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 = (1, 0, 1), Q = (1, 2, −2), and R = (1, −1, 1) 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 = (0, 2, −3) and the vector from P to R is R − P = (0, −1, 0). Hence N = (Q − P) × (R − P) = (−3, 0, 0) is a normal vector for the plane we want. Using \frac{−1}{3}N and the point P, equation is

x = 1.

(Note that this could also be derived by observation without any calculations!) B) (15) Give a parametrization of the line segment from R to Q (in that direction),

including the proper range of t-values. Which t-value gives the midpoint?

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

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

The midpoint is obtained with t = 1/2. (The line segment is parametrized with a constant speed curve, so over the interval [0, 1], the midpoint is obtained for t = 1/2.)

C) (10) Compute the angle between the vectors v = Q − P and w = R − P.

Solution: Q − P = (0, 2, −3) and R − P = (0, −1, 0) from part A. The angle satisfies

cos(θ) = \frac{(0, 2, −3) · (0, −1, 0)}{\sqrt{13} · 1} = \frac{−2}{\sqrt{13}}.

The angle is

θ = cos^{−1} (\frac{−2}{\sqrt{13}}) ≐ 2.16 radians.

II. II. Let α(t) = (u(t), v(t)) and β(t) = (w(t), z(t)) be parametric curves in R^{2}.

A) (10) Show that if α(t) and β(t) are differentiable, then

\frac{d}{dt}(α(t) · β(t)) = α′(t) · β(t) + α(t) · β′(t),

(where · denotes the dot product). Solution: We have

α(t) · β(t) = u(t)w(t) + v(t)z(t).

So by the product and sum rules for derivatives:

\frac{d}{dt}(α(t) · β(t)) = u′(t)w(t) + u(t)w′(t) + v′(t)z(t) + v(t)z′(t)
= (u′(t)w(t) + v′(t)z(t)) + (u(t)w′(t) + v(t)z′(t))
= α′(t) · β(t) + α(t) · β(t).

This is what we wanted to show. B) (10) Let β(t) be a parametric curve in R^{2}. Suppose that β(t) · β(t) = 1 (dot product)

for all t. Show that the vector β′(t) is orthogonal to the vector β(t) for all t. Solution: Using part A and the commutativity of the dot product, we see

0 = \frac{d}{dt}(β(t) · β(t)) = 2β(t) · β′(t).

Therefore β(t) · β′(t) = 0, which shows that they are orthogonal.

III. All parts of this question refer to

α(t) = (cos(2t) cos(t), cos(2t) sin(t))

called a 4-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/4. There are two t in the range 0 ≤ t < π where α(t) = (0, 0): t = π/4, 3π/4. These all come from zeroes of the function cos(2t).

B) (15) Find a parametrization of the tangent line to the rose curve at t = π/4.

Solution: We have

α(π/4) = (cos(π/2) cos(π/4), cos(π/2) sin(π/4)) = (0, 0).

Then by the product rule in each component,

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

Hence

α′(π/4) = (−\sqrt{2}, −\sqrt{2}).

The tangent line is

\{(−s\sqrt{2}, −s\sqrt{2}) | s ∈ R\}.

C) (10)Give a parametrization of a circle that will fit completely inside the “leaf” of α(t) in quadrants I and IV. (Any circle that does that is OK.)

Solution: If we place the center at (1/2, 0), then to fit inside the “leaf” in quadrants I and IV, the radius must be small, something like 1/10. Something like

β(t) = (1/2 + cos(t)/10, sin(t)/10)

is a reasonable answer.

Extra Credit (10) Identify and sketch the slices of the surface

{(x, y, z) ∈ R^{3} | x^{2} − y^{2} + z^{2} = 1}

in the planes x = 1 and x = 2.

Solution: The slice in the plane x = 1 is given by −y^{2} + z^{2} = 0. This is the union of two lines: x = 1, y = z, and x = 1, y = −z. The slice in the plane x = 2 is given by −y^{2} + z^{2} = −3, or \frac{y^{2}}{3} − \frac{z^{2}}{3} = 1. This is a hyperbola opening in the direction of the positive and negative y-axes.