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1) A parallelogram has (ordered) vertices at the points A, B, C, and D.A = (2, −1, −5) B = (1, 0, −1) C = (1, 2, 3)

(a) What are the coordinates of D?

(b) What is the interior angle at B?

(c) What is the area of the parallelogram?

(d) Find an equation of the plane containing the parallelogram.

2) Suppose that the point P = (1, −1, 2) lies on the surface S. Also, suppose that the curves

~r(t) =

1 − 2t, −1 + t, 2 − t + t

2

, ~s(t) =

t

2

, 1 − 2t

2

, 3t − 1

both lie on S. Find an equation of the tangent plane to S at P.2

3) Consider the upper half-cone z = f(x, y) = p

x

2 + y

2

. We will calculate the tangent plane through

(3, 4, 5).

(a) Let ~u =

1, 0, fx(3, 4)

, ~v =

0, 1, fy(3, 4). [Here fx denotes the partial derivative ∂z∂x .] Find thetangent plane P through3, 4, f(3, 4).

(b) Use the tangent plane to estimate f(2.9, 4.2).

4) Let ~u = h1, 5, 6i, ~v = h9, −3, 4i, and ~w = h2, 7, 7i. Evaluate

a) ~u · ~w

b) ~u × ~w

c) ~v × (~u + ~w)

5) Find the line and angle of intersection of the two planes x − 3y − 4z = 8 and 4x + 2y = −7.

6) The voltage V across a circuit is given by Ohm’s law

V = IR

where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance is given by

1R=1R1+1R2

Suppose the current is 4 amps and increases at 10−2 amps/sec, R1 is 5 ohms and decreases at 0.5 ohm/sec, and R2 is 3 ohms and increases at 0.1 ohm/sec. Calculate the rate at which the voltage is changing.

7) Below is a contour map of Lawrence. The red dot is Snow Hall (SH) and the green dot is Java Break

(JB), the best coffee shop in town; the legend is scaled in feet.

(a) This map shows in two ways that the University of Kansas is on a large hill. Explain how.

(b) What is the change in elevation between SH and JB?

(c) Find the distance as the crow flies between SH and JB and show there is approximately an average grade of 2% on the walk between them.

8) One of the most powerful inequalities in mathematics is the Cauchy-Bunyakovsky-Schwarz (CBS)inequality. It states: if ~u, ~v ∈ Rn, then|~u · ~v| ≤ ||~u|| ||~v|| Further, equality is achieved only when ~u and ~v are parallel, i.e. when ~u = c~v for some constant c. Here are several problems that explore the power and scope of CBS.

a) Suppose you want to maximize f(x, y, z) = 8x + 4y + z subject to x

2 + y

2 + z

2 = 1. One way is to use Lagrange multipliers; another is to use CBS. Identify two choices for ~u and ~v and use CBS to bound f2.

b) This problem is adapted from the famous Putnam competition: Show there is no sequence (an) such that for every positive integer m a

m

1 + a

m

2 + a

m

3 + · · · =

X∞

k=1

a

m

k = m

Hint: define a new sequence (bn) = (a2n) and use CBS9) Let f(x, y) = 4 + x2 + y − 3xy − y25, for −3 ≤ x ≤ 3 and −3 ≤ y ≤ 3.

(a) Sketch a contour plot f(x, y) = c for c = −4, . . . , 4 [all nine integer contours].

(b) Find a unit vector ~u so that the directional derivative of f along ~u at (1, 0) is zero.

10) The study of differential equations is a fundamental branch of mathematics. They pose interesting mathematical problems and are often used to model physical phenomena. A partial differential equation (or PDE) is a differential equation involving functions of multiple variables and their partial derivatives.

(a) Verify that u(x, y, z) = e

3x+4y

cos(5z) satisfies the PDE uxx + uyy + uzz = 0.

(b) Let g be a differentiable function. Assuming y > 0, verify that u(x, y) = (−x + y)gpy2 − x

2

x + y

satisfies the PDE

yux + xuy = −2u, u(0, y) = g(y).

11) Find the absolute maximum and absolute minimum values of f(x, y) = −56y + 36xy − 4x2y − 2y2 +3xy2 − x2y2 on the closed triangular region whose vertices are (0, 0), (0, 28), and (7, 0).10Bonus) In mathematics, sometimes the best way to analyze an object is as a special case of a moregeneral object. For instance, consider the two-variable real polynomial

p(x, y) = x

3 − 3x

2

y

2 + xy4 − y + 4

p has degree 5; however, not all terms have degree 5. Suppose we homogenized p by introducing a new variable z and defined a polynomial P(x, y, z) such that:

• All terms of P have degree 5

• P(x, y, 1) = p(x, y)

(a) Determine P(x, y, z).

(b) Verify that if λ is a constant, P(λx, λy, λz) = λ

5P(x, y, z) but p(λx, λy) 6= λ

5p(x, y).

(c) Verify that ∇P · (x, y, z) = 5P.

(d) In the projective plane, two points (a : b : c), (a0: b0: c0) are homothetic if there is some nonzero λsuch that (a : b : c) = (λa0: λb0: λc0). Group the following triples into different groups such that each all the points in each group are homothetic and different groups represent different points.

{(0 : 0 : 1), (1 : 0 : 1), (0 : 0 : π), (2/3 : 0 : 2/3), (2 : −3 : 4)(4 : 4 : 0), (5 : 5 : 5), (−π : −π : 0), (4 : −6 : 8), (π : 0 : π)}11Bonus) It turns out that showing a function of multiple variables f(x1, x2, . . . , xn) is differentiable is somewhat difficult. In practice, it is often easier to show a stronger condition: if each partial derivative ∂f∂xi , i = 1, . . . , n, is continuous in a disc around p = (a1. . . . , an), then f is differentiable at p = (a1, . . . , an). Put differently: if f is continuously differentiable at p, it is differentiable at p. However, just as in the one-variable case, there are functions that are differentiable but not continuously differentiable.

Consider the function f : R

2 → R

f(x, y) = (

yx2

sin(x

−1

), x 6= 0

0, x = 0

(a) Show that if y 6= 0, fx is not continuous at x = 0, in that lima→0 fx(a, y) doesn’t exist.

(b) Nevertheless, show that f is differentiable when x = 0 (this is clearly the only problematic case),with f0(0, y) = 0. To be precise, show thatlim

(h,k)→(0,0)

f(h, y + k) − f(0, y) − 0

√

h

2 + k

2

2

= 0