Math 8 Practice Exam Problems - Solutions

Disclaimer: This set of problems is meant neither to indicate the length nor composition of the actual exam. These are merely problems which were considered for inclusion on your exam, but for one reason or another were rejected. On the other hand, they should provide some flavor of the type of problems we considered.

1.

Find the Taylor series (about $x=0$) for $$f(x) = \begin{cases} {\sin x}/{x} & \textrm{if } x \ne 0\\ 1 & \textrm{if } x = 0 \end{cases}$$

We really need not be too concerned by the complicated definition. The question is really asking us to produce a series for $\sin x/x$ which (at first blush) seems not even defined at $x=0$. Well, it's not, but is has a removable discontinuity, so we can make it continuous (and even infinitely differentiable) by defining $f$ as above. In any case, the point of the problem is simply to observe that the Taylor series for $\sin x$ about $x=0$ is given by (with infinite radius of convergence) $$\sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$, so that
$$\sin x/x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k+1)!} = 1 - x^2/3! + x^4/5! - x^6/7! + \cdots$$.

2.

Find the radius and endpoints of the interval of convergence of the power series $$\sum_{n=0}^\infty \frac{n (x-3)^n}{7^n}$$.

To find the radius of convergence, we use the ratio test: We compute the ratio of the $(n+1)$st term to the $n$th term, and take the limit of the absolute value as $n \to \infty$. Setting this limiting value less than 1, and solving allows us to deduce the radius as well as the endpoints of the interval of convergence.

$$\lim_{n\to \infty}\left\vert \frac{(n+1)(x-3)^{n+1}}{7^{n+1}}\cd... ...ert\frac{(n+1)}{n}\cdot \frac{x-3}{7}\right\vert = \frac{\vert x-3\vert}{7} < 1$$ implies $\vert x-3\vert < 7$ so the radius of convergence is 7, and the endpoints of the interval of convergence are $-4$ and $10$. We have not verified whether or not these endpoints are actually contained in the interval of convergence; simply, that they delimit it.

3.

$$\sum_{n=10}^\infty \left(\frac{-2}{3}\right)^n =$$ ? This is a geometric series in which the first term is $$a = \left(\frac{2}{3}\right)^{10}$$ and constant ratio is $r = -2/3$, so the sum of the series is $$\frac{a}{1-r} = \frac{(2/3)^{10}}{1 - (-2/3)}$$.

4.

Find an equation of the line through the point $(1,2,3)$ orthogonal to the plane $x - 3y + 5z = 1$.

The normal to the plane is $\langle 1, -3, 5\rangle$ so the equation of the line is
${\mathbf r}= \langle 1, 2, 3\rangle + t\langle 1, -3, 5\rangle$.

5.

Are the lines ${\mathbf x}= \langle 1,2,3 \rangle + t \langle 4, -4, 6\rangle$ and ${\mathbf r}= \langle 3+s, 2s, 6+3s\rangle$ skew, parallel, or intersecting?

We solve the parametric equations simultaneously:

$$x = 1 + 4t = 3 + s$$

$$y = 2 - 4t = 2s$$

$$z = 3 + 6t = 6 + 3s$$

Since the direction vector $\langle 4, -4, 6\rangle$ is not parallel to $\langle 1, 2, 3\rangle$, the lines are either skew or intersect. Solving the first two equations simultaneously, yields $s = 0$ and $t = 1/2$. Since these values are consistent with the third equation, we conclude that the lines intersect. The point of intersection is $(3,0,6)$ (corresponding to $s = 0$ and $t = 1/2$).

6.

Consider the matrix $A = \begin{pmatrix} 1& 0&1&0\\ 0&1&0&2 \end{pmatrix}$.

Show that $A {\mathbf x}= {\mathbf b}$ is solvable for all ${\mathbf b}\in { \mathbbR}^2$, and find all solutions to $A{\mathbf x}= \begin{pmatrix}1\\ 2\end{pmatrix}$. What is the dimension of the solution space of $A {\mathbf x}= {\mathbf 0}$?

