18.02 Practice Exam 4 -- May , 1999 2:05-2:55

Directions: Suggested time: 3 hours.

No calculators or notes. There are 20 questions; each counts 12.5 points.

1.

$P: (0,0,0)$ $Q: (0,1,1)$ $R: (1,0,1)$ are three points in space. Find the angle between the vectors $PQ$ and $PR$.

2.

$P: (1,1,0)$ $Q: (2,1,1)$ and $R: (3,2,-1)$ are three points.

(a)

Find the cross product $PQ \times PR$.

(b)

Using (a), get the equation of the plane through the three points, in the form $ax+by+cz=1$.

3.

For what value(s) of the constant $c$ will the system of equations $Ax=0$ have a non-zero solution, if $A$ is the matrix shown and $x$ is the column vector?

$$\begin{eqnarray*} A= \left[ \begin{array}{ccc} 1 &0 &c\\ 2 &1 &-1\\ 1 &2 ... ...[ \begin{array}{c} x \\ y \\ z \end{array} \right] \, . \end{eqnarray*}$$

4.

Express in terms of $A$ and $B$ the two diagonals of the parallelogram shown, and then prove using vector algebra that if the two diagonals are perpendicular, the sides of the parallelogram are equal.

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5.

The position vector of a moving point $P$ is $r=(\cos t)i+ (\sin t) j + tk$.

(a)

At what point $(a,b,c)$ does $P$ pass through the surface $z= \pi (x^2 + y^2)$?

(b)

What is its speed at that time?

6.

For the function $w=x^2 + xy^2$, and the point $P: (1,1)$, find

(a)

the value of grad $w$ at $P$;

(b)

the directional derivative $dw/ds$ at $P$ in the direction of $A=3i-4j$;

(c)

starting at $P$, approximately how far should you go in the direction of $A$ in order to increase the value of $w$ by $.01$ ?

7.

$P$, $V$ and $T$ for an ideal gas are related by $V=kT/P$, where $k$ is a constant.

(a)

Take $k=2$, and give an approximate formula telling how $\Delta V$ varies with $\Delta p$ and $\Delta T$ when $P=2$ and $T=1$.

(b)

Continuing part (a), when $P=2$ and $T=1$, is $V$ numerically more sensitive to $P$ or $T$? (Indicate brief reason.)

8.

By using Lagrange multipliers, find the point on the plane $x+2y+z-1=0$ which is closest to the origin. (Minimize the square of the distance from the origin. If you don't use Lagrange multipliers, do it some other way for 6 points credit; L.M.s are the easiest way.)

9.

Let $w=f(x,y)$ and $r, \theta$ be the usual polar coordinates. By using the chain rule, find the $2 \times 2$ matrix $A$ (such that the entries of $A$ are explicitly given functions)

$$\begin{eqnarray*} \left( \begin{array}{c} w_r\\ w_{\theta} \end{array}\righ... ...A \left( \begin{array}{c} w_x\\ w_y \end{array}\right) \, . \end{eqnarray*}$$

10.

For the surface $x^2-y^2+2z^2=8$, find the tangent plane at $(1,1,2)$, in the form

$$ax + by + cz =1 \, .$$

11.

Evaluate by changing the order of integration:

$$\int^2_0 \int^4_{x^2} x e^{-y^2} \, dy \, dx \, .$$

12.

Set up an iterated integral in polar coordinates for the moment of inertia about the origin of the triangular plate shown. Take the density $=1$. Do not evaluate the integral.

13.

Let $F=xy \, i+j$. Find the work done by $F$ going over the quarter-circular path shown, going from $(1,0)$ to $(0,1)$.

14.

(a)

Express the field $F=x(x-2y) i + (2y-x^2)j$ in the form $F= \triangledown f$, for some function $f(x,y)$. Use a systematic method; show work.

(b)

Use this to evaluate $\int_C F \cdot \, dr$ over the portion of the ellipse given by the graph of $x^2+4y^2=4$ and lying in the upper half-plane; integrate in the direction left to right.

15.

