Mathematics 241 – Multivariable Calculus Solutions for Final Examination – December 14, 2013

I. In this problem, P = (1, 0, 1), Q = (−2, 3, 2), and R = (1, 2, 0).

A) Find the equation of the plane containing the points P, Q, R in R^{3}.

Solution: The displacement vector from P to Q is v = Q − P = (−3, 3, 1) and the vector from P to R is w = R − P = (0, 2, −1). For the plane we can take N = (−3, 3, 1) × (0, 2, −1) = (−5, −3, −6). Then the equation of the plane is 0 = N · (x − 1, y − 0, z − 1) = −5x + 5 − 3y − 6z + 6, or 5x + 3y + 6z = 11.

B) At what point does the line containing P, Q meet the xy-plane?

Solution: The line is (1, 0, 1)+(−3, 3, 1)t = (1−3t, 3t, 1+t). This meets the xy-plane when z = 1 + t = 0, so t = −1. The point of intersection is (4, −3, 0).

C) If v is the displacement vector from P to Q and w is the displacement vector from P to R, at what angle do v, w meet?

Solution: The angle θ satisfies cos θ = \frac{v·w}{∥v∥∥w∥} = \frac{5}{\sqrt{19}\sqrt{5}}. So

θ = cos^{−1} (\sqrt{5/19}) ≐ 1.032 radians.

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

α(t) = (\frac{cos(t)}{sin^{2}(t) + 1}, \frac{cos(t) sin(t)}{1 + sin^{2}(t)})

defined for all t ∈ [0, 2π], called a lemniscate. A) Is α(t) a simple closed curve? (Hint: Thinking of α(t) as the position of a moving

object as a function of time, are there different times t ∈ [0, 2π) at which the object is at the location (x, y) = (0, 0)?)

Solution: We have α(0) = (\frac{1}{2}, 0) = α(2π), so this is a closed curve. However, cos(t) = 0 = cos(t) sin(t) for t = \frac{π}{2} and t = \frac{3π}{2} in the given interval. Since there are two such t, the answer to the first question is NO.

B) What is the tangent vector to the curve at t = π?

Solution: By the quotient rule in each component,

α′(t) = \frac{1}{(1 + sin^{2}(t))^{2}} (−(1 + sin^{2}(t)) sin(t) − 2 cos^{2}(t) sin(t),
(1 + sin^{2}(t))(cos^{2}(t) − sin^{2}(t)) − 2 cos^{2}(t) sin^{2}(t))

When t − π, we get α′(π) = (0, 1) C) A thin wire has the shape of the portion of the curve α for t ∈ [0, 1]. What integral

would you use to compute its arc length. (You do not need to evaluate!)

Solution: The arc length would be computed by

M = ∫_{0}^{1} ds = ∫_{0}^{1} ∥α′(t)∥dt

III. All parts of this problem refer to f(x, y) = (x + 1)^{2} + y^{2}.

A) Sketch the level curves of f(x, y) for the values c = 1, 4, 9.

Solution: The level curves of f are circles with center at (−1, 0) the radii are r = 1, 2, 3 respectively.

B) At the point (1, 2), in which direction is f increasing the fastest? Express your answer as a unit direction vector.

Solution: This is in the direction of the gradient vector ∇f(1, 2). The gradient vector is ∇f(x, y) = (2(x + 1), 2y) at a general point. So ∇f(1, 2) = (4, 4). The unit vector in this direction is \frac{1}{4\sqrt{2}}(4, 4) = (\sqrt{2}/2, \sqrt{2}/2).

C) Find the points on the curve g(x, y) = \frac{x^{2}}{4} + y^{2} = 1 at which f(x, y) takes its largest and smallest values. What is true about the vectors ∇f and ∇g at your points?

Solution: Using the Lagrange multiplier method, we must solve

\begin{aligned}2(x + 1) & = λx/2 \\ 2y & = 2λy \\ \frac{x^{2}}{4} + y^{2} & = 1 \end{aligned}

From the second equation, y = 0 or λ = 1. If y = 0, the constraint equation gives x = ±2, so we obtain two points (±2, 0). If λ = 1, then from the first equation, 2(x + 1) = x/2, so x = −4/3. Then from the constraint equation we get y = ±\sqrt{5}/3. To determine which of these give maximum and minimum values, we substitute into f(x, y):

f(2, 0) = 9 (maximum)
f(−2, 0) = 1
f(−4/3, ±\sqrt{5}/3) = 1/9 + 5/9 = 2/3 (minimum)

The points we found here are the points where the level curve of f passing through that point and the constraint curve are tangent.