We observe that this matrix is already in row-reduced echelon form, and since there are two nonzero rows in the echelon form, we conclude that the dimension of the solution space to $A {\mathbf x}= {\mathbf 0}$ is $4 - 2 = 2$, and that the rank of $A$ is two that is, that the column space has dimension two. Since the column space is a subspace of ${ \mathbbR}^2$ of dimension 2, the column space is all of ${ \mathbbR}^2$, that is $A {\mathbf x}= {\mathbf b}$ is solvable for all ${\mathbf b}\in { \mathbbR}^2$. To find all solutions to $A{\mathbf x}= \begin{pmatrix}1\\ 2\end{pmatrix}$, we could row reduce the augmented matrix obtained by adding the column $\begin{pmatrix}1\\ 2\end{pmatrix}$ to $A$, row-reducing and solving the resulting system, or simply observe that ${\mathbf x}= \begin{pmatrix}1\\ 2\\ 0\\ 0\end{pmatrix}$ is a particular solution. Since the matrix $A$ is already in row-reduced form, we solve the homogeneous systems follows: Writing the matrix equation as a system, we see $x_1 + x_3 = 0$ and $x_2 + 2x_4 = 0$, so the general solution to the homogeneous is $\langle x_1, x_2, x_3, x_4\rangle = x_3\langle -1, 0, 1, 0\rangle + x_4\langle 0,-2,0,1\rangle$. Thus the general solution to the nonhomogeneous system is $\langle x_1, x_2, x_3, x_4\rangle = x_3\langle -1, 0, 1, 0\rangle + x_4\langle 0,-2,0,1\rangle + \langle 1, 2, 0, 0\rangle$.

7.

Find the volume of the parallelepiped determined by the vectors ${\mathbf u}= \langle 1,2,3\rangle$, ${\mathbf v}= \langle 2,0,1 \rangle$, and ${\mathbf w}=\langle 3,0,4\rangle$.

This is simply $\vert{\mathbf u}\bullet ({\mathbf v}\times {\mathbf w})\vert = \vert\langle 1,2,3\rangle \bullet \langle 0,-5,0\rangle \vert = 10$.

8.

An aircraft flies 200 kph in still air. There is a wind from the north at 100kph. The pilot wants to fly due east. In what direction should the pilot fly, and what is the ground speed of the aircraft?

Let ${\mathbf u}= \langle a,b\rangle$ be the direction in which the plane should fly. Let ${\mathbf w}= \langle 0, -100\rangle$ represent the wind. The desired flight direction is ${\mathbf v}= \langle c, 0\rangle$. We have ${\mathbf u}+ {\mathbf w}= {\mathbf v}$, or $\langle a, b-100\rangle = \langle c,0\rangle$, so $a = c$ and $b = 100$.

One way to proceed is: Since $\vert{\mathbf u}\vert = 200$ we have $a^2 + b^2 = 200^2$ or $a = \pm 100\sqrt 3$, clearly ${\mathbf u}= \langle 100\sqrt3, 100\rangle$ is in the correct direction (some easterly compnent). The ground speed is $\vert{\mathbf v}\vert$ which is $100\sqrt3$ km/h. The direction of flight makes an angle of 30 degrees with the ``positive $x$ axis''.

9.

Find the inverse of the matrix $A = \begin{pmatrix}1&2\\ 2&5\end{pmatrix}$ and use it to solve the system $A{\mathbf x}= \begin{pmatrix}3\\ 4\end{pmatrix}$. The matrix $\begin{pmatrix}1&2&1&0\\ 2&5&0&1\end{pmatrix}$ row reduces to $\begin{pmatrix}1&0&5&-2\\ 0&1&-2&1\end{pmatrix}$, so $A^{-1} = \begin{pmatrix}5&-2\\ -2&1\end{pmatrix}$, and $A{\mathbf x}= \begin{pmatrix}3\\ 4\end{pmatrix}$ implies that ${\mathbf x}= A^{-1}\begin{pmatrix}3\\ 4\end{pmatrix} = \begin{pmatrix}7\\ -2\end{pmatrix}$. The inverse for the matrix can also be found by formula.