Vector Calculus   Solutions to Sample Final Examination #1

1.

Let $f (x, y) = e ^{ x y} \sin (x + y)$.

(a)

In what direction, starting at $(0, \pi / 2)$, is f changing the fastest?

(b)

In what directions starting at $(0, \pi / 2)$ is f changing at 50% of its maximum rate?

(c)

Let ${\bf c} (t)$ be a flow line of ${\bf F} = \nabla f$ with ${\bf c} (0) = (0, \pi/2 )$. Calculate

$$\left. \frac{d }{d t} [ f (c (t)) ] \right\vert _{ t = 0} .$$

Solution

(a)

f is changing fastest in the direction of $\nabla f (0, \pi /2)$. But and so

$$\nabla f (0, \pi /2) = \frac {\pi}{2} {\bf i}.$$

Thus f is increasing fastest in the direction i (and decreasing fastest in the direction - i).

(b)

If n is a unit vector, f is changing at the rate

$$\nabla f (0, \pi / 2) \cdot {\rm {\bf n}} = \frac {\pi}{2} {\bf n} \cdot {\bf i}$$

in the direction n. The maximum value is $\pi / 2$, so the rate is $50 \%$ of its maximum when

$$\frac {\pi}{2} {\bf n} \cdot {\bf i} = \frac {\pi}{2} \cdot \frac {1}{2}$$

i.e.,

$${\bf n} \cdot {\bf i} = \frac {1}{2}$$

This means n makes an angle $\theta$ with i where $\cos \theta = 1/2$, or $\theta = \pm \pi / 3$ or $\pm 60$ degrees. Note that this defines two directions (if this were in space and not the plane, we would get a cone).

(c)

By the chain rule, and since ${\bf c}'(t) = {\bf \nabla }f({\bf c}(t))$,

2.

Let $f : { \mathbb R}^3 \rightarrow { \mathbb R}^3$ be a given mapping and write f (x, y, z) = (u (x, y, z) , v (x, y, z), w (x, y, z)). Let $g : { \mathbb R}^3 \rightarrow { \mathbb R}^3$ be defined by g (u,v,w) = (u - v, u + w, w + v ) and let $h = g \circ f$.

(a)

Write a formula for the derivative matrix ${\bf D} h$.

(b)

Show that ${\bf D} h$ cannot have rank 3 at any point (x, y, z).

(c)

Show that ${\bf D} h$ has an eigenvalue zero at every (x, y, z).

Solution

(a)

By the chain rule,

(b)

We claim that this matrix has determinant zero. Its determinant is the product of the determinants of the two factors. But

$$\begin{array} {ccccc} \nonumber \left\vert \begin{array} {c... ... & 1 \\ 0 & 1 & 1 \end{array} \right\vert &=& 0\\ \end{array}$$

Thus, Dh (x, y, z) is not invertible, so it must have nullity $\neq 0$, so rank $Dh(x, y, z) \neq 3$.

(c)

Since nullity $\neq 0$ some non-zero vector must get sent to zero; this is an eigenvector with eigenvalue zero.

3.

Extremize f (x, y, z) = x subject to the constraints

$$x^2 + y^2 + z^2 = 1 \quad \mbox{and} \quad x + y + z = 1.$$

Solution We are to extremize f(x, y , z) = x subject to

$$\begin{array} {ccc} \nonumber g _1 = x ^2 + y ^2 + z ^2 -1 = 0 &{\rm and}& x + y + z -1 = 0.\\ \end{array}$$

Using the method of Lagrange multipliers, this means

i.e.,

Subtracting (2) and (3) gives

$$2\lambda _1 (y - z) = 0$$

and so either $\lambda _1 = 0$ or y=z. But $\lambda _1 = 0$ is not consistent with (1) and (2). Hence, $\lambda _1 \neq 0$ and so y=z. Thus we have

Substituting (9) in (8) gives

(1-2y) ² + 2y ² =1

i.e.,

1-4y+4y ² + 2y =1

i.e.,

-2y + 3y ² = 0.

Thus, either y=0, or y=2/3. If y=0 then x=1 (and $\lambda _2 = 0, \lambda _1 = 1/2)$ and if y=2/3 then x=-1/3.

Therefore, the solutions are

$$(1, 0, 0) \, {\rm and} \, (-1/3, 2/3, 2/3)$$

The former maximizes f while the latter minimizes it.

4.

(a)

Evaluate

$$\int \!\!\! \int \!\!\! \int _D \exp [(x^2 + y^2 + z^2) ^{ 3/2}] \, d x\, d y\, d z$$

where D is the region defined by $1 \leq x^2 + y^2 + z^2 \leq 2$and $z \geq 0$.

(b)

Sketch or describe the region of integration for

$$\int^1_0 \int^x_0 \int^y_0 f (x, y, z) d z\, d y\, d x,$$

and interchange the order to $dy\, dx \, dz$.

