Mathematics 241, section 1 – Multivariable Calculus Solutions for Final Examination – December 14, 2010

I. In this problem,

P = (1, −1, 3), Q = (0, 3, 1), and R = (−4, 1, 2).

A) (10) Find the equation of the plane containing the points P, Q, R in R^{3}.

Solution: A normal vector is

N = ((−4, 1, 2) − (1, −1, 3)) × ((0, 3, 1) − (1, −1, 3))
= (−5, 2, −1) × (−1, 4, −2) = (0, −9, −18)

Since any scalar multiple of N is also normal to the plane, we can also use N = (0, 1, 2), and then the equation becomes

(0, 1, 2) · (x − 1, y + 1, z − 3) = 0

or

y + 2z − 5 = 0.

B) (10) At what point does the line containing P, Q meet the xy-plane?

Solution: The line containing P and Q can be parametrized as

α(t) = (1, −1, 3) + t(−1, 4, −2) = (1 − t, −1 + 4t, 3 − 2t).

This meets the xy plane when z = 0 so 3 − 2t = 0, or t = 3/2. The corresponding point is

α(3/2) = (−1/2, 5, 0).

1. C) (5) If v is the vector from P to Q and w is the vector from P to R, at what angle do v, w meet?

Solution: The angle is θ satisfying

cos θ = \frac{v · w}{∥v∥∥w∥} = \frac{15}{\sqrt{30}\sqrt{21}} = \frac{5}{\sqrt{70}}.

This says

θ = cos^{−1} (\frac{5}{\sqrt{70}}) .

II. Suppose you follow a flow line of the vector field ∇f for f(x, y) in the xy-plane.

A) (5) 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.

Solution: Uphill – the gradient vector always points in the direction of maximum rate of increase.

B) (5) What does the vector field ∇f look like near a local maximum of f? Near a local minimum of f? Make rough sketches by hand to illustrate.

Solution: Near a local maximum, all the arrows from the vector field ∇f will be pointing “in” – the local maximum is a sink of the vector field. Near a local minimum, all the arrows will be pointing “out” – the local minimum is a source of the vector field.

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

A) (5) Sketch the contours of f(x, y) for the values c = −1, 0, 1.

Solution: The contours for c = ±1 are hyperbolas with asymptotes along the lines y = x − 1 and y = −x + 1. The c = −1 contour opens up and down, and the c = 1 contour opens left and right. The c = 0 contour is the union of the two asymptotes.

B) (10) At the point (1, 2), in which direction is f increasing the fastest? Express your answer as a unit direction vector.

Solution: The gradient vector ∇f(1, 2) points in this direction. ∇f(x, y) = (2(x − 1), −2y) so we have ∇f(1, 2) = (0, −4). The unit vector in this direction is u = (0, −1).

IV. (20) After an ill-fated “three hour tour” goes awry, you are stranded on a island at the point with coordinates (1, 1). Fortunately, you have a radio transmitter with you. Unfortunately, it has a limited range – its signal can only be received at distances less than or equal to 2/3 from its position. You know that there is a Coast Guard patrol boat that makes a circuit of the path x^{2} + y^{2} = 4 every day, and they always carry a radio receiver and listen for transmissions. Will the patrol boat ever get within 2/3 of your position and receive your signal? (Note: Minimizing the distance from a point (a, b) is the same as minimizing the function f(x, y) = ∥(x, y) − (a, b)∥^{2}.)

Solution: One solution of this problem uses the method of Lagrange multipliers to find the point on the path of the patrol boat closest to (1, 1). If the distance from the closest point to (1, 1) is less than 2/3, then the signal will be heard and you will be rescued. Using the hint, we want to minimize

f(x, y) = ∥(x, y) − (1, 1)∥^{2} = (x − 1)^{2} + (y − 1)^{2}.

The constraint curve is g(x, y) = x^{2} + y^{2} − 4 = 0. The Lagrange equations are

2(x − 1) = 2λx
2(y − 1) = 2λy
x^{2} + y^{2} − 4 = 0.

We can eliminate λ between the first two equations by the usual method (multiply the first by y and the second by x, then equate the left sides), yielding

y(x − 1) = x(y − 1) ⇒ y = x.

In the constraint equation, this gives two points (\sqrt{2}, \sqrt{2}) and (−\sqrt{2}, −\sqrt{2}). The values of f at these points are

f(\sqrt{2}, \sqrt{2}) = 2(\sqrt{2} − 1)^{2} = 6 − 4\sqrt{2}
f(−\sqrt{2}, −\sqrt{2}) = 2(−\sqrt{2} − 1)^{2} = 6 + 4\sqrt{2}.