How should the constants $a$ and $b$ be related if for any simple closed curve $C$

$$\oint_C ay \, dx + bx \, dy = \hbox{area inside } C \quad ?$$

(Give the most general relation; indicate reasoning.)

16.

The solid shown has as its sides the vertical right circular cone, with vertex at the origin and $60$-degree vertex angle. Its top is a portion of the sphere of the radius $2$, and its bottom is horizontal and flat, intersecting the cone at a point having distance $1$ from the origin. (Cross-section is picture.)

Set up (but do not evaluate) an iterated triple integral in spherical coordinates giving the gravitational attraction of the solid on a unit mass placed at the origin. (Take the density $=1$, and the gravitational constant $G=1$.)

17.

Let $F=xi+yj+zk$, and $S$ be the closed cylindrical surface pictured; its sides are the cylinder $x^2+y^2=1$; its top and bottom are horizontal, at heights $a$ and $O$, respectively.

(a)

Find the flux of $F$ over the top and bottom discs.

(b)

Using part (a) and the divergence theorem, find the flux of $F$ across the side cylinder.

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18.

Referring to problem 17, find the flux of $F$ across the side cylinder by evaluating a surface integral directly (i.e., without using the divergence theorem).

19.

For what value(s) of the constants $a$ and $b$ will the line integral

$$\int^Q_P (axy+yz) \, dx + (x^2 +2y+xz)\, dy + (y^2+bxy) \, dz$$

be independent of the path? (Show work.)

20.

By using Stokes' theorem, prove that $\oint_C ydx+zdy+xdz=0$ around any simple closed curve lying in the plane $x-2y+z=3$.

Brief Solutions:

1.

$$\vec{PQ} = \hat{\bf j} + \hat{k} \, , \quad \vec{PR} = \hat{{\bf i}} + \hat{k}$$

so

$$\vec{P}Q \cdot \vec{PR} = 1 \, , \quad \left\vert \vec{PQ} \right\vert = \sqrt{2} = \left\vert \vec{PR} \right\vert$$

hence $\cos \gamma = 1/2 \, , \gamma =$ angle between $\vec{PQ}$ and $\vec{PR}$. Since angle is less that $90^{\circ}$, angle $= 30^{\circ}$.

2.

$$\vec{PQ} = \hat{{\bf i}} + \hat{k} \, , \quad \vec{PR} = 2 \hat{{\bf i}} + \hat{\bf j} - \hat{k}$$

(a)

$\vec{PQ} \times \vec{PR} = \hat{k} + \hat{\bf j} + 2 \hat{\bf j} - \hat{{\bf i}} = - \hat{{\bf i}} + 3 \hat{\bf j} + \hat{k}$.

(b)

Since this is a normal to the plane, it must be

$$-x+3y+z=d$$

substituting $d=2$ so plane is

$$-\frac{x}{2} + \frac{3y}{2} + \frac{3}{2} =1 \, .$$

3.

It will have a non-zero solution if and only if

$$\det A=1+4c-c+2=0 \qquad \hbox{i.e.} \quad c=-1 \, .$$

4.

Diagonals are $A-B$ and $A+B$. These are perpendicular if $(A-B) \cdot (A+B) =0$, so

$$\left\vert A \right\vert^2 = \left\vert B \right\vert^2$$

hence $\left\vert A \right\vert = \left\vert B \right\vert$ and the side lengths are equal.

5.

$$\vec{r} = \cos t \hat{{\bf i}} + \sin t \hat{\bf j} + t \hat{k}$$

(a)

Passes through $z= \pi (x^2 + y^2)$ if

$$z=t= \pi \quad \hbox{i.e. at } (a,b,c)=(-1,0, \pi) \, .$$

(b)

$$\frac{d \vec{r}}{dt} = - \sin t \hat{{\bf i}} + \cos t \hat{\bf j} + \hat{k} = - \hat{\bf j} + \hat{k} \hbox{ at } t= \pi$$

so speed $= \sqrt{2}$.

6.

