Mathematics 241 – Multivariable Calculus Final Examination – December 11, 2007

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

A) (10) 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) (10) 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) (5) 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.

A) (7.5) Let F : R_{u,v}^{2} → R_{x,y}^{2} be F (u, v) = (x(u, v), y(u, v)) = (u^{2} − v^{2}, 2uv) and let g : R_{x,y}^{2} → R be g(x, y) = sin(x) cos(y). Find the derivative matrix D(g ∘ F) by direct substitution and differentiation.

Solution: The function (g ∘ F)(u, v) = sin(u^{2} − v^{2}) cos(2uv), so computing partial derivatives by the product rule

\frac{∂(g ∘ F)}{∂u} = 2u cos(u^{2} − v^{2}) cos(2uv) − 2v sin(u^{2} − v^{2}) sin(2uv)
\frac{∂(g ∘ F)}{∂u} = −2v cos(u^{2} − v^{2}) cos(2uv) − 2u sin(u^{2} − v^{2}) sin(2uv)

and the derivative matrix is the 1 × 2 matrix with these entries. B) (7.5) Compute D(g ∘ F) for the functions in part A using the Chain Rule and show

you get the same result as in part A.

Solution: The Chain Rule says D(g∘F) = D(g)(F (u, v))D(F)(u, v) (matrix product). So we compute D(g) = ( cos(x) cos(y) − sin(x) sin(y) ) and

D(F) = ( \begin{matrix}2u & −2v \\ 2v & 2u \end{matrix} ) .

Hence the product matrix to be computed is

( cos(u^{2} − v^{2}) cos(2uv) − sin(u^{2} − v^{2}) sin(2uv) ) ( \begin{matrix}2u & −2v \\ 2v & 2u \end{matrix} )

which gives a 1 × 2 matrix with first entry

2u cos(u^{2} − v^{2}) cos(2uv) − 2v sin(u^{2} − v^{2}) sin(2uv)

and second entry

−2v cos(u^{2} − v^{2}) cos(2uv) − 2u sin(u^{2} − v^{2}) sin(2uv)

as in the matrix D(g ∘ F). C) (10) Now let z = g(x, y) for a general function g – not specifically g(x, y) as above in

part A, and x = u^{2} − v^{2}, y = 2uv as in the definition of F above. Show that

\frac{1}{4(u^{2} + v^{2})} ((z_{u})^{2} + (z_{v})^{2}) = (z_{x})^{2} + (z_{y})^{2}.

Solution: Now we use the expanded form of the Chain Rule: We have z_{u} = z_{x}x_{u}+z_{y}y_{u} and z_{v} = z_{x}x_{v} + z_{y}y_{v} so

(z_{u})^{2} + (z_{v})^{2} = (z_{x}x_{u} + z_{y}y_{u})^{2} + (z_{x}x_{v} + z_{y}y_{v})^{2}
= (2uz_{x} + 2vz_{y})^{2} + (−2vz_{x} + 2uz_{y})^{2}

= 4u^{2}(z_{x})^{2} + 8uvz_{x}z_{y} + 4v^{2}(z_{y})^{2} + 4v^{2}(z_{x})^{2} − 8uvz_{x}z_{y} + 4u^{2}(z_{y})^{2}

= 4(u^{2} + v^{2})((z_{x})^{2} + (z_{y})^{2})

Dividing by 4(u^{2} + v^{2}) yields the desired equation.

III. All parts of this problem refer to f(x, y) = (x + 1)^{2} + y^{2}. A) (5) 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) (10) 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) (15) 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) (10) 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) (10) 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) (20) 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

D^{2}(f)(1/2, 0) = ( \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

D^{2}(f)(−1/2, 0) = ( \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) (5) 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) (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: 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) (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: 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 slug having the shape of the region in R^{3} with x^{2} + y^{2} ≤ 4 and −1 ≤ z ≤ 1 has density δ(x, y, z) = 5 + x at all points. Determine the coordinates of its center of mass.