Since 6 − 4\sqrt{2} ≐ .343 < (2/3)^{2} ≐ .444, your radio signal will be received by the patrol boat and you will be rescued!

Notes:

1. That the closest point to (1, 1) on the circle x^{2} + y^{2} = 4 is the point (\sqrt{2}, \sqrt{2}) can also be seen by elementary geometry. I gave full credit for solutions that showed this without using Lagrange multipliers.

2. The problem has a number of other correct solutions as well. Another way that many of you thought of is to find whether the circle (x − 1)^{2} + (y − 1)^{2} = 4/9 (the boundary of the region in which the signal from your receiver can be heard) intersects the path of the patrol boat x^{2} + y^{2} = 4. This can be determined by elementary algebra – the two circles intersect at approximately (1.122, 1.655) and (1.655, 1.122), so there is an arc on the path of the patrol boat consisting of points from which the signal can be heard.

V. Let f(x, y) = ye^{−x^{2}−2y^{2}}.

A) (10) Find the equation of the tangent plane to the graph z = f(x, y) at the point (1, 1, e^{−3}).

Solution: We have f_{x} = −2xye^{−x^{2}−2y^{2}} and f_{y} = e^{−x^{2}−2y^{2}}(1 − 4y^{2}). So the tangent plane is

z = f(1, 1) + f_{x}(1, 1)(x − 1) + f_{y}(1, 1)(y − 1)
= e^{−3} − 2e^{−3}(x − 1) − 3e^{−3}(y − 1).

B) (10) Find all the critical points of f(x, y).

Solution: We have f_{x}, f_{y} from part A. Setting those equal to zero, we obtain from f_{x} = 0 that x = 0 or y = 0. However f_{y} = 0 implies that y = ±1/2. So there are only two critical points (0, ±1/2).

C) (20) Use the Second Derivative Test to determine the type of each critical point you found in part B.

Solution: We continue to compute the second-order partial derivatives.

f_{xx} = y(−2 + 4x^{2})e^{−x^{2}−2y^{2}}
f_{xy} = x(−2 + 8y^{2})e^{−x^{2}−2y^{2}}
f_{yy} = (−12y + 16y^{3})e^{−x^{2}−2y^{2}}

At (0, 1/2),

A = f_{xx}(0, 1/2) = −e^{−1/2}
B = f_{xy}(0, 1/2) = 0
C = f_{yy}(0, 1/2) = −4e^{−1/2}.

Hence AC − B^{2} > 0 and A < 0, so f has a local maximum at (0, 1/2).

Similarly, at (0, −1/2),

A = f_{xx}(0, −1/2) = +e^{−1/2}
B = f_{xy}(0, −1/2) = 0
C = f_{yy}(0, −1/2) = +4e^{−1/2}.

Hence AC − B^{2} > 0 and A > 0, so f has a local minimum at (0, −1/2).

VI. The region R in R^{2} is the set of points

R = {(x, y) ∈ R^{2} | x^{2} + y^{2} ≤ 25, y ≥ x, x ≥ 0, and y ≤ 4}

and let f(x, y) be some continuous function defined on R.

A) (5) Sketch R.

Solution: This is the region inside the circle of radius 5 centered at (0, 0), above the line y = x, to the right of the y-axis, but below the line y = 4:

The line y = 4 intersects the circle at (3, 4), and the line y = x intersects at (5\sqrt{2}/2, 5\sqrt{2}/2). Note that this has y = 5\sqrt{2}/2 ≐ 3.54 < 4. So the boundary of the region we want includes a small arc of the circle x^{2} + y^{2} = 25 between the points (3, 4) and (5\sqrt{2}/2, 5\sqrt{2}/2).

B) (10) 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: This way, we must split the interval of y-values at y = 5\sqrt{2}/2, where the line y = x crosses the circle:

∫_{0}^{5\sqrt{2}/2} ∫_{0}^{y} f(x, y) dx dy + ∫_{5\sqrt{2}/2}^{4} ∫_{0}^{√\overline{25−y^{2}}} f(x, y) dx dy.

C) (10) 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: This one also must be split at x = 3 where the line y = 4 crosses the circle.

∫_{0}^{3} ∫_{x}^{4} f(x, y) dy dx + ∫_{3}^{5\sqrt{2}/2} ∫_{x}^{√\overline{25−x^{2}}} f(x, y) dy dx.