Solution

(a)

We use spherical coordinates

(b)

The region for

$$\int ^1 _0 \int ^x _0 \int ^y _0 f(x,y,z) dzdydx$$

is that under the plane z = y and over the region in the plane bounded by the x-axis, the line x = y and the line x = 1. (The student should draw a figure here).

(c)

In the order dydxdz, the integral is easiest to write down by consulting the figure drawn in the previous part; one gets

$$\int ^1 _0 \int ^1 _z \int ^1 _x f(x,y,z) dydxdz.$$

5.

Let ${\bf G} (x, y) = (x e ^{ x^2 + y^2} + 2 xy) {\bf i} + (ye ^{ x^2 + y^2} + x^2) {\bf j}$.

(a)

Show that ${\bf G} = {\bf \nabla} f$ for some f; find such an f.

(b)

Use (a) to show that the line integral of ${\bf G}$ around the edge of the triangle with vertices (0,0) , (0,1), (1,0) is zero.

(c)

State Green's theorem for the triangle in (b) and a vector field ${\bf F}$ and verify it for the vector field ${\bf G}$ above.

Solution

(a)

If ${\bf G}(x,y)=P {\bf i} + Q {\bf j}, P = xe^{x^2 + y ^2} + 2xy, Q = ye ^{x ^2 + y ^2} + x ^2$, note that

$$\frac {\partial P}{\partial y} = 2xy e ^{x^2 + y^2} + 2x = \frac {\partial Q}{\partial x},$$

and so ${\bf G}$ is a gradient. Writing

$$P = \frac {\partial f}{\partial x}, \quad \mbox{and} \quad Q = \frac {\partial f}{\partial y}$$

we see that

f(x,y) = e ^{x 2 + y 2} + x ² y.

(b)

Let C be the boundary of the triangle T (the student should draw a figure of T.) Since the integral of a gradient around any closed curve is zero in general, it is zero in this particular case.

(c)

Green's Theorem states, in this case that

$$\int _C Pdx + Qdy = \int \int \left( \frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right) dx\,dy$$

We computed that $\frac {\partial Q}{\partial x} = \frac {\partial P}{\partial y}$ above, so the right hand side is zero as well.

6.

Let W be the three dimensional region under the graph of $f (x, y) = \exp (x^2 + y^2)$and over the region in the plane defined by $1 \leq x^2 + y^2 \leq 2$.

(a)

Find the volume of W.

(b)

Find the flux of the vector field ${\bf F} = (2 x - x y) {\bf i} - y {\bf j} + yz {\bf k}$ out of the region W.

Solution

(a)

The volume of W is

(b)

The flux of F is, by Gauss' theorem

7.

Let C be the curve x² + y² = 1 lying in the plane z = 1. Let ${\bf F} = (z - y) {\bf i} + y {\bf k}$.

(a)

Calculate $\nabla \times {\bf F}$.

(b)

Calculate ${\displaystyle \int _C {\bf F} \cdot d {\bf s} }$ using a parametrization of C and a chosen orientation for C.

(c)

Write $C = \partial S$ for a suitably chosen surface S and, applying Stokes' theorem, verify your answer in (b) .

(d)

Consider the sphere with radius $\sqrt{2}$ and center the origin. Let $S ^\prime$ be the part of the sphere that is above the curve (i.e., lies in the region $z \geq 1$), and has C as boundary. Evaluate the surface integral of $\nabla \times {\bf F}$over $S ^\prime$. Specify the orientation you are using for $S ^\prime$.

Solution

(a)

We evaluate the curl by writing out the expression for the curl as a cross product of $\nabla$ and F:

$$\begin{array} {ccccc} {\bf \nabla \times {\bf F} = \left\ver... ...array} \right\vert = {\bf i} + {\bf j} + {\bf k} } \end{array}$$

(b)

Parameterize C by $x=\cos \theta, y=sin \theta, z=1, 0 \leq \theta \leq 2\pi$, where the orientation is counter-clockwise as viewed from above. Then

$$\int _C {\bf F} \cdot {\bf d}s = \int ^{2\pi} _0 (1 -\sin \t... ...n \theta)d\theta = \int ^{2\pi} _0 \sin ^2 \theta d\theta = \pi$$

(since the average of $\sin ^2 \theta$ is 1/2).

(c)

Let us choose S to be the disk $x ^2 + y ^2 \leq 1, z=1$. Then

$$\int _C {\bf F} \cdot {\bf d}s = \int \int _S {\bf \nabla } ... ...} \times dS = \int \int _S {\bf k} \cdot {\bf k} dx\,dy = \pi.$$

(d)

Let the orientation of S'' be given by the outward normal. The student should draw a figure that shows $\partial S' = C$. Then,

$$\int \int _S'' ({\bf \nabla } \cdot {\bf F}) \times dS = \int _C {\bf F} \times {\bf d}s = \pi.$$