Math 223
Final Exam, Spring 1997
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1. No, f(x,y) is not linear because the increments of f for the same changes in x and y is not the same in either the rows or the columns.

2. False. If $\vec{u}$ and $\vec{v}$ and perpendicular vectors, the angle between them is $\frac{\pi}{2}$. The dot product is defined as

$$\vec{u} \cdot \vec{v} = \Vert \vec{u}\Vert \Vert\vec{v}\Vert\cos (\theta )$$

Therefore if the vectors are perpendicular, $\cos(\theta)$ is 0, and so is the dot product.

3. True. $\nabla f = 0\vec{\imath} + 0\vec{\jmath}$ so $f_{\vec{u}} = \nabla f \cdot \vec{u} = 0$ for all $\vec{u}$.

4.Since $x=\sin(t)$ and $y = \cos(t)$ for $0 \leq t \leq \pi$, we can make the following table:

t	x	y
0	0	1
$\frac{\pi}{2}$	1	0
$\pi$	0	-1

which shows the answer is c.

5. Because f(x,y) depends only on x, the contours on the xy-plane will be straight lines up and down, parallel to the y-axis as shown in Figure a. Constant x gives constant z.

6. The effect of the vector field on the top (y > 0) is larger and in the direction of the orientation of the curve, than the effect on the bottom (y<0), which is opposite to the orientation of the curve.

7. Using the curl test, $(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})\vec{k} = \vec{k} \neq \vec{0}$. So $\vec{F}$ is not conservative.

8. True, the functions are just multiples of each other.

9. Since PV = RT, $p = \frac{RT}{V}$ so,

a.

$$\frac{\partial P}{\partial T} = \frac{R}{T}$$

and

$$\frac{\partial P}{\partial V} = - \frac{RT}{V^2}$$

b. For the given conditions:

$$R = \frac{PV}{T} = \frac{(1atm)(.0245m^3)}{298 ^oK} = 8.22 \times 10^{-5}\frac{atm-m^3}{^oK}$$

so

$$\frac{\partial P}{\partial V} = \frac{-RT}{V^2} = - 40.82 \frac{atm}{m^3}$$

which tells us that as V increases by one cubic meter, P will decrease by about 41 atmospheres, starting at the given conditions, with T held constant.

10.

a. The equation of the line is

$$y = -\frac{2}{3}x + \frac{11}{3}$$

so the integral is

$$\int_1^4 \int_1^{y = -\frac{2}{3}x + \frac{11}{3}} f(x,y)\/dy\/dx$$

b. We know x² + y² = r² so this suggests polar coordinates:

$$\int\int e^{-x^2 = y^2}\/dx\/dy =\int_0^{2\pi}\int_0^{\sqrt{2}} e^{-r^2}r\/dr\/d\theta$$

By u-substitution, with u=-r² get

$$-\frac{1}{2}\int\int e^u du d\theta$$

$$= \left. -\frac{1}{2}\int e^{u} \right\vert _0^2 d\theta = -\frac{1}{2} \int_0^{2\pi} (e^{-4} - 1)\/d\theta = \pi - \pi e^{-4}$$

11.

a. Stokes' Theorem says that the circulation of a vector field around the closed boundary of a capping surface is the same as the flux through the surface. In symbols:

$$\int_{\cal{S}} \nabla \times \vec{F}\cdot d\vec{A} = \oint_{\cal{C}=\partial{\cal{S}}} \vec{F} \cdot d \vec{r}$$

where:

• $\cal{S}$ is a piecewise smooth orientated surface
• $\cal{C}$ is the piecewise smooth boundary of $\cal{S}$, consistently orientated, a closed curve.

b. In

$$\int_{\cal{S}} \nabla \times \vec{F}\cdot d\vec{A} = \oint_{\cal{C}=\partial{\cal{S}}} \vec{F} \cdot d \vec{r}$$

let $\cal{C}$ be the circle boundary and $\cal{S}$ be the circle itself, Then $\vec{n}$ is $\vec{k}$ to be consistent with the orientation of the circle boundary given by the problem. Thus the curl of $\vec{F}$and the normal vector are antiparallel and the required circulation is negative.

12. To be supplied.

13.

a.

$$\vec{F} = \frac{x\vec{\imath}}{(x^2 + y^2 = z^2)^{\frac{3}{2}... ...ac{3}{2}}} + \frac{z\vec{k}}{(x^2 + y^2 = z^2)^{\frac{3}{2}}}$$

so

$$\nabla \cdot\vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$$

and

$$\frac{\partial F_1}{\partial x} = \frac{(x^2+y^2+z^2)^{\frac{3}{2}} - 3x^2(x^2+y^2+z^2)^{\frac{1}{2}}}{(x^2+y^2+z^2)^3}$$

$$\frac{\partial F_2}{\partial y} = \frac{(x^2+y^2+z^2)^{\frac{3}{2}} - 3y^2(x^2+y^2+z^2)^{\frac{1}{2}}}{(x^2+y^2+z^2)^3}$$

$$\frac{\partial F_3}{\partial z} = \frac{(x^2+y^2+z^2)^{\frac{3}{2}} - 3z^2(x^2+y^2+z^2)^{\frac{1}{2}}}{(x^2+y^2+z^2)^3}$$

And these add to 0.

b. Since the sphere is of radius 1, and centered at (0,0,2), the Divergence Theorem can be used, since the field $\vec{F}$ is defined everywhere within the sphere. Thus

$$\int_{\cal{S}} \vec{F}\cdot \/d\vec{A} = \int_{\cal{V}}0\/dv = 0$$

14.

a.

$$\nabla f(1,0) = \frac{\partial f}{\partial x} +\frac{\partial... ...= (3e^{x^2y}(1+2x@y)) \vec{\imath} + (3x^3e^{x^2y})\vec{\jmath}$$

therefore $\nabla f(1,0) = 3\vec{\imath} - 3\vec{\jmath}$

b. The function f increases most rapidly in the direction $3\vec{\imath} - 3\vec{\jmath}$ and the rate of change is 0 along the contour, in the directions $\pm( 3\vec{\imath} - 3\vec{\jmath})$.