IV. Let f(x, y) = xe^{−2x^{2}−y^{2}}.

A) Find the equation of the tangent plane to the graph z = f(x, y) at the point (1, 1, e^{−3}).

Solution: We must compute the partial derivatives to start:

\begin{aligned}f_{x} & = (1 − 4x^{2})e^{−2x^{2}−y^{2}} \\ f_{y} & = −2xye^{−2x^{2}−y^{2}}. \end{aligned}

At (x, y) = (1, 1), f_{x}(1, 1) = −3e^{−3}, and f_{y}(1, 1) = −2e^{−3}, so the tangent plane is

z = e^{−3} − 3e^{−3}(x − 1) − 2e^{−3}(y − 1).

B) Find all the critical points of f(x, y).

Solution: The critical points are the solutions of f_{x} = 0 and f_{y} = 0. Using the formulas for f_{x}, f_{y} from part A, we see that f_{x} = 0 when x = ±1/2 and f_{y} = 0 when x = 0 or y = 0 (Note: the exponential factor is never zero.) Hence the simultaneous solutions are the two points (±1/2, 0).

C) Use the second derivative test (Hessian criterion) to determine the type of each critical point you found in part B.

Solution: Now we need the second-order partial derivatives as well:

\begin{aligned}f_{xx} & = (16x^{3} − 12x)e^{−2x^{2}−y^{2}} \\ f_{xy} & = (1 − 4x^{2})(−2y)e^{−2x^{2}−y^{2}} \\ f_{yy} & = −2x(1 − 2y^{2})e^{−2x^{2}−y^{2}} \end{aligned}

So at (1/2, 0) the Hessian matrix is

( \begin{matrix}−4e^{−1/2} & 0 \\ 0 & −e^{−1/2} \end{matrix} )

The determinant is 4e^{−1} > 0 and the upper left entry is < 0 so this is a local maximum. At (−1/2, 0) the Hessian matrix is

( \begin{matrix}4e^{−1/2} & 0 \\ 0 & e^{−1/2} \end{matrix} )

The determinant is 4e^{−1} > 0 and the upper left entry is > 0 so this is a local minimum.

V. A region R in R^{2} is the set of points satisfying x^{2} + y^{2} ≥ 1, y ≥ x, x ≥ 0, and y ≤ 4.

A) Sketch the region R.

Solution: This is the region outside the unit circle with center (0, 0), to the right of the y-axis, below the horizontal line y = 4, and above the line y = x.

B) Set up the limits of integration of iterated integral(s) to compute ∫ ∫_{R} f(x, y) dA integrating with respect to x first, then y.

Solution: The circle intersects the line y = x at (\sqrt{2}/2, \sqrt{2}/2). From there to the top of the circle at y = 1, the left boundary of the region is part of the circle. For y > 1, though, the left boundary is part of the y-axis so we have to split the integral at y = 1:

∫_{\sqrt{2}/2}^{1} ∫_{√\overline{1−y^{2}}}^{y} f(x, y) dx dy + ∫_{1}^{4} ∫_{0}^{y} f(x, y) dx dy.

C) Now reverse the order of the variables and set up iterated integral(s) to compute the same integral, but integrating with respect to y first, then x.

Solution: We also need to split the integral this way since the bottom boundary changes at x = \sqrt{2}/2. The region extends all the way to x = 4 on the right, where the line y = 4 intersects y = x:

∫_{0}^{\sqrt{2}/2} ∫_{√\overline{1−x^{2}}}^{4} f(x, y) dy dx + ∫_{\sqrt{2}/2}^{4} ∫_{x}^{4} f(x, y) dy dx.

VI. (20) The metal making up a solid half-cone in the shape of

H = {(x, y, z) ∈ R^{3} | z^{2} ≥ x^{2} + y^{2}, 0 ≤ z ≤ 1, y ≥ 0}

has density δ(x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}} at all points. Determine its total mass. (Hint: a wise choice of coordinate system will simplify this one immensely!)