$w=x^2 + xy^2$

(a)

$\vec{\triangledown} w = (2x+y^2) \hat{{\bf i}} + 2xy \hat{\bf j} = 3 \hat{{\bf i}} + 2 \hat{\bf j}$ at $(1,1)$.

(b)

$$\begin{eqnarray*} \frac{dw}{ds} &=& \vec{\triangledown} w \cdot \vec{n} \, , \... ... \hbox{ in direction of } 3 \hat{{\bf i}} - 4 \hat{\bf j} \, . \end{eqnarray*}$$

(c)

Approximately $.01 \times 5 = .05$.

7.

(a)

$$\begin{eqnarray*} \Delta V \cong \frac{\partial V}{\partial T} \Delta T + \fra... ... \hbox{so } \Delta V \cong \Delta T - \frac{1}{2} \Delta P \, . \end{eqnarray*}$$

(b)

More sensitive to $\Delta T$.

8.

Lagrange multiplier

$$x+2y+z-1- \gamma (x^2+y^2+z^2)$$

$$\begin{eqnarray*} \begin{array}{clrcl} \partial_x : & 1-2x \gamma & =0 & \gamma... ...\\ \partial z : & 1-2z \gamma & =0 & x+4x+x-1=0 \\ \end{array}\end{eqnarray*}$$

$$x= 1/6 \, , y=1/3 \, , z=1/6 \, .$$

Closest point is $(\frac{1}{6}\, , \frac{1}{3} \, , \frac{1}{6})$.

NB It is easier to look at the normal $\hat{{\bf i}}+2 \hat{\bf j}+ \hat{k}$ and see where the line in this direction meets the plane

$$x=t \, , y=2t \, , z=t \quad : \quad t+4t+t=1 \, , t=1/6$$

Point is $(\frac{1}{6}\, , \frac{1}{3} \, , \frac{1}{6})$.

9.

$w=f(x,y) \, , x=r \cos \theta \, , y=r \sin \theta$

$$\begin{eqnarray*} \begin{array}{lll} w_r = & \frac{\partial f}{\partial x} \fra... ...heta} \qquad & w_y= \partial f / \partial y \\ [1ex] \end{array}\end{eqnarray*}$$

$$\begin{eqnarray*} {w_r \choose w_{\theta}} = A {w_x \choose w_y} \, , A= \left(... ...} \quad & \frac{\partial y}{\partial \theta} \end{array}\right) \end{eqnarray*}$$

$$\begin{eqnarray*} \hbox{so } A= \left( \begin{array}{cc} \cos \theta \quad & \... ...\ -r \sin \theta \quad & r \cos \theta \end{array}\right) \, . \end{eqnarray*}$$

10.

$f=x^2-y^2+2z^2-8$. Normal to surface is

$$\begin{eqnarray*} \vec{\triangledown} f = 2x \hat{{\bf i}}-2y \hat{\bf j} + 4z ... ...hat{{\bf i}} - 2 \hat{\bf j}+ 8 \hat{k} \hbox{ at } (1,1,2) \, . \end{eqnarray*}$$

Tangent plane is $\vec{N} \cdot(x \hat{{\bf i}}+y \hat{\bf j}+z \hat{k} = \hat{N} \cdot ( \hat{{\bf i}} + \hat{\bf j} + 2 \hat{k} )$

$$\begin{eqnarray*} 2x-2y+8z=2-2+16=16 \\ \frac{x}{8} - \frac{y}{8} + \frac{z}{2} =1 \, . \end{eqnarray*}$$

11.

$$\begin{eqnarray*} \int^2_0 \int^4_{x^2} x e^{-y^2} \, dy \, dx \\ &=& \int^4_0... ...^2} \right]^4_0 \\ &=& \frac{1}{4} \left( 1-e^{-16} \right)\, . \end{eqnarray*}$$

12.

Twice integral for upper half:

$$\begin{eqnarray*} 2 \times \int^{\pi / 4}_0 \int^{1/ \cos \theta}_0 r^2 r \, dr \, d \theta \, .\\ (x=r \cos \theta =1) \end{eqnarray*}$$

13.

