqrt{3}}{2}, −\frac{1}{2}) , s ∈ R.

C) (10) Give a parametrization of a circle surrounding the “leaf” of α(t) in quadrants I and IV, but not completely enclosing either of the other leaves. (Any circle that does that is OK.)

Solution: If we place the center at (1, 0), then to enclose the “leaf” in quadrants I and IV, but not the other two “leaves,” we can take a radius greater than or equal to 1, but strictly less than the distance from (1, 0) to the point α(π/3), which is \sqrt{3}. Note that r = 1 is in that range. Something like

β(t) = (1 + cos(t), sin(t))

is a reasonable answer.

Extra Credit (10) Let β(t) be a parametric curve in R^{3}. Suppose that β(t) · β(t) = 1 (dot product) for all t. Show that β′(t) is orthogonal to β(t) for all t.

Solution: By the product formula for derivatives, it follows that

(β(t) · γ(t))′ = β′(t)γ(t) + β(t) · γ′(t)

for any vector valued functions β and γ. If γ = β and β(t) · β(t) is constant, then this shows

2β(t) · β′(t) = 0

for all t. This shows that β(t) and β′(t) are orthogonal for all t.