This is a solution of the second exam in MA114 Sections 001 to 006.
Question statements may not be verbatim. Each problem is 10 points.

Complete the following statements. You do not need to show work for this problem.

By definition `sum_(i=1)^oo a_i = lim_(n->oo)`

`sum_(i=1)^oo a_i = lim_(n->oo) sum_(k=1)^n a_k`, i.e. the infinite
sum is the limit of the partial sums.

For which values of `r` does the series `sum_(i=0)^oo r^i` converge absolutely?

It converges absolutes when `|r| < 1`.

For which values of `p` does the series `sum_(i=0)^oo 1/(i^p)` converge absolutely?

It converges absolutes when `p > 1`.

The alternating series `sum_(n=1)^oo a_n` converges if `lim_(i->oo) |a_i| = 0` and

The additional condition is that `|a_i| >= |a_(i+1)|` for all sufficiently large `i`.

According to the ratio test, `sum_(n=1)^oo a_n` is absolutely convergent if

The condition is that `lim_(n->oo) |(a_(n+1))/(a_n)| < 1`.

The partial sum `S_8 = sum_(j=1)^8 a_j` of the series `sum_(j=1)^oo (-1)^j j is

The sum partial sum is `-1+2-3+4-5+6-7+8 = 4`.

Use the alternating series test to find an `n` such that `S_n = sum_(i=1)^n ((-1)^i)/i` is within 0.05 of `S = sum_(i=1)^oo ((-1)^i)/i`. Justify your answer.

The alternating series test is applicable because the series is alternating,
the absolute value of thee terms is decreasing and approaches 0 as `i` approaches `oo`. The test includes the truncation error estimate
of `|a_(n+1)| = 1/(n+1)`. This is at most 0.05 when `n+1 > 1/0.05 = 20` and
this is true when `n` is 19 or larger.

SHOW YOUR WORK!

Find the radius of convergence of the power series `sum_(n=1)^oo (3x^n)/(n^4+9n^2)` and its interval of convergence.

and so the radius of convergence is 1 and the interval of convergence is
`-1 <= x <= 1`. (The justification for the endpoints is given below.)

Analyze the behavior at the left endpoint of the interval of convergence.

The left endpoint is at `x = -1` and the series there is
`S = sum_(n=0)^oo (3(-1)^n)/(n^4+9n^2)`. We will show that this is
absolutely convergent by replacing the terms with their absolute values. One
has for the new terms:
`0 <= 3/(n^4+9n^2) = 3/(n^2)(1/(n^2+9n^2)) <= 3 (1/n^2)`. By the comparison test applied to 3 times the convergent p-series `sum_(n=1)^3 1/(n^2)`, it follows that the series is absolutely convergent at `x=-1`.

Analyze the behavior at the right endpoint of the interval of convergence.

The right endpoint is at `x=1`. In the last part, we already saw that
the series converges.

SHOW YOUR WORK. A car travels counterclockwise around a circular track
at a constant speed, circling the track six (6) times per hour. The track has a radius of 2 miles. If the track is assumed to be the graph of `x^2+y^2 = 4` and at time `t = 0` the car is at the point (0, 2).

Find the position of the car at time t hours expressing your answer parametrically.

The parametric equations for the circle are `x = 2 cos(theta), y = 2 sin(theta)`. The angle `theta` can be calculated as initial value plus angular velocity times `t`, i.e. `theta = pi/2 + (2pi)6t`. So the parametric equations
are:

`x = 2cos(pi/2 + 12pi t)` and `y = 2 sin(pi/2 + 12 pi t)`

Find the position of the car at time `t = 1/4`.

Substitute into the parametric equations to get
`x = 2 cos(pi/2 + 12 pi (1/4)) = 2 cos((3pi)/2) = 0` and
`y = 2 sin(pi/2 + 12 pi (1/4)) = 2 sin((3pi)/2) = -2`.

SHOW YOUR WORK!

Find the radius of convergence of the power series `sum_(n=1)^oo (3(7-5x)^n)/n`.

and so the radius of convergence is 1/5 and the interval of convergence
is at least `|7/5-x| < 1/5` or `6/5 < x < 8/5`. The interval contains only the
right endpoint as is justified below. So the interval is actually `(6/5,8/5]`.

Analyze the behavior at the left endpoint of the interval of convergence.

The left endpoint is at `x = 6/5` where `7 - 5x = 1` The
series is `sum_(n=1)^oo 3/n` which is three times the harmonic series
which we know diverges.

Analyze the behavior at the right endpoint of the interval of convergence.

The right endpoint is at `x = 8/5` where `7-5x = -1`. The series
is `sum_(n=1)^oo 3((-1)^n)/n`. This is an alternating series where the
absolute values of the terms are clearly decreasing and have limit 0 as
`n-> oo`. So the series converges.

K/li>

Use the integral test to determine whether the series `sum_(n=1)^oo (2n)/(n^2+1)` is convergent or divergent. You must use the integral test and must show that it applies.

The terms are `a_n = f(n)` where `f(x) = (2n)/(n^2+1)`. The
function f(x) is continuous and positive for `x >=1`. Furthermore,

because of sum telescopes and the sum of all the intermediate terms is zero.
So the sum of the series is `lim_(n->oo) S_n = lim_(n->oo) 1 - 1/(n+1) = 1`.

SHOW YOUR WORK. A 10 foot ladder leans against the right wall of a narrow passageway with the bottom of the ladder touching the left wall as shown in the figure. The right wall is moving to the right (i.e. the passageway is getting wider) at a constant speed of 3 feet per minute. At time `t=0`, the passageway is 4 feet wide. Give parametric equations describing `P(t)` the point of contact of the ladder with the right wall for times between `t=0` and `t=3` minutes. The diagram represents the time `t=0`.

Since the wall is moving at 3 feet per minute from an initial value of
4 feet, the abscissa of `P(t)` is `x = 4 + 3t` feet at time t minutes. The
Pythagorean Theorem gives `x^2 + y^2 = 10^2` where `y >= 0` is the
ordinate of `P(t)`. So, `y = sqrt(100 - (4+3t)^2)` feet at time t minutes.

SHOW YOUR WORK. The curve C is described parametrically by `x(t) = 7 sin(t)`, `y(t) = 3 + 4t+t^2`.

Calculate a parametric form for the tangent line to C at the point corresponding to `t=1`.

One has `x'(t) = 7 cos(t)` and `y'(t) = 4 + 2t`. So, the tangent
line goes through the point `(x(1), y(1)) = (7sin(1), 8)` and its slope is
`(y'(1))/(x'(1)) = 6/(7cos(1))`. The line can therefore be parameterized as
`x(t) = x(1) + t x'(1) = 7 sin(1) + 7t cos(1)`, `y(t) = y(1) + t y'(1) = 8 + 6t`.