$\vec{F}=xy \hat{{\bf i}} + \hat{\bf j}$. Parameterization

$$\begin{eqnarray*} x= \cos t \, , y= \sin t \quad 0 \leq t \leq \pi/2 \, .\\ \i... ..._0 s^2 \, ds \qquad s= \sin t \\ &=& 1 - \frac{1}{3} = 2/3 \, . \end{eqnarray*}$$

14.

(a)

$$\begin{eqnarray*} \vec{F} = F_1 \hat{bf i} + F_2 \hat{\bf j} \, , F_1=x(x-2y) \... ... \frac{\partial f}{\partial y} = -x^2+g' =2y-x^2 \, , g=y^2 \\ \end{eqnarray*}$$

Thus $\vec{F}$ is the gradient $\frac{x^3}{3} -x^2y+y^2$.

(b)

$$\begin{eqnarray*} \int_c \vec{F} \cdot d \vec{r} = f(P_1)-f(P_0) \hbox{ if } \... ...ox{ so } \\ \int_c \vec{F} \cdot d \vec{r} = \frac{16}{3} \, . \end{eqnarray*}$$

15.

By Green's theorem

$$\oint_C ay \, dx + bx \, dy = \int\int_S (b-a) \, dx \, dy \, .$$

For this to be $1/$ the area, must have $b-1=1$ always.

16.

Question is confusing without picture. I think it should be:

In spherical coordinate $Gm=1$)

$$\begin{eqnarray*} \int^{2 \pi}_0 \int^{\pi /3}_0 \int^2_{1/2 \cos \varphi} \cos \varphi \sin \varphi \, d \rho \, d \varphi \, d \theta \\ \end{eqnarray*}$$

Bottom plane in $z= \frac{1}{2} = \rho \cos \varphi$.

17.

(a)

$$\begin{eqnarray*} \int\int_T \vec{F} \cdot \vec{n} \, dS = \int\int_T z \, dS =... ...through top} \\ \int\int_B \vec{F} \cdot \vec{n} \, dS =0 \, . \end{eqnarray*}$$

(b)

$$\div \vec{F} = 3 \hbox{ so } \quad \int\int\int_D \div \vec{F} \, dV= 3 \pi a$$

where $D$ is the interior of the cylinder. Thus the flux through the sides is $2 \pi a$.

18.

$$\int\int_S \vec{F} \cdot \vec{n} \, dS = \int\int_S (x^2+y^2) \, dS$$

For the side surface $\vec{n} = x \hat{\bf i} + y \hat{\bf j} \, , x^2+y^2=1$ so the flux through the sides in the area, $2 \pi a$.

19.

(Note this has been corrected.) The line integral is independent of the path only if

$$\vec{F}=(axy+yz) \hat{{\bf i}} + (x^2+2yz+xz) \hat{\bf j} +(y^2+bxy) \hat{k}$$

is a gradient.

$$\begin{eqnarray*} \frac{\partial F_1}{\partial y} &=& ax+z \, , \frac{\partial ... ...bx \Rightarrow b=1 \\ [1ex] \hbox{So only if } a=2 \, , b=1 \, . \end{eqnarray*}$$

20.

By Stokes' theorem

$$\begin{eqnarray*} \oint_C y \, dx + z \, dy + x \, dz = \int\int_S \cos \vec{F}... ...{array}\right\vert \\ = - \hat{{\bf i}} - \hat{\bf j} - \hat{k} \end{eqnarray*}$$

Since $d \vec{S} = \vec{n} \, d \vec{S}$, $\vec{n} =( \hat{{\bf i}}-2 \hat{j}+ \hat{k})/ \sqrt{6}$

$${\rm curl} \vec{F} \cdot \vec{n} = 0 \hbox{ so } \int\int_S {\rm curl} \vec{F} \cdot d \vec{S} =0$$

for any surface contained in $x-2y+z=3$. Thus

$$\oint_C y \, dx + z \, dy +x \, dz =0 \, .$$