Solution: We will set up the triple integrals to compute the coordinates of the center of mass using cylindrical coordinates, since the slug is just a cylinder with axis along the z-axis. The total mass is

M = ∫_{0}^{2π} ∫_{0}^{2} ∫_{−1}^{1}(5 + r cos θ)rdz dr dθ

= 2 ∫_{0}^{2π} ∫_{0}^{2} 5r + r^{2} cos θ dr dθ

= 2 ∫_{0}^{2π} \frac{5}{2}r^{2} + \frac{1}{3}r^{3} cos θ|_{0}^{2} dθ

= 2 ∫_{0}^{2π} 10 + \frac{8}{3} cos θ dθ

= 20θ + \frac{8}{2} sin θ|_{0}^{2π}

= 40π.

Since the density does not depend on y or z, by the symmetry of the cylinder, it can be seen that \overline{y} = \overline{z} = 0. To compute the x-coordinate,

\overline{x} = \frac{1}{40π} ∫_{0}^{2π} ∫_{0}^{2} ∫_{−1}^{1} r cos θ(5 + r cos θ)r dz dr dθ

= \frac{1}{40π} · 2 ∫_{0}^{2π} ∫_{0}^{2} 5r^{2} cos θ + r^{3} cos^{2} θ dr dθ

= \frac{1}{20π} ∫_{0}^{2π} \frac{5}{3}r^{3} cos θ + \frac{1}{4}r^{4} cos^{2} θ|_{0}^{2} dθ

= \frac{1}{20π} ∫_{0}^{2π} \frac{40}{3} cos θ + 4 cos^{2} θ dθ

= \frac{1}{20π} (\frac{40}{3} sin θ + 2θ + cos(2θ))|_{0}^{2π}

= \frac{1}{20π} (0 + 4π + 1 − 0 − 0 − 1)

= \frac{1}{5}.

The values of \overline{y} and \overline{z} can also be computed directly as above, replacing the x = r cos θ by y = r sin θ and z respectively.

VII.

A) (10) 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) = (M(x, y), N(x, y)) is a C^{1} vector field on D, then

∮_{∂D} Mdx + Ndy = ∫ ∫_{D} N_{x} − M_{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} Mdx + Ndy = ∫_{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} N_{x} − M_{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.

C) (5) We can think of a vector field on R^{2} as a vector field on R^{3} by making the last component equal to zero at all points. Compute curl(F) for F = (M(x, y), N(x, y), 0) and relate your result to the statement of Green’s Theorem.

Solution: We have

curl(F) = ∇ × (M(x, y), N(x, y), 0)

Since M, N do not depend on z and because of the zero in the last component,

curl(F) = (0, 0, N_{x}(x, y) − M_{y}(x, y))

The integrand in the double integral in Green’s theorem is just the last component here.

Comment: In fact, Green’s theorem is a special case of a more general result called Stokes’ theorem in which the plane region is replaced by a surface S in R^{3}. Stokes’ theorem then asserts the equality of the line integral of F over the boundary of S and the surface integral of curl(F) over S. The surface integral involves integrating curl(F) · N. where N is a normal vector to the surface. The normal vector for D in R^{2} is just k = (0, 0, 1) at each point, and curl(F) · k is exactly the integrand of the double integral in Green’s theorem.

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) (5) How does a nondegenerate critical point of a harmonic function fit into our classi-

fication? Is it a local maximum, local minimum, or a saddle point? Explain how you can tell from the Hessian criterion.

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) (5) 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 (20)

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 uphill by the most direct route – recall ∇f(a, b) gives the direction in which f is increasing the fastest. The gradient vector field near a local maximum will have all arrows pointing in toward the critical point (flow lines will converge toward the maximum). Near a local minimum, the gradient vector field will be pointing away from the crirical point (flow lines will be diverging away from the minimum).

Have a peaceful and joyous holiday season!