Solution: We will set up the triple integral to compute the mass using spherical coordinates, since the spherical equation of the boundary cone is ϕ = 1. The restriction y ≥ 0 says 0 ≤ θ ≤ π. The plane z = 1 is ρ cos ϕ = 1, so ρ = sec ϕ. In spherical coordinates, the density is just d = ρ. So the total mass is

M = ∫_{0}^{π} ∫_{0}^{π/4} ∫_{0}^{sec ϕ} ρ^{3} sin ϕ dρ dϕ dθ

= ∫_{0}^{π} ∫_{0}^{π/4} ∫_{0}^{sec ϕ} ρ^{3} sin ϕdρ dϕ dθ

= ∫_{0}^{π} ∫_{0}^{π/4} \frac{ρ^{4}}{4} |_{0}^{sec ϕ} sin ϕ dϕ dθ

= ∫_{0}^{π} ∫_{0}^{π/4} \frac{sin ϕ}{4 cos^{4} ϕ} dϕ dθ (u^{−4}du)

= π \frac{1}{12 cos^{3}(ϕ)}|_{0}^{π/4}

= \frac{π}{12}(2\sqrt{2} − 1)

VII.

A) State Green’s Theorem.

Solution: If D is a region in R^{2} bounded by a finite collection of simple closed curves, ∂D is the positively-oriented boundary of D, and F(x, y) = (F_{1}(x, y), F_{2}(x, y)) is a C^{1} vector field on D, then

∮_{∂D} F · T ds = ∮_{∂D} F_{1}dx + F_{2}dy = ∫ ∫_{D} \frac{∂F_{2}}{∂x} − \frac{∂F_{1}}{∂y} dA.

B) (10) Let F(x, y) = (x − y^{2}, x^{2} + y). Verify that Green’s Theorem holds for the region D = {(x, y) ∈ R^{2} : x^{2} + y^{2} ≤ 9}.

Solution: Using the standard parametrization (x, y) = (3 cos(t), 3 sin(t)) of the boundary circle of D,

∮_{∂D} F_{1}dx + F_{2}dy = ∫_{0}^{2π}(3 cos(t) − 9 sin^{2}(t))(−3 sin(t))
+ (9 cos^{2}(t) + 3 sin(t))(3 cos(t)) dt
= 27 ∫_{0}^{2π} sin^{3}(t) + cos^{3}(t) dt

= 27 (−\frac{2}{3} cos(t) − \frac{1}{3} sin^{2}(t) cos(t) + \frac{2}{3} sin(t) + \frac{1}{3} cos^{2}(t) sin(t))|_{0}^{2π}

= 0.

The double integral over D is

∫ ∫_{D}(F_{2})_{x} − (F_{1})_{y} dA = ∫ ∫_{D} 2x + 2y dA.

This can be evaluated in a number of ways. Switching to polar coordinates, for instance,

= ∫_{0}^{2π} ∫_{0}^{3} 2r^{2}(cos θ + sin θ) dr dθ = 0

since both ∫_{0}^{2π} cos θ dθ = 0 and ∫_{0}^{2π} sin θ dθ = 0.

VIII. A function f(x, y) is said to be harmonic on an open set U in R^{2} if it satisfies the equation

f_{xx} + f_{yy} = 0

at all points in U. A) How does a nondegenerate critical point of a harmonic function fit into our classifica-

tion? Is it a local maximum, local minimum, or a saddle point? Explain how you can tell from the second derivative test.

Solution: Every nondegenerate critical point of a harmonic function is a saddle point because the Hessian matrix is

D^{2}(f) = ( \begin{matrix}f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{matrix} ) = ( \begin{matrix}f_{xx} & f_{xy} \\ f_{xy} & −f_{xx} \end{matrix} ) .

The determinant here is −(f_{xx})^{2} − (f_{xy})^{2} < 0.

B) If f is harmonic, what is true about the line integral of the vector field

F(x, y) = (−f_{y}, f_{x})

around any simple closed curve in U?

Solution: Let D be the region bounded by the simple closed curve. By Green’s theorem, the integral is equal to

∫ ∫_{D}(f_{x})_{x} − (−f_{y})_{y} dA = ∫ ∫_{D} f_{xx} + f_{yy} dA = 0.

Extra Credit (10)

Suppose you follow a flow line of the vector field −∇f for f(x, y) in the xy-plane. As you traverse the flow line in the increasing t-direction, is the corresponding path on the graph z = f(x, y) going uphill or downhill? Explain. What does the vector field −∇f look like near a local maxmimum of f? near a local minimum of f?

Solution: You are always going downhill by the most direct route – recall ∇f(a, b) gives the direction in which f is increasing the fastest. The negative gradient vector field near a local maximum will have all arrows pointing away from the critical point (flow lines will diverge from the maximum). Near a local minimum, the negative gradient vector field will be pointing toward from the crirical point (flow lines will be converging toward the minimum).

Have a peaceful and joyous holiday season!