VII. (20) The metal making up a thin plate with the shape of the region in R^{2} with x^{2} + y^{2} ≤ 4 has density δ(x, y) = 6 + y at all points. Determine the coordinates of its center of mass.

Solution: Since the plate has the shape of a circular disk, we will set up all the integrals in polar coordinates. First the total mass is

M = ∫ ∫_{R} 6 + y dA = ∫_{0}^{2π} ∫_{0}^{2}(6 + r sin θ)r dr dθ
= ∫_{0}^{2π} 12 + \frac{8}{3} sin θ dθ
= 24π.

We can see \overline{x} = 0 since the plate and the density function are symmetric under x ↦ −x. Finally

\overline{y} = \frac{1}{24π} ∫_{0}^{2π} ∫_{0}^{2} r sin θ(6 + r sin θ)r dr dθ
= \frac{1}{24π} ∫_{0}^{2π} 16 sin θ + 4 sin^{2} θ dθ
= \frac{4π}{24π}
= \frac{1}{6}.

VIII. Consider the following triple integral:

∫_{−3}^{3} ∫_{−\sqrt{9−x^{2}}}^{√\overline{9−x^{2}}} ∫_{−\sqrt{9−x^{2}−y^{2}}}^{√\overline{9−x^{2}−y^{2}}} 1 + x dz dy dx

A) (5) Describe the solid over which you are integrating here.

Solution: It is the solid sphere of radius 3 centered at (0, 0, 0) in R^{3}. B) (20) Using any convenient coordinate system, evaluate this integral.

Solution: Spherical coordinates are probably the best choice:

∫_{0}^{2π} ∫_{0}^{π} ∫_{0}^{3}(1 + ρ sin ϕ cos θ)ρ^{2} sin ϕ dρ dϕ dθ

= ∫_{0}^{2π} ∫_{0}^{π} 9 sin ϕ + \frac{81}{4} sin^{2} ϕ cos θ dϕ dθ
= ∫_{0}^{2π} 18 + \frac{81π}{8} cos θ dθ
= 36π.

This can also be done relatively easily in cylindrical coordinates, of course:

∫_{0}^{2π} ∫_{0}^{3} ∫_{−\sqrt{9−r^{2}}}^{√\overline{9−r^{2}}}(1 + r cos θ)r dz dr dθ.

A) (10) State Green’s Theorem.

Solution: If α is a differentiable simple, closed, positively oriented curve enclosing a region R and F (x, y) = (u(x, y), v(x, y)) is a vector field whose component functions u, v have continuous first order partial derivatives on the region R, then

∫_{α} F · T ds = ∫ ∫_{R} v_{x} − u_{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} by computing both sides of the equation in the theorem and showing that they are equal.

Solution: The line integral can be computed using α(t) = (3 cos(t), 3 sin(t)). This gives:

∫_{α} F · T ds = ∫_{0}^{2π}(3 cos(t) − 9 sin^{2}(t), 9 cos^{2}(t) + 3 sin(t)) · (−3 sin(t), 3 cos(t)) dt
= ∫_{0}^{2π} 27 sin^{3} t + 27 cos^{3} t dt
= 0.

The double integral can be computed using polar coordinates:

∫ ∫_{R} 2x + 2y dA = ∫_{0}^{2π} ∫_{0}^{3} 2r^{2}(cos θ + sin θ) dr dθ
= ∫_{0}^{2π} 18(cos θ + sin θ) dθ
= 0.

Extra Credit

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. (Here f_{x}, f_{y}, f_{xx}, f_{yy} are the partial derivatives of the harmonic function f with respect to the indicated variables.)

A) (5) How does a nondegenerate critical point of a harmonic function fit into our classification? Is it a local maximum, local minimum, or a saddle point? Explain how you can tell from the Second Derivative Test.

Solution: Since f_{yy} = −f_{xx} at all critical points,

f_{xx}f_{yy} − (f_{xy})^{2} = −(f_{xx})^{2} − (f_{xy})^{2} < 0

at every nondegenerate critical point. They are all saddle points(!) B) (5) If f is harmonic, what is true about the total flux of the vector field

F(x, y) = ∇f(x, y)

across any simple closed curve in U? Explain. Solution: The divergence of ∇f = (f_{x}, f_{y}) is

(f_{x})_{x} + (f_{y})_{y} = f_{xx} + f_{yy} = 0.

Hence

∫_{α} F · N ds = ∫ ∫_{R} f_{xx} + f_{yy} dA = 0

by Green’s